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Quantitative Aptitude

Divisibility Rule for 7 & 13

N =4 !+5 !+6 ...

Question

N= 4! + 5!+6!.........200!. Find the last two digits of N

A

00

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B

13

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C

04

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D

54

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E

24

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Solution

The correct option is C 04
After 10!, the factorials end with 0 as the last two digits. We thusneed to find the last two digits from 4! To 9! Only

4!+5!+6!....9!= 24+120 + 720 + 5040 + 40320 + 362880 = ___04

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