Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
For an integer $n, n!=n(n-1)(n-2) \ldots 3.2.1$. (P.C.S., 2008) Then, $1!+2!+3!+\ldots+100!$ when divided by 5 leaves remainder
(a)0
(b)1
(c)2
(d)3
Answer
Answer (as printed): D
Explanation
Every number from 5 onwards is completely divisible by 5. $\therefore(5!+6!+7!+\ldots+100!)$ is completely divisible by 5. And, $(1!+2!+3!+4!)=(1+2+3 \times 2 \times 1+4 \times 3 \times 2 \times 1)=(1+2+6+24)=33$. Clearly, 33 when divided by 5 leaves a remainder 3. Hence, $(1!+2!+3!+4!+5!+\ldots+100!)$ when divided by 5 leaves a remainder 3.
Explanation as extracted from the printed page; notation may be imperfect.