I'm not adding so many times

Number Theory Level 3

( x + 1 ) + ( x + 2 ) + ( x + 3 ) + … + ( x + 2015 ) \large (x+1) + (x+2) + (x+3) + \ldots + (x+2015)

What is the minimum positive integer value of x x such that the expression above is a perfect square?


The answer is 1007.

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Akshay Yadav by Akshay Yadav , 20, India

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2 solutions

Akshat Sharda
Jun 13, 2015

We are given, (x+1)+(x+2)+(x+3)+....+(x+2015) =2015x+[{2015(2016)}÷2]

Taking 2015 common , =2015[x+{2016÷2}] =2015(x+1008)

We are also given that the number obtained is a perfect square. Therefore, 2015(x+1008)=n² Here, 2015(x+1008) can only be a perfect square when , x+1008=2015 x = 2015-1008 x = 1007

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Can u explain the first step? How did u equate the given expression to 2015x+[{2015(2016)÷2]

Harshdeep Singh - 5 years, 11 months ago

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Actually he has applied the formula for sum of natural numbers,

S = n(n+1)/2

Where n is number till you want to find the sum.

Akshay Yadav - 5 years, 11 months ago
Figel Ilham
Jun 13, 2015

The equation can be written as 2015 ( x + 1008 ) = k 2 2015(x+1008) = k^2 We know that 2015 = 5 × 13 × 31 2015=5\times 13\times 31 To minimize the value, x + 1008 x+1008 must be 5 × 13 × 31 = 2015 5\times 13\times 31 = 2015 , thus x + 1008 = 2015 ⇒ x = 2015 − 1008 = 1007 x+1008=2015 \Rightarrow x = 2015-1008=1007

DONE.

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