NMTC 2014 Problem-2

Number Theory Level 3

If a , b , c , d a,b,c,d are positive integer such that a 5 = b 4 a^{5}=b^{4} , c 3 = d 2 c^{3}=d^{2} . The numerical value of d − b d-b = ?; Given c − a = 19 c-a=19


The answer is 757.

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Krishna Ar by Krishna Ar , 21, India

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2 solutions

Rohit Pant
Mar 17, 2015

Let a^5=b^4=k^20 and c^3=b^2=m^6 a=k^4, b=k^5, c=m^2, d=m^3 m^2-k^4=19 so, m+k^2=19 and m-k^2=1 m=10, k=3 d-b=1000-243=757

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Even I did the same +1 !!

Akshat Sharda - 5 years, 7 months ago

What have you done in d=m^3m^2 ..... I am not able to understand pls explain this step again (that 'd' one )

Chirayu Bhardwaj - 5 years, 3 months ago
Nathan Ramesh
Aug 23, 2014

Hi, this is also 1985 AIME #7

Note that a and c must both be perfect squares (actually a is a perfect fourth) and factor c-a as a difference of squares. One factor must be 19 and the other must be 1. You can solve this to find the values of a and c and then find the values of d and b for an answer of 757.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow..thanks for adding that in..show's that familiarity with questions too matter :D

Krishna Ar - 6 years, 9 months ago

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@Krishna Ar well, this is a duplicate of the problem by Joshua Ong .

mathh mathh - 6 years, 9 months ago

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Aw! Yes! :(

Krishna Ar - 6 years, 9 months ago

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