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# If x = 1! + 2! + 3! + ... + n! then what is the remainder

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If x = 1! + 2! + 3! + ... + n! then what is the remainder  [#permalink]

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Updated on: 18 Aug 2014, 08:59
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95% (hard)

Question Stats:

40% (01:49) correct 60% (01:55) wrong based on 149 sessions

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

(1) n > 8
(2) n < 10

Source: 4Gmat

Originally posted by alphonsa on 18 Aug 2014, 08:04.
Last edited by Bunuel on 18 Aug 2014, 08:59, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: If x = 1! + 2! + 3! + ... + n! then what is the remainder  [#permalink]

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18 Aug 2014, 09:08
2
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If x = 1! + 2! + 3! + ... + n! then what is the remainder when x is divided by 25?

The least n for which n! is a multiple of 25 is 10 --> 10! will have two 5's, so it will be divisible by 25. For any greater n, n! will have at least two 5's so will also be divisible by 25.

(1) n > 8. So, n is 9 or greater than 9. The remainder of 1! + 2! + 3! + ... + n! will come from the first 9 terms: 1! + 2! + 3! + ... + 9! because all possible other terms 10!, 11!, ... will be divisible by 25. Sufficient.

(2) n < 10. If n = 1, you get one remainder and if n = 2, you get another. Not sufficient.

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Re: If x = 1! + 2! + 3! + ... + n! then what is the remainder  [#permalink]

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18 Aug 2014, 08:56
1
STAT1
for any value of n greater than 8 the remainder will be the same as 10! on-wards all factorials will be divisible by 25 as the factorials will be multiples of 5 and 10, so 10! on-wards all factorials will yield 0 as remainder. So, Remainder will be same as what we get by dividing 1!+... 9! by 25
So, A is SUFFICIENT

STAT2
Is INSUFFICIENT as for n=1 and n=2 we get 1 and 3 as remainders

Hope it helps!
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Re: If x = 1! + 2! + 3! + ... + n! then what is the remainder  [#permalink]

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06 Sep 2014, 05:38
Bunuel wrote:
If x = 1! + 2! + 3! + ... + n! then what is the remainder when x is divided by 25?

The least n for which n! is a multiple of 25 is 10 --> 10! will have two 5's, so it will be divisible by 25. For any greater n, n! will have at least two 5's so will also be divisible by 25.

(1) n > 8. So, n is 9 or greater than 9. The remainder of 1! + 2! + 3! + ... + n! will come from the first 9 terms: 1! + 2! + 3! + ... + 9! because all possible other terms 10!, 11!, ... will be divisible by 25. Sufficient.

(2) n < 10. If n = 1, you get one remainder and if n = 2, you get another. Not sufficient.

WHOA..!!
i dint notice that concept of 9! before marking A..! just went on to assume that all other factors greater than 8 would be divisible by 25..!Narrow escape..!would have got trolled if the number was 7 instead of 8..!
Thanks..!
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Re: If x = 1! + 2! + 3! + ... + n! then what is the remainder  [#permalink]

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18 Feb 2016, 19:09
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) n > 8
(2) n < 10

In the original condition, there is 1 variable(n), which should match with the number of equations. So you need 1 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), in n=9,10,11,12..., there is 5 more than twice after n=10. That is, the remainder when dividing with 25 is 0 and you only need to figure out when n=9, which is unique and sufficient.
For 2), as n=9,8..., there are many remainders, which is not unique and not sufficient.

 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If x = 1! + 2! + 3! + ... + n! then what is the remainder  [#permalink]

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04 Jan 2018, 19:26
alphonsa wrote:
If x = 1! + 2! + 3! + ... + n! then what is the remainder when x is divided by 25?

(1) n > 8
(2) n < 10

Source: 4Gmat

Hi..
25 means 5^2, so which factorial has 5^2 in it.. 10!, One in 5 and other in 10.
So 10! And above do not effect the Remainder as all of them give 0 as Remainder.
Now let's see what each statement tells us.
1) n>8
So n can be 9 or 10 or....
As we have seen 10! And above do not effect the Remainder, all terms above 8 will give same Remainder as 1!+2!+...+9!...
1!+2!+....9!+10!+11!... Divided by 25 means 1!+2!+..9!+0+0
So sufficient as we will get same Remainder for every value of n>8

2) n<10
n as 2 will give 3 as Remainder..
n as 3 will give 1+2+6=9
Insufficient

A
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Re: If x = 1! + 2! + 3! + ... + n! then what is the remainder  [#permalink]

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08 Jan 2018, 19:50
Could anyone please explain to me how statement 1 is sufficient. I was able to understand that when we add up to 10!, the sum would be divisible by 25 (5^2- one from 5 and another from 10). I was wondering if we add the sum upto 9!, where do we get the second 5 from? One will come from the 5 but I am confused regarding the second 5. Please help! Would greatly appreciate it!
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Re: If x = 1! + 2! + 3! + ... + n! then what is the remainder  [#permalink]

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08 Jan 2018, 20:00
csaluja wrote:
Could anyone please explain to me how statement 1 is sufficient. I was able to understand that when we add up to 10!, the sum would be divisible by 25 (5^2- one from 5 and another from 10). I was wondering if we add the sum upto 9!, where do we get the second 5 from? One will come from the 5 but I am confused regarding the second 5. Please help! Would greatly appreciate it!

Hi..
Adding up to 10! Is not necessarily div by 25..
But 10! And beyond 11! or 12! Are divisible individually by 25....
So the sum beyond 9! that is 10!+11!+12!+..... will always give 0 as remainder... This means 10! And beyond will not effect the remainder..

Now none from 1! to 9! is div by 25 so each will give different remainder...
So when you know that n>8....
n can be 9!, So remainder can be found of 1!+2!+...9!
If n is 10, remainder will come from 1!+2!+...9!+10! Which will be same as 1!+2!+...9! as 10! Will give you 0 as remainder..

Hope it helps
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Re: If x = 1! + 2! + 3! + ... + n! then what is the remainder  [#permalink]

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26 Sep 2019, 02:31
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Re: If x = 1! + 2! + 3! + ... + n! then what is the remainder   [#permalink] 26 Sep 2019, 02:31
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