Bunuel
A and B when divided by 56 leave remainders of 48 and 32 respectively. When divided by 44, both leave a remainder of 24. If the sum of A and B is divisible by the sum of the divisors, then find the minimum value of A + B.
(A) 1,200
(B) 1,800
(C) 2,600
(D) 3,400
(E) 3,800
If we focus on A + B, we know A + B when divided by 44 must leave a remainder of \(24 + 24\) -> \(48\) -> \(4\) since we can divide out that extra 44. If we divide A + B by 56 then the remainder must be \(48 + 32\) -> \(80\) -> \(24\). So our goal is to find a number that can be satisfy both forms, \(44*i + 4\) and \(56*j + 24\), where i and j are integers.
Equate these and we can simplify this down to \(11*i + 1 = 14*j + 6\). In an attempt to find a chain of integer solutions, try adding 14's onto 6 to find a multiple of 11 plus 1 since multiples of 11 are easier to count. If we scroll through the numbers we have 6, 20, 34, and bingo stop there since 34 is a multiple of 11 plus 1. Then (i, j) = (3, 2) is a viable solution. And by adding 14's to the i and 11's to the j values, we will get additional (and infinitely many) solutions.
Now let's go back to the original number, the first satisfying number would be \(44*3 + 4 = 136\) from our i = 3 case. We can get more solutions by adding 14 to the i values or 11 to the j values, hence 56*11 = 560 + 56 = 616 increments. We want a number that is divisible by 56 + 44 = 100, start with 136 and check how many 16's we need to add to reach a whole hundred. We are missing 64 = 4 * 16, so add four more of 616's.
Finally the answer would be \(616*4 + 136 = 2464 + 136 = 2600.\)
Ans: C