Question

If $a,b,c$ and $d$ are natural numbers such that $a^3=b^6,c^3=d^4,$ and $d-a=61,$ then the smallest value of $c-b$ is:

• A. 617
• B. 122
• C. 239
• D. 593

Answer 1

The correct option is A 617

a,b,c,d are natural numbers.

Also,

$a^3=b^6=x^6\Rightarrow a=x^2;b=x;x\in integers$

$c^3=d^4=y^{12}\Rightarrow c=y^4;d=y^3;yinintegers.$

Given,

$d-a=614$

$\Rightarrow y^3-x^{12}=61$

So, y & all ohers

Since, y& x are all integers,we can determine their values

$\therefore y=5,x=8$

$thereforea=\ast =8^2=64$

$c=5^4-625$

$d=5^3=125$

$\therefore c-b=625-8=617$