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805+ (Hard)|   Algebra|   Exponents|      
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Bunuel
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If a^b + b^a = 145 and a, b are positive integers, then how many such 19 Nov 2020, 03:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

22% (02:13) correct 78% (02:11) wrong based on 195 sessions

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If \(a^b + b^a = 145\) and a, b are positive integers, then how many such ordered pairs (a, b) are possible?

A. 0
B. 2
C. 4
D. 6
E. 8
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C
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UsamaGuddar
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If a^b + b^a = 145 and a, b are positive integers, then how many such 19 Nov 2020, 03:41
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(12,1) (144,1) are the 2 ordered pairs that come to my mind. But it’s taking longer than I’d expect. Any structured approach?

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Nipungupta9081
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If a^b + b^a = 145 and a, b are positive integers, then how many such 19 Nov 2020, 04:02
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UsamaGuddar
(12,1) (144,1) are the 2 ordered pairs that come to my mind. But it’s taking longer than I’d expect. Any structured approach?

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Hey bro , i think ( 12,1 ) wrong here

12^2 = 144
1^2 or 12 = 1

but you are using A, B and adding C also which is 2.
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chipsy
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If a^b + b^a = 145 and a, b are positive integers, then how many such 19 Nov 2020, 04:04
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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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If a^b + b^a = 145 and a, b are positive integers, then how many such 19 Nov 2020, 04:09
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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 22 Nov 2020, 02:16
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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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If a^b + b^a = 145 and a, b are positive integers, then how many such 28 Nov 2020, 19:39
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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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AKukreja
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If a^b + b^a = 145 and a, b are positive integers, then how many such 08 May 2024, 09:16
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Hey, Bunuel! Is there any structured way to solve this if we don't want to use Hit & Trial method?
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If a^b + b^a = 145 and a, b are positive integers, then how many such 09 May 2024, 05:06
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AKukreja
Hey, Bunuel! Is there any structured way to solve this if we don't want to use Hit & Trial method?
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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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