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vikrantr93
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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
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Find the last digit
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manpsingh87
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sum of square of first n natural no. is given as n(n+1)(2n+1)/6;--------1)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
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
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