Not so simple

Algebra Level 2

Let x be the sum of first n natural numbers and y be the sum of cubes of first n natural numbers, if it is known that:- $x + y = 240$ Find out the value of n

The answer is 5.

4 solutions

Rohit Sachdeva
Sep 3, 2014

$x=n(n+1)/2$

$y=n^{2} (n+1)^{2}/4 = x^{2}$

$x+y = x+x^{2} = 240$

$(x+16)(x-15)=0$

$x=15$

$n(n+1)/2=15$

$(n+6)(n-5)=0$

$\boxed{n=5}$

Ariijit Dey
Aug 24, 2014

x=n(n+1)/2;y=[(n^2)(n+4)^2]/2 so x=y^2. solving the equations, x=15; therefore, n(n+1)=15 (n^2)+6n-5n-30=0 (n-5)(n+6)=0; n=5, as n cannot be negative.

How did you get the formula of x and y ?

Mohaned Khaled Serag El-Den - 6 years, 9 months ago

http://mathschallenge.net/library/number/sum of cubes Really interesting proof

Simona Vesela - 6 years, 9 months ago

for cubes the formula is (n^2)[(n+1)^2]/4. so the value of y is not correct. so, x= n(n+1)/2; y= (n^2)[(n+1)^2]/4 therefore x(1+x)= 240. from this conclusion how can you get the answer please deduce.

monami samajdar - 6 years, 9 months ago

Mohammad Saif
Oct 3, 2014

I wrote this program in java to find the answer:-

public class finder { public void main() { int x=0,y=0;int c=0; for(int i=1; ;i++) { x=x+i; c=1; while(c==1) { y=y+(i i i); if(x+y==240) { System.out.println(i); break; } c++; } if(x+y==240) break; } } }

Subodh Khandelwal
Sep 1, 2014

we can also do hit and try method.. as if we start adding cubes from 1 then 1+8+27+64+125= 225, that gives us y, and after that 1+2+3+4+5=15, that gives us x so x+y=15+225=240 but it will work only for short terms..

Not so simple

The answer is 5.

4 solutions

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