JEE Novice - (21)

Algebra Level 3

If ' $n$ ' is the smallest positive integer such that $(2001+n)$ is the sum of the cubes of the first ' $m$ ' natural numbers , then find the value of $(m-n)$ .

This question is a part of JEE Novices .

The answer is -15.

2 solutions

Vishwak Srinivasan
May 31, 2015

The sum of cubes of first $m$ natural number is a perfect square . The smallest value of $n$ for which $2001 + n$ is a perfect square is $24$ .

Hence,

$(\dfrac{m(m+1)}{2})^2 = 2025 \Rightarrow \dfrac{m(m+1)}{2} = 45$

$m(m+1) = 90 \Rightarrow m = 9$

$m-n = \boxed{-15}$

Jesse Nieminen
May 27, 2015

sum of cubes from 1^3 to m^3 is (m^2(m+1)^2)/4 = (m(m+1)/2)^2

first square number after 2001 is 2025 so n would be 24

Let's see if 2025 fits

(m(m+1)/2)^2 = 2025

m(m+1)/2 = 45

m(m+1) = 90

m^2 + m - 90 = 0

m = 9

so m - n = 9 - 24 = -15

Better solution:

sum of cubes from 1^3 to m^3 is (m^2(m+1)^2)/4 = (m(m+1)/2)^2

2001 + n must be a square number

so x^2 = 2001 + n

and m(m+1) = 2x

x^2 > 2001 x > 44

m(m+1) = 2x

m^2 + m - 2x = 0

m = (-1 +- sqrt(1 + 8x))/2

m is positive integer so

m = (-1 + sqrt(1 + 8x))/2

sqrt(1 + 8x) is odd if it is an integer (odd * odd = odd)

sqrt(1 + 8x) is integer if and only if 1 + 8x is square number so

y^2 = 8x + 1

x > 44 so x >= 45

y^2 >= 8*45 + 1 = 361

y >= sqrt(361) = 19

n is smallest possible so

x must be smallest possible so

y must be smallest possible so

y = 19 so

x = 45

so n = 24

m = (-1 + y) / 2 = 9

m - n = 9 - 24 = -15

Jesse Nieminen - 6 years ago

You can also notice that we have,

$m(m+1)=\sqrt{8004+4n}$

Prasun Biswas - 6 years ago

Yes, but I think that makes it more complicated.

Jesse Nieminen - 6 years ago

JEE Novice - (21)

This question is a part of JEE Novices .

The answer is -15.

2 solutions

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