Find the last digit

This topic has expert replies
User avatar
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Thu Feb 03, 2011 11:51 pm
Thanked: 1 times

Find the last digit

by vikrantr93 » Wed May 25, 2011 1:12 am
Find the last digit of the number 1^2 + 2^2 + 3^2 + 4^2 + ... + 99^2

a. 0
b. 1
c. 2
d. 5
e. 3

Legendary Member
Posts: 586
Joined: Tue Jan 19, 2010 4:38 am
Thanked: 31 times
Followed by:5 members
GMAT Score:730

by rohu27 » Wed May 25, 2011 1:38 am
concentrate on only the last digits..

1+4+9+6+5+6+9+4+1+0.... - this cycle repeats for evry 10 numebrs.
1+4+9+6+5+6+9+4+1+0=45*10=450
so last digit is 0
pick A

User avatar
Master | Next Rank: 500 Posts
Posts: 436
Joined: Tue Feb 08, 2011 3:07 am
Thanked: 72 times
Followed by:6 members

by manpsingh87 » Wed May 25, 2011 1:41 am
vikrantr93 wrote:Find the last digit of the number 1^2 + 2^2 + 3^2 + 4^2 + ... + 99^2

a. 0
b. 1
c. 2
d. 5
e. 3
sum of square of first n natural no. is given as n(n+1)(2n+1)/6;--------1)
here n=99; so submitting n=99 in 1 we have;
99*100*199/6; as the numerator is multiple of 10; therefore its last digit will be 0, hence a
O Excellence... my search for you is on... you can be far.. but not beyond my reach!