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If a, b, x and y are prime numbers and a^x + b^y = 129, what is the va [#permalink]

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14 Sep 2014, 15:33

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If a, b, x and y are prime numbers and a^x + b^y = 129, what is the value of x+y?

(A) 5 (B) 6 (C) 10 (D) 13 (E) 15

Can someone explain how to approach this question systematically? Ie. what are your first couple steps in handling this question and how do you break it down?

Re: If a, b, x and y are prime numbers and a^x + b^y = 129, what is the va [#permalink]

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26 Sep 2014, 22:59

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beef001 wrote:

If a, b, x and y are prime numbers and a^x + b^y = 129, what is the value of x+y?

(A) 5 (B) 6 (C) 10 (D) 13 (E) 15

Can someone explain how to approach this question systematically? Ie. what are your first couple steps in handling this question and how do you break it down?

a^x + b^y = 129 (odd) all primes except 2 are odd. but if a and b are odd primes then odd+odd = even so one of a or b has to be 2. 2^x + b^y = 129 2^2 + 5^3 = 129
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Re: If a, b, x and y are prime numbers and a^x + b^y = 129, what is the va [#permalink]

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20 Sep 2016, 17:13

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Re: If a, b, x and y are prime numbers and a^x + b^y = 129, what is the va [#permalink]

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26 Aug 2017, 11:47

thefibonacci wrote:

beef001 wrote:

If a, b, x and y are prime numbers and a^x + b^y = 129, what is the value of x+y?

(A) 5 (B) 6 (C) 10 (D) 13 (E) 15

Can someone explain how to approach this question systematically? Ie. what are your first couple steps in handling this question and how do you break it down?

a^x + b^y = 129 (odd) all primes except 2 are odd. but if a and b are odd primes then odd+odd = even so one of a or b has to be 2. 2^x + b^y = 129 2^2 + 5^3 = 129

Will leave it here).

All prime numbers that can fit: 2, 3, 5, 7, 11. It is more comfortable to check first the closest to result (129) numbers.

First option fits: 11^2 + 2^3 = 129

I also think there is no systematic approach in such cases, but you can use different prompts to solve them. In this case yo can use that: 1. One on of them has to be even, because 129 is odd, and all prime numbers are odd except of 2. So we have to have odd + even ---> one of them is 2. 2. 11^2 is the closest to 129, so it is in our best interests to check it first. 2^3 fits. No one saidthat a, b, x, y have to be different.