Expert: Sherry Wallin - 8/3/2009

Question
Sherry,

This is a question from my last math competition. I have the answer key to the questions, but I'm not sure how to go about solving this one. I've redone it many times, but I still haven't been able to get to the right answer. Please help :)

Given the system: 3x+8y=23778
                 17x+ky=14952
Let 'k' and 'x' both be positive integers and let 'y' be an integer. Then if k<76, there are two consecutive positive odd integers for 'k' that satisfy the given conditions. Find the sum of those two consecutive positive odd integers.

Thanks in advance,
Ankai

Answer
Ankai~
    There are probably many ways to approach this but this is the way I would approach the problem: I would solve the system of equations using elimination. Multiply the first equation by -17 getting -51x -136y = -404226 and multiply the second equation by 3 getting 1x +3ky = 44856. Notice now the x's cancel out leaving you with (3k -136)y = -359370 -> y = -359370/(3k - 136)
Now find the prime factors of 359370 = 2*3^3*5*11^3,so the only numbers that will divide -359370 are combinations of those prime factors. A starting list is: 2,3,5,9,11,27,121,1331. I say starting list because you might need to examine other combinations like 3*5, 3*11, 5*9, 5*11.
Since you are looking for two values for k that are  consecutive *odd* integers both less than 76, eliminate any combination of the divisors with the 2 since they will be even and notice all other combinations would bear a number greater than 76. So our list is now:
3,5,9,11,15,27,33,45,55, 121, 1331. There are only two pairs of consecutive odd integers:
3,5 and 9, 11. 3 and 5 work and 9 and 11 do not bear integers for y. So the answer is 8. Don't FORGET you are looking for the sum of those odd integers for k!

Math Prof

Let me know if this explanation is helpful. Thanks.