Perfect.

Number Theory Level 3

For positive integers $a$ and $b$ , it is given that both $21ab^{2}$ and $15ab$ are perfect squares. Find $min(a+b)$

Details
A perfect square is square of an integer.

Note- I take no credit for this question. This has appeared in $IMC , RMTC$ and many other contests too.

The answer is 56.

2 solutions

Nanayaranaraknas Vahdam
Sep 12, 2014

Given that $21ab^2$ is a perfect square. Since $b^2$ is a perfect square, $21a$ must also be a perfect square.

$21=3\times 7$ . Thus, $a$ must also be equal to $3\times7=21$ in order for it to be a perfect square. Thus, $a=21$

Now, $15ab=(15)(21)b$ is also a perfect square.

$15\times21=3^2\times5\times7$ . Now, this must be multiplied by $5\times7$ at least to make it a perfect square. Thus, $b=35$

Therefore, $a+b=\boxed{56}$

Exactly!! Quite easy!

Kartik Sharma - 6 years, 9 months ago

Expected this stupid comment or the comment-overrated of yours over here -_-

Krishna Ar - 6 years, 9 months ago

Oh, is it that? I wrote just what I thought! Not the right place? Should I delete it?

Kartik Sharma - 6 years, 9 months ago

Niranjan Khanderia
May 14, 2018

$For~p~and~q~as~integers,\\ Let~~21ab^2=p^2*b^2.~~\therefore~21*a=p^2~~\implies~a_{min}=21.\\ Let~~15ab=9q^2.~~\therefore~15*21*b=9q^2~~\implies~35b=q^2.~~\implies~b_{min}=35.\\ (a+b)_{min}=21+35=\Huge~~56.$

Perfect.

The answer is 56.

2 solutions

0 pending reports