Made Easy Test Series 2019: General Aptitude - Numerical Ability

Question

If $N=1!+3!+5!+7!…...199!$ then remainder obtained when N is divided by $120$ is 

• A. 7
• B. 9
• C. 11
• D. 13

Comments

If it was asked that the series is divided by 6. Then what would be the remainder 1?

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Yes

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Thanks, I got the concept.:)

Answer 1

Look, any number which is in the form of (a*120) i.e. multiple of 120 will be divisible by 120 right ?

Here, 1!=1

3!= 6

5!= 120 (5*4*3*2*1)

7!= (7*6)*120 i.e. multiple of 120

Now, take an example for understanding purpose (30+1)/2 => reminder will be 1. since 30 is divisible by 2 but 1 is not .

similarly, here after (1!+3! ) all are divisible by 120.

hence ans. is 1+6= 7

Comments

Nice I understood.

But going by the procedure, everything is also a multiple of $3!$, so the remainder can also be $1$.

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Such a simple explanation to a problem that looked almost impossible to solve. I love this forum ! <3