NMTC

Number Theory Level 2

For natural number $n$ , it is given that $(n+20)+(n+21)+(n+22)+\ldots +(n+100)$ is a perfect square. What is the least value of $n$ ?

The answer is 4.

3 solutions

Shivam Jadhav
May 11, 2015

$(n+20)+(n+21)+.....+(n+100) = 81n+4860 = 81(n+60)$ and for $81(n+60)$ to be a perfect square $n=4,21,40....etc$ . But we need to find $n_{least}$ , therefore $\boxed{n=4}$ .

Could you explain me why $n=4,21,40....etc$ ?

Omkar Kulkarni - 6 years, 1 month ago

@Omkar Kulkarni 81 is a perfect square right?Now we need that expression to be a perfect square as a whole,thus to find the least possible integer $n$ we need to confirm that $60+n$ is also a perfect square.Now the $60+n$ can be all perfect squares that are greater than $60$ like $64 , 81 , 100 , ...$ and so forth.Now for $n$ to be least plug $(n+60) = 64$ and hence the answer is $n=4$ I hope I explained well enough.

Arian Tashakkor - 6 years, 1 month ago

Ahh okay. I get it now! Thanks.

Omkar Kulkarni - 6 years, 1 month ago

@Omkar Kulkarni – Moreover, any perfect square is of the form 4K or 4K + 1 where K is a non - negative integer.

Aditya Sky - 5 years, 6 months ago

The main idea used here is the following fact (where $x$ and $y$ are non-zero integers):

$x^2y\textrm{ is a perfect square}\iff y\textrm{ is a perfect square}$

Prasun Biswas - 5 years, 11 months ago

Yeah exact same method i also had

Aditya Kumar - 5 years, 1 month ago

Noel Lo
Jun 12, 2015

The total would be $(100-20+1)(\frac{n+20+n+100}{2}) = 81(n+60)$

Since $81$ is a perfect square, $n+60$ must be a perfect square too. For $n$ to be as small as possible,

$n+60 = 64$

$n = \boxed{4}$

Kshitij Ghoshal
Sep 17, 2015

(N+20) + (N+21)......+(N+100) => 81n + 4860 => 81(N+60) Therefore , for N+60 to be the least perfect square, N= 4 I.e. 81(N+60) = 81x64 Sqrt (81x64) = 9x8 = 72 Thus answer is 4.

NMTC

The answer is 4.

3 solutions

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