10 times n

For which positive integer n n do we have

1 + 2 + 3 + 4 + + n = 10 × n ? 1 + 2 + 3 + 4 + \ldots + n = 10 \times n ?


The answer is 19.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

7 solutions

Ryan Soedjak
Dec 28, 2013

n ( n + 1 ) 2 = 10 n n = 19. \frac{n(n+1)}2=10n\implies n=19.

will u plz tell me how it is

yamini latha - 7 years, 5 months ago

Log in to reply

The sum of n n consecutive integers (from 1 1 to n n ) is equal to n ( n + 1 ) 2 \frac{n(n+1)}{2} . Therefore, we have, n ( n + 1 ) 2 = 10 n n 2 + n = 20 n n 2 19 n = 0 \frac{n(n+1)}{2} = 10n \Rightarrow n^2 + n =20n \Rightarrow n^2-19n=0 Solving gives n = 0 , 19 n=0, 19 however n n is a positive integer so n = 19 n=\fbox{19}

Oliver Welsh - 7 years, 5 months ago
Dang Anh Tu
Jun 9, 2014

Please read this note, you will understand:

https://brilliant.org/discussions/thread/discuss-more-about-pyramid-investigations-by-arron/

Partha Ghosh
Apr 2, 2014

n(n+1)/2 =10n => n=19 (ans)

Sunil Pradhan
Mar 16, 2014

1 + 2 + 3 + ... + n = 10 × n n(n+1)/2 = 10n (n+1)/2 = 10 n + 1 = 20

n = 19

Prasun Biswas
Feb 22, 2014

We should first know about the identity for the series : 1 + 2 + 3 + . . . . + n = n ( n + 1 ) 2 1+2+3+....+n=\frac{n(n+1)}{2} . So, on applying this in the problem, we get ----

1 + 2 + 3 + . . . . . . + n = 10 × n 1+2+3+......+n=10\times n

n ( n + 1 ) 2 = 10 n \implies \frac{n(n+1)}{2}=10n

n + 1 2 = 10 n + 1 = 20 n = 19 \implies \frac{n+1}{2}=10 \implies n+1=20 \implies n=\boxed{19}

Josh Grotstein
Feb 18, 2014

The sum of all integers from 1 to N = ((N+1)/2)*N

If we set that = to 10 * N, then...

(N+1)/2 = 10, and

N+1 = 20, and

N =19

Mesbah Tanvir
Dec 29, 2013

n(n+1)/2=10n

just solve it ... ans 19

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...