remainder

Algebra Level 3

Find the remainder when 1 ! + 2 ! + 3 ! + 4 ! . . . . . . . . . . . . . . . . . + 100 ! 1! + 2! +3! + 4! .................+100! is divided by 24.

6 7 10 9

Aashish Patel by Aashish Patel , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

4! and all terms after that are divisible by 24. So the remainder is 1!+2!=3!=9.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

not sure how to apply the ! , but if I pair up the numbers 1 + 99 , 2 + 98, . . . 49+51, I get 49 pairs * 100 + 50 + 100 = 5050 /24 = 210 with remainder 10

rob funicella - 5 years, 11 months ago

Log in to reply

It is not 1+2+ . . . 100. It is 1!+2!+3!+ . . . 100!. You have missed factorial.

Niranjan Khanderia - 5 years, 11 months ago

Log in to reply

Thanks, That's a factorial horse of a different color!

rob funicella - 5 years, 11 months ago
Ameya Daigavane
May 8, 2015

Notice that, if n >= 4, 24 divides n! as n! = n * (n -1)*...(4) * (3) * (2) * (1) . Hence, the remainder is 1! + 2! + 3! = 9.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

How are we supposed to check the divisibility of all n!, when n>4? Even when n=4 then it is also divisible by 24.

Mehdia Nadeem - 5 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...