Latest TCS Aptitude Question SOLUTION: 1!+2!+3!+4!+....+50! Divided by 5! Remainder will be

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1!+2!+3!+4!+....+50! Divided by 5! Remainder will be

Asked In TCS vishal kumar singh (10 years ago)

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Read Solution (6) :

Bholanath   (10 years ago)

from 5! onward all are divisible by 5!
remaining 1!+2!+3!+4!=33
so, 33/5!=33/120=33 Ans.

Is this solution Helpfull? Yes (45) | No (4)  

k.m  (10 years ago)

remainder is 33.

Is this solution Helpfull? Yes (5) | No (1)  

somya  (10 years ago)

Plz give the solution of this-
1!+2!+3!+4!+....+50! ???

Is this solution Helpfull? Yes (3) | No   

prarit jain  (10 years ago)

its 1+2+6+24=33 answer rest 6!7!...... 50! are divisible

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monica  (10 years ago)

33/5 rem =3

Is this solution Helpfull? Yes | No (10)  

sreebhuvan  (10 years ago)

ANS=10

(1!+2!+3!)+4!+.....+50!=(1+2+6)+(24+120+.....+50!)
=(10)+(24*(1+5+(5*6)+(5*6*7)+....))
SECOND TERM 24*(1+(5*60)+....)IS DIVIDED BY 24 DUE TO IT HAS MULTIPLIER HAS 24

SO REMAINDER FROM FIRST TERM IS REMAINDER OF WHOLE PROBLEM SO =10%24=10

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