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I think that if I meet this on the real Gmat, I must waste 10 minute, even 1 hour 15 minutes more!

There's a easier way to crack this...

1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 Now, all figures after 5! will have 0 in their unit digit (720 on 6!, 5040 on 7! etc.) so, all that we have to look into is the unit's place in the answer!

1+2+6+4 = 13 so, the answer should end with 3. Pick your answer! _________________

I think that if I meet this on the real Gmat, I must waste 10 minute, even 1 hour 15 minutes more!

how about 15 seconds??? answer is B just look at the unit digit 1!=1 2!=2 3!=6 4!=4 5!=0..after this all numbers will have a 2 and 5..and thus unit digit for all of them is 0..

1+2+6+4=13..luckily 13, 3 is only one of the ans choices..

hello, Can somebody explain /explain whats wrong with my method:

[(1+10!)/2] *10 ( average * number of terms)

i am getting 5 as last unit digit ???

You can apply the formula above: \(Sum=\frac{first+last}{2}*# \ of \ terms\), the mean multiplied by the number of terms for evenly spaced set (aka arithmetic progression). But as in the sequence given (1!, 2!, 3!, ..., 10!) the difference between any two successive terms is not the same then we don't have evenly spaced set (arithmetic progression) and thus can not apply this formula.

As for the solution: What is 1!+2!+...+10! ? A. 4,037,910 B. 4,037,913 C. 4,037,915 D. 4,037,916 E. 4,037,918

For any integer \(n\) more than or equal to 5 the units digit of \(n!\) will be zero as \(n!\), in this case, will contain at least one 2 and 5 which when multiplied will give a trailing zero. So terms from 5! to 10! will have zero as their units digit. 1!+2!+3!+4!=1+2+6+24=33 so the whole sum will have 3 as the units digit. Only option B offers a number with 3 as its units digit.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Taking units digits of 1! To 4! ie 1+2+6+4+0=13(Because only the units digits are diff in options).After that 0 will be at the unit's place for 5!,6!,7!,8!,9!,10!.Therefore,the last two digits will be 13. Ans.B

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

hello, Can somebody explain /explain whats wrong with my method:

[(1+10!)/2] *10 ( average * number of terms)

i am getting 5 as last unit digit ???

You can apply the formula above: \(Sum=\frac{first+last}{2}*# \ of \ terms\), the mean multiplied by the number of terms for evenly spaced set (aka arithmetic progression). But as in the sequence given (1!, 2!, 3!, ..., 10!) the difference between any two successive terms is not the same then we don't have evenly spaced set (arithmetic progression) and thus can not apply this formula.

As for the solution: What is 1!+2!+...+10! ? A. 4,037,910 B. 4,037,913 C. 4,037,915 D. 4,037,916 E. 4,037,918

For any integer \(n\) more than or equal to 5 the units digit of \(n!\) will be zero as \(n!\), in this case, will contain at least one 2 and 5 which when multiplied will give a trailing zero. So terms from 5! to 10! will have zero as their units digit. 1!+2!+3!+4!=1+2+6+24=33 so the whole sum will have 3 as the units digit. Only option B offers a number with 3 as its units digit.

Answer: B.

Hello Bunuel,

so in that case, 1!+2!.....1500! will also have the unit digit as 3 only.....

I can remember this concept and apply to any similar questions

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Re: What is 1! + 2! + ... + 10! ?
[#permalink]
05 Feb 2016, 07:40

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