Latest Elitmus Aptitude Question SOLUTION: What is the maximum remainder when (104)^n is divided with 51

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(#M40030075) ELITMUS Numerical Ability question Keep an EYE

What is the maximum remainder when (104)^n is divided with 51

Asked In Elitmus Sagar Dasari (11 years ago)

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shubhi garg  (11 years ago)

since 104/51=2, hence 2^n can give max 50, (as no. is divided by 51 so max remainder can be 50), thus 2^5=32 is max what we get

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LOKESH SAMANT  (11 years ago)

(2)^n
remainder=32

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saikumar  (11 years ago)

if
N value Reminder
1 104/51 2
2 104*104/51*51 2*2
3 104*104*104/51*51*51 2*2*2
4 104*104*104*104/51*51*51*51 2*2*2*2
5 104*104*104*104*104/51*51*51*51*51 2*2*2*2*2
6 104*104*104*104*104*104/51*51*51*51*51*51 2*2*2*2*2*2=64 further divisible by 51 reminder is 13

so maximum possible reminder is : 32

Thank you

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naga sarath chand  (11 years ago)

ans is 32.-maximum

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Tarun Sharma  (11 years ago)

104%51= 2.

so we have 2^n /9
the max no that 9 leaves as remainder is 8.
so ans =8.

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Anu   (11 years ago)

How 32?Please explain

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mahima tiwari  (10 years ago)

For maximum firstly we will find the value of n we can find it by 51=3×17 so we will get maximum value [104/17] or [104/3]+[104/3^2] and so on and we will get n=6 so 104^6/51=2^6/51=64/51=13(ans)

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kundan prasad  (9 years ago)

Maximum Reminder is 32 and minimum remainder is 1

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Amit Kumar Singh  (7 years ago)

32 would be the answers

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