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austin: Master | Next Rank: 500 Posts; Posts: 110; Joined: Thu Oct 23, 2008 9:12 am; Location: India; Thanked: 9 times; Followed by:1 members

Excellent DS question

by austin » Thu Nov 27, 2008 4:44 am

If x = 1! + 2! + 3! +....... + n!, what is the remainder when x is divided by 25?

1. n>8
2. n<10

Will post OA in a while....

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Source: Beat The GMAT — Data Sufficiency | Thu Nov 27, 2008 4:44 am

rohangupta83: Legendary Member; Posts: 541; Joined: Thu May 31, 2007 6:44 pm; Location: UK; Thanked: 21 times; Followed by:3 members; GMAT Score:680

by rohangupta83 » Thu Nov 27, 2008 6:09 am

imo A

I haven't solved it but here is my approach

Statement I

n>8

10! and beyond, all numbers are divisible by 25.

10! = 10 *9*8*7*6*5*4*3*2*1
or 10! = 5*2*9*8*7*6*5*4*3*2*1

and 11! = 11*10!

and so on..

so basically if we find the remainder for the sum of the series 1! + 2! +3!...9!

that'll be the remainder for the entire series.

This will make statement I sufficient

Statement II

n<10

now, the remainder may vary because n could be 1, 2, 3, 4, 5, 6, 7, 8 or 9

lets say n = 4

1! + 2+ 3! + 4! = 1+2+6+24 = 33
33/25 ------ remainder = 8

now n=6
1! +2!+3!+4!+5!+6!
1+2+6+24+120+720 = 840+33 = 873
873/25---------remainder = 23

Hence, Statement II is insufficient

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rohangupta83: Legendary Member; Posts: 541; Joined: Thu May 31, 2007 6:44 pm; Location: UK; Thanked: 21 times; Followed by:3 members; GMAT Score:680

by rohangupta83 » Thu Nov 27, 2008 6:10 am

ok, maybe I did solve it in the end

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vikz_316: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Tue Sep 16, 2008 12:29 am; Thanked: 1 times

by vikz_316 » Thu Nov 27, 2008 6:19 am

Yes that looks like the correct solution Rohan. Looks like you did solve it!

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austin: Master | Next Rank: 500 Posts; Posts: 110; Joined: Thu Oct 23, 2008 9:12 am; Location: India; Thanked: 9 times; Followed by:1 members

by austin » Thu Nov 27, 2008 8:21 am

OA is A

Rohan is spot on...

The highest power of 10 in 10! is 2 => 10! ends with 00
Same with 11!, 12!, 13!, 14!..

The highest power of 10 in 15! = 3 => 15! ends with 000

The last two digits of 15! is 00

All n!; for n > 10 ends with 00

1!+2!+...+9! gives a certain remainder (which is 13 FYI)...when this sum is added with 10! or higher factorials, the last 2 digits remain 13...

Thus, the remainder is always 13...

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