Problem No. 54

Number Theory Level 4

Find the number of ordered triples ( x , y , z ) (x, y, z) of positive integers satisfying ( x + y ) z = 64. (x+y)^{z} = 64.


The answer is 74.

Benedict Dimacutac by Benedict Dimacutac , 21, Philippines

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2 solutions

2 6 = 4 3 = 8 2 = 6 4 1 = 64. So there are four groups that give 64. Option available for each group is , (N-1). N= 2,4,8,64 . 2 − 1 = 1.............................1 O p t i o n 4 − 1 = 3.............................3 O p t i o n s 8 − 1 = 7............................7 O p t i o n s 64 − 1 = 63..........................63 O p t i o n s 1 + 3 + 7 + 63 = 74 2^6=4^3=8^2=64^1=64.\\ \text{So there are four groups that give 64.}\\ \text{Option available for each group is , (N-1). N= 2,4,8,64 .}\\ 2~-1=~1.............................1~~Option\\ 4~-1=~3.............................3~~Options\\ 8~-1=~7............................7~~Options\\ 64-1=63..........................63~~Options\\ 1+3+7+63=~~~~~~~\Large \color{#D61F06}{74}

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is there any other way to get the solution

Rokshana alam - 5 years, 7 months ago

Shouldn't the answer be twice this many since the triples are ordered? (5 + 3)^2 is distinct from (3 + 5) ^2 for the purposes of this exercise (for example).

Alex Hume - 5 years, 7 months ago

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That has been a part of total. Say 8. 1+7, 2+6, 3+5, 4+4, 5+3, 6+2, 7+1 these are 7 as given above..

Niranjan Khanderia - 5 years, 7 months ago

This was a pretty simple one.

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