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Re: If a^b + b^a = 145 and a, b are positive integers, then how many such [#permalink]
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3^4 + 4^3 = 145

so , a=3 , b=4 or a=4 , b=3

Option B
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Re: If a^b + b^a = 145 and a, b are positive integers, then how many such [#permalink]
Nipungupta, absolutely true! Thanks for pointing out

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If a^b + b^a = 145 and a, b are positive integers, then how many such [#permalink]
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I got 1^144 + 144^1 as the only option, otherwise its taking too long, but since I got one and finding the second is so time consuming, also that the next option is 2 I went for it. Is there a faster and easier way to do this?

I picked B.
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Re: If a^b + b^a = 145 and a, b are positive integers, then how many such [#permalink]
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There are 4 pairs (144,1), (1,144), (4,3) and (3,4)
Correct answer is C
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Re: If a^b + b^a = 145 and a, b are positive integers, then how many such [#permalink]
Hey, Bunuel! Is there any structured way to solve this if we don't want to use Hit & Trial method?
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Re: If a^b + b^a = 145 and a, b are positive integers, then how many such [#permalink]
AKukreja wrote:
Hey, Bunuel! Is there any structured way to solve this if we don't want to use Hit & Trial method?



One way to quickly narrow it down, which would probably be quicker than any algebraic approach is to assume:

a=b

So then it becomes:

2a^a = 145 and

a^a = 72.5

So this bounds "a" as follows:

3^3 = 27
4^4 = 256

So a<4

So the "average" value of a and b is less than 4.

This quickly narrows down the options to be examined

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Re: If a^b + b^a = 145 and a, b are positive integers, then how many such [#permalink]
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