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Maths Puzzle
1(1!)+2(2!)+3(3!)+4(4!)+5(5!)+..........................................100(100!) if it is divided by 42 then what is the remainder?
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- In case of
1(1!)+2(2!)+3(3!)+4(4!)+5(5!)+..........................................100(100!) , all terms after 7! are divisible by 42.
so we consider remainder for terms
1(1!)+2(2!)+3(3!)+4(4!)+5(5!)+6*(6!) only when divided by 42.
1(1!)+2(2!)+3(3!)+4(4!)+5(5!)+6*(6!)= 1+4+18+96+600+4320 = 5039 = 119 41/42
so 41 is the remainder.
- 11 years agoHelpfull: Yes(3) No(0)
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