Max@Math Revolution wrote:If 77n is divisible by 3, 7, and 77 for a positive integer n, which of the following is also divisible by 3, 7, and 77?
A. 77n+231 B. 11n+231 C. 77n+321 D. 7n+231 E. 11n
* A solution will be posted in two days.
I came up with (A), not sure how to offer a good mathematical proof.
My thought process went as follows - to be divisible by 3, the sum of the digits of a number must itself be divisible by 3. So we know the sum of the digits of 77n must be divisible by 3. I figured it was unlikely that this would met in the case of dividing n by 7 or 11, as in (B), (D), and (E), for all possible values of n, so I wanted to focus on (A) and (C). I saw the digits of 231 sum to 6, and so clearly it is divisible by 3 - and we already know 77n is divisible by 3. Therefore, we know we can factor a 3 out of both 231 and 77n, so 77n+231 must be divisible by 3. Unfortunately, the same is true of 321. So this alone did not rule out either (A) or (C).
So I now needed to determine which of 231 or 321 were divisible by 77. 231 is divisible by 77 - it equals 3. Pretty simple arithmetic. And if its divisible by 77, it must also be divisible by 7 (since 77 = 7 * 11). Therefore, we know we can factor a 3, 7, or 77 out of 77n and out of 231, therefore the sum 77n + 231 must be divisible by 3, 7 and 77. And so I selected choice (A). Not 100% positive though.
800 or bust!
