Hướng dẫn divisible by 3 c++ - chia hết cho 3 c ++

Đưa ra một số, nhiệm vụ là chúng tôi chia số cho 3. Số đầu vào có thể lớn và có thể không thể lưu trữ ngay cả khi chúng tôi sử dụng int.examples dài dài: & nbsp; & nbsp;
Examples: 
 

Input  : n = 769452
Output : Yes

Input  : n = 123456758933312
Output : No

Input  : n = 3635883959606670431112222
Output : Yes

Vì số đầu vào có thể rất lớn, chúng tôi không thể sử dụng N % 3 để kiểm tra xem một số có chia hết cho 3 hoặc không, đặc biệt là bằng các ngôn ngữ như C/C ++. Ý tưởng này dựa trên thực tế sau. & NBSP;
 

Một số được chia cho 3 nếu tổng các chữ số của nó chia hết cho 3.

Minh họa: & nbsp; & nbsp; 
 

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Điều này hoạt động như thế nào? & NBSP; 

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Dưới đây là việc thực hiện thực tế trên: & nbsp;
 

C++

#include

using namespace std;

int check(string str)

{

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

int #include0

{

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

____10#include7#include8 #include9using0#include5

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Java

using6 using7

{

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

int0int1

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

int0{8#include8

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

int0{8

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Python3

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

{5

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

____10

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

____10

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

C#

using

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

using6

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

{

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

int0

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

int0

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

int0

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

PHP

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

{

____10

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

____10

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Các

____10

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Is

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

JavaScript

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

{

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

#include04#include4#include5

int #include0

#include12

#include13

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

____10#include7#include8 #include9using0#include5O(1), as we are not using any extra space. 

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

C++

#include14

using namespace std;

int #include0

{

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

____10#include7#include8 #include9using0#include5

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Java

Java

using6 using7

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

#include55{

____10#include22 #include64#include65#include5

____10{5 #include69__777 int8std;1#include73

#include74{8#include8

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

#include74{8

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

#include55

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Python3

#include92

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Is

#include55

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

#include55

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

C#

using

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

check(string str)4 using6 using20

#include55namespace0 check(string str)4 check(string str)6 using25

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

#include74using35#include8

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

#include74using35

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

#include55

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

JavaScript

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

int0using61#include8

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Let us consider 1332, we can write it as
1332 = 1*1000 + 3*100 + 3*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 can be written as : 
1*1 + 3*1 + 3*1 + 2 = 9
The above expression is basically sum of
all digits.

Since 9 is divisible by 3, answer is yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

int0using61

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
             = 9
Since sum is divisible by 3,
answer is Yes.

Độ phức tạp về thời gian: O (1) vì nó đang thực hiện các hoạt động liên tục O(1) as it is doing constant operations

Không gian phụ trợ: O (1): O(1)

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