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

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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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New post 14 Sep 2014, 14:33
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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?
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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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New post 26 Sep 2014, 21:59
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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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New post 14 Sep 2014, 16:10
Hi, It should not be difficult, since simple testing 2 pares of primes gives the result.
a(x) + b(y) = 129
11(2) + 2(3) = 129
x+y = 5

I think, that such questions do not have system approach, as they generally test the skills of correct Picking of numbers...
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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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New post 15 Sep 2014, 00:56
In fact 5^3+2^2=129 too...thus yielding x+y=5.
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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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New post 26 Aug 2017, 10:47
2
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.
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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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New post 13 Jan 2019, 23:34
Bunuel
Can process of elimination remove some options? Is there a different method than guessing combinations of 2^x+b^y?
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Re: If a, b, x and y are prime numbers and a^x + b^y = 129, what is the va &nbs [#permalink] 13 Jan 2019, 23:34
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If a, b, x and y are prime numbers and a^x + b^y = 129, what is the va

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