How could this happen?

Number Theory Level 4

Find the smallest positive integer $n$ such that $7n + 55$ divides $n^2 -71$ .

The answer is 57.

1 solution

Agniprobho Mazumder
Sep 27, 2014

$n^2 - 71 = k*(7n + 55)$

$n^2 -7kn - 71 - 55k = 0$

$n = \dfrac{7k\pm \sqrt{(7k )^2 -4\times (-55k-71 )}}{2}$

$n = \dfrac{7k\pm \sqrt{(7k+16 )^2 + (28-4k)}}{2}$

Now, for n to be minimum and a positive integer, $28-4k=0$

or, $k=7$

$n=\dfrac{7k + 7k +16}{2}$ (+ as n is a positive integer)

$n=7k + 8 = \boxed{57}$

can you explain more deeply ,especially when it comes sqrt() ?

Alexander Freeman - 6 years, 8 months ago

basically, consider the constant term = -71 -55k . In the square root, all that he has done is factorize the value of b^2 -4ac. Then, he continues with the problem. And he shouldve written (7k+16)^2 instead of (7k+16)2. if you want confirmation, calculate (7k+16)^2 + (28 - 4k ) , you will get the value of b^2-4ac.

Kunal Jadhav - 6 years, 8 months ago

Yes, how is 7*7+16 a perfect square?

Trevor Arashiro - 6 years, 8 months ago

Dude, it's already square root of $(7k+16)^2$

Aditya Raut - 6 years, 7 months ago

@Agniprobho Mazumder , I have edited the LaTeX of this solution and added the missing step, please note.

Aditya Raut - 6 years, 7 months ago

How could this happen?

The answer is 57.

1 solution

0 pending reports