How many numbers are there with 3 digits such that if 4 is one of the digits
|
Let a,b,c be digits such that the six digit number $abcabc$ has 4 prime factors and only one prime factor out of the four has a power of 3 (say $\mathrm{k}^3$ ). If there are $\mathrm{n}$ such numbers find $n$. Show
$abcabc$ can be re-written as $7 \times 11 \times 13 \times(100a+10b+c)$. Hence $3$ of those primes are $7,11,13$ and so the prime with power of $3$ has to be $100a+10b+c$. Let it be equal to $k^{3}$, then $100 \leq k^{3} \leq 999$. Hence $k=5,6,7,8,9$ and from there we get $5$ cases for $a,b,c$. But the answer key says there are $23$ such numbers. What did I get wrong /missed? How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two? Solution 1We can find the number of increasing arithmetic sequences of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences. This gives us a total ofsequences. There areto permute these, for a total of. However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There arenumbers which start with zero, so our answer is. Solution 2Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit givespossible 3-digit numbers; otherwise,possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not. Case 1: 0 is not in the number. Then there are  Case 2: 0 is in the number. Then there areways to choose the largest digit (2, 4, 6, or 8), and each choice generates3-digit numbers, givingnumbers. Thus the total is. (by scrabbler94) Video Solution by OmegaLearnhttps://youtu.be/0W3VmFp55cM?t=2012 ~ pi_is_3.14 https://www.buzzsprout.com/56982/episodes/415913 starts with a variation on this question (plus solution) See AlsoThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Question 1159034: How many even numbers are there with three digits such that if 5 is one of the digits in a number then 7 is the next digit in that number in the numbers 0,1,2,3,4,5,6,7,8,9? Answer by Edwin McCravy(19274) You can put this solution on YOUR website! We include all 3-digit even numbers that don't contain a 5 at all.
Case 1: the number does not contain a 5 at all.
We choose the first digit 8 ways [from {1,2,3,4,6,7,8,9}]
We choose the second digit 9 ways [from {0,1,2,3,4,6,7,8,9}]
We choose the third digit 5 ways [from {0,2,4,6,8}].
That's 8∙9∙5 = 360 ways
Case 2: the first digit is a 5 and the second digit is a 7.
They are 570, 572, 574, 576, and 578.
Total 360+5 = 365.
[Note: the 2nd digit cannot be 5 because the third digit must be even, and
thus cannot be 7]
EdwinExecution Info
Messages
URIs/URLs
Conversions
Calendars
Numbers
Strings
Objects
Booleans
Arrays
Lists
Sets
Maps
Aggregates
IDs
How many 3 digit numbers can you form such that if one of the digits is 5?Total numbers=Total ways=81+72+72=225.
How many 3 digit even numbers can you form such that if one of the digits is 3 then the 3 is followed by 5?Therefore, Total no. of even numbers =8×9×5+5=365.
How many 3 digit numbers exist which can be divisible by 4?∴ there are 225 3-digit numbers which are completely divisible by 4.
How many even numbers are there with 3 digits such that?The total 3 digit numbers are 999 including preceding zeros and there are 999/2 even numbers. so total three digit even numbers are 499.
|
