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1!+2!+3!+.....+50! divided by 5! remainder will be?
Read Solution (Total 3)
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- (1!+2!+3!+.....+50!)/5!
(1!+2!+3!+4!)+(5!+6!+...+50!)/5!
from 5! onwards each term is divisible by 5!, so only (1!+2!+3!+4!)/5! gives the required remainder
Remainder=(1!+2!+3!+4!)/5! = 33/5! gives a remainder 33
ans: 33
- 10 years agoHelpfull: Yes(65) No(0)
- (1!+2!+3!+.....+50!)/5
(1!+2!+3!+4!)+(5!+6!+...+50!)/5
from 5! onwards each term is divisible by 5!, so only (1!+2!+3!+4!)/5! gives the required remainder
Remainder=(1!+2!+3!+4!)/5! = 33/5! gives a remainder 33
ans: 33
- 10 years agoHelpfull: Yes(9) No(0)
- (1!+2!+3!+.....+50!)/5!
(1!+2!+3!+4!)+(5!+6!+...+50!)/5!
from 5! onwards each term is divisible by 5!, so only (1!+2!+3!+4!)/5! gives the required remainder
Remainder=(1!+2!+3!+4!)/5! = 33/5! gives a remainder 33
ans: 33 - 10 years agoHelpfull: Yes(1) No(0)
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