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# What is 1! + 2! + ... + 10! ?

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Re: What is 1! + 2! + ... + 10! ? [#permalink]
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sondenso wrote:
What is 1!+2!+...+10! ?

4,037,910
4,037,913
4,037,915
4,037,916
4,037,918

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

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..
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Re: What is 1! + 2! + ... + 10! ? [#permalink]
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 ???
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Re: What is 1! + 2! + ... + 10! ? [#permalink]
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sondenso wrote:
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

Notice that the units digit of each answer choice is different; rather than perform the arithmetic as indicated, let’s determine the units digit of each factorial:

1! = 1

2! = 2

3! = 6

4! = 24

After 4!, all other factorials end in zero.

So, the units digit of the answer is 1 + 2 + 6 + 4 = 13, or 3.

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Re: What is 1! + 2! + ... + 10! ? [#permalink]
tinki wrote:
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 ???

That formula can only be used in case of an arithmetic progression. This question is not an AP question.
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Re: What is 1! + 2! + ... + 10! ? [#permalink]
Bunuel wrote:
tinki wrote:
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: $$Sum=\frac{first+last}{2}*number \ of \ terms$$, the mean multiplied by the number of terms, for an 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, so we don't have an evenly spaced set (arithmetic progression) and thus cannot apply this formula.

Check this thread for more: https://gmatclub.com/forum/math-number- ... 88376.html and also this thread: https://gmatclub.com/forum/sequences-pr ... 01891.html

As for the solution:

What is the value of $$1! + 2! + ... + 10!$$?

A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918

For any integer $$n$$ greater than or equal to 5, the units digit of $$n!$$ is zero. This is because, for $$n \geq 5$$, $$n!$$ will contain at least one 2 and at least one 5 as factors. When multiplied together, these produce a trailing zero. Hence, the terms from $$5!$$ through $$10!$$ each have zero as their units digit. Adding up the remaining terms, we'd have $$1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$$. Therefore, we see that the entire sum will have a units digit of 3 + 0 = 3. Only option B provides a number with 3 as its units digit.

Alternatively, we can do the following:

$$1! + 2! + ... + 10! = 1 + (2! + ... + 10!)$$ = odd + even = odd.

$$1! + 2! + ... + 10! = 3 + (3! + ... + 10!)$$ = integer divisible by 3.

Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.

$$1! + 2! + ... + 10! = 3 + (3! + ... + 10!)$$ = integer divisible by 3.

Question: How do we know 3 + {even} = divisible by 3??­
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Re: What is 1! + 2! + ... + 10! ? [#permalink]
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RahulJain293 wrote:
Bunuel wrote:
tinki wrote:
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: $$Sum=\frac{first+last}{2}*number \ of \ terms$$, the mean multiplied by the number of terms, for an 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, so we don't have an evenly spaced set (arithmetic progression) and thus cannot apply this formula.

Check this thread for more: https://gmatclub.com/forum/math-number- ... 88376.html and also this thread: https://gmatclub.com/forum/sequences-pr ... 01891.html

As for the solution:

What is the value of $$1! + 2! + ... + 10!$$?

A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918

For any integer $$n$$ greater than or equal to 5, the units digit of $$n!$$ is zero. This is because, for $$n \geq 5$$, $$n!$$ will contain at least one 2 and at least one 5 as factors. When multiplied together, these produce a trailing zero. Hence, the terms from $$5!$$ through $$10!$$ each have zero as their units digit. Adding up the remaining terms, we'd have $$1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$$. Therefore, we see that the entire sum will have a units digit of 3 + 0 = 3. Only option B provides a number with 3 as its units digit.

Alternatively, we can do the following:

$$1! + 2! + ... + 10! = 1 + (2! + ... + 10!)$$ = odd + even = odd.

$$1! + 2! + ... + 10! = 3 + (3! + ... + 10!)$$ = integer divisible by 3.

Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.

$$1! + 2! + ... + 10! = 3 + (3! + ... + 10!)$$ = integer divisible by 3.