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Hi @LeeTaiTai! This question was pushed behind by all the others.
Only just saw it. Hope it is not too late!

The question is testing:
- prime numbers: 2,3,5,7,11,13,17,19,...
- every number can be factorized into a product of primes
- divisibility conditions

Note that: 34 = 2 * 17, 51 = 3 * 17 and 6 = 2*3.
So we get 51y = 6z - 34x
        3*17y = 2(3z - 17x), which has a factor of 2.
So y = 2.

Next     34x = 6z - 51y
       2*17x = 3(2*z - 17y), which has a factor of 3.
So x = 3.

Finally z = (34x + 17y)/6 = (34*3+ 51*2)/6 = 34.
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