Elitmus
Exam
Numerical Ability
Number System
Let N=4831*4833*4835.What is the remainder when N divided by 24?
(A)9
(B)23
(C)21
(D)11
Read Solution (Total 9)
-
- 4831%24=7
4833%24=9
4835%24=11
7*9*11%24=21ans.
C is the ans. - 10 years agoHelpfull: Yes(41) No(0)
- @Shashi is correct
- 10 years agoHelpfull: Yes(4) No(2)
- 21..would b the ans
- 10 years agoHelpfull: Yes(4) No(0)
- Ans.21
remainders of all numbers 4831,4833 & 4835 are 7,9 & 11 resp.
The remainder of exp. 7*9*11/24 will be same as exp. given in question.i.e.21 - 10 years agoHelpfull: Yes(2) No(0)
- Ans is a
31*33*35= 35805
805%24=153
153%24=9 - 10 years agoHelpfull: Yes(0) No(4)
- (7*9*11)%24=21
- 10 years agoHelpfull: Yes(0) No(0)
- individually divide all the no n get the result
like 7*9*11/24
agaun same procedure is follow
n final result is 21
- 10 years agoHelpfull: Yes(0) No(0)
- ans is 21
4831%24=7,4833%24=9,4835%24=11
7*9*11=693
693%24=21
- 10 years agoHelpfull: Yes(0) No(0)
- these kind of easy questions cannot be asked in elitmus
- 5 years agoHelpfull: Yes(0) No(0)
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