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vipulgoyal
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If N is a positive integer, what is the last digit of 1! + 2! + ... +N!?
1) N is divisible by 4
2) (N^2 + 1)/5 is an odd integer.
OE
Generally the last digit of 1! + 2! + ... +N! can take ONLY 3 values:
A. N=1 --> last digit 1;
B. N=3 --> last digit 9;
C. N=any other value --> last digit 3 (N=2 --> 1!+2!=3 and for N=4 --> 1!+2!+3!+4!=33, N\geq{4} the terms after N=4 will end by 0 thus not affect last digit and it'll remain 3).
So basically question asks whether we can determine which of three cases we have.
(1) N is divisible by 4 --> N is not 1 or 3, thus third case. Sufficient.
(2) (N^2 + 1)/5 is an odd integer --> N is not 1 or 3, thus third case. Sufficient.
Answer: D.
my query ; In option B, why n cant be 2 ??
1) N is divisible by 4
2) (N^2 + 1)/5 is an odd integer.
OE
Generally the last digit of 1! + 2! + ... +N! can take ONLY 3 values:
A. N=1 --> last digit 1;
B. N=3 --> last digit 9;
C. N=any other value --> last digit 3 (N=2 --> 1!+2!=3 and for N=4 --> 1!+2!+3!+4!=33, N\geq{4} the terms after N=4 will end by 0 thus not affect last digit and it'll remain 3).
So basically question asks whether we can determine which of three cases we have.
(1) N is divisible by 4 --> N is not 1 or 3, thus third case. Sufficient.
(2) (N^2 + 1)/5 is an odd integer --> N is not 1 or 3, thus third case. Sufficient.
Answer: D.
my query ; In option B, why n cant be 2 ??
