Latest Exam Aptitude Question SOLUTION: What is the trick for solving following remainder questions? 1. 123123123123/11 = rem 2. 989898/101= rem? 3.(11111*22222*33333)/5 = rem?

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(#M40038226) EXAM Numerical Ability question number System Keep an EYE

What is the trick for solving following remainder questions?
1. 123123123123/11 = rem
2. 989898/101= rem?
3.(11111*22222*33333)/5 = rem?

Asked In Exam suraj (9 years ago)

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SUDHIR SHARMA  (9 years ago)

(1). When 123123 is divided by 11 we get 11193 as quitent and 0 as remainder.

It means that 123123 is completely divisible by 11 , if next six digits are

also same then that term will also be divisible by 11 completely.

i.e. when we divide 123123123123 by 11 we get 0 as a remainder.

(2). 9898 is completely divisible by 101 , it means remainder will be 98

(3). (11111*22222*33333)/5
using remainder theorem,
when we diide 11 by 5 we have 1 as remainder.
Similiarly when we divide 11111 by 5 we get 1 as remainde.

Now, dividing 22 by 5 gives 2 as remainder,
similiarly 22222 will give 2 as remainder.

And 33333 will give 3 as remainder.

now, 1*2*3/5 = 6/5 = 1

Therefore 1 is the remainder

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suraj   (9 years ago)

hello sudhir, thnx for explaining
i think in last 11111*22222*33333 they are already divided by 5 seprately and remainder comes out to be 1*2*3 = 6
6 should be remainder ??
or i am mistaking somewhere?

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