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24 + 6 + 2 + 1 = 33 last digit is 3 If x! and x>=5 then the number will always end in zero

Essentially you will always be adding 4! + 3! + 2! + 1! and always have a value of 3 as the last digit

Generally the last digit of \(1! + 2! + ... +N!\) can take ONLY 3 values: A. N=1 --> last digit 1; B. N=3 --> last digit 9; C. N=any other value --> last digit 3 (N=2 --> 1!+2!=3 and for N=4 --> 1!+2!+3!+4!=33, \(N\geq{4}\) the terms after N=4 will end by 0 thus not affect last digit and it'll remain 3).

So basically question asks whether we can determine which of three cases we have. (1) N is divisible by 4 --> N is not 1 or 3, thus third case. Sufficient.

(2) (N^2 + 1)/5 is an odd integer --> N is not 1 or 3, thus third case. Sufficient.

Even if the generalization doesn't occur to you, take some values to see if there is a pattern: 1! = 1 1! + 2! = 3 1! + 2! + 3! = 9 1! + 2! + 3! + 4! = ...3 1! + 2! + 3! + 4! + 5! = ....3 (because 5! is 120 so it doesn't contribute in last digit) As we go higher, 6!, 7! etc, each will end with 0 because it will have a 5 and a 2. So we only have to rule out N=1 and N=3.
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Re: If N is a positive integer, what is the last digit of 1! + [#permalink]

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07 Mar 2011, 06:30

If N is a positive integer, what is the last digit of 1! + 2! + ... + N! ?

1. N is divisible by 4 2. {N^2 + 1}{5} is an odd integer

(C) 2008 GMAT Club - m15#15

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient * Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient * BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient * EACH statement ALONE is sufficient * Statements (1) and (2) TOGETHER are NOT sufficient

\(1! + 2! + ... + N!\) ends with 3 if \(N \gt 3\) . Starting from 5! each term in the sum ends with 0, so if \(N \gt 3\) , the sum looks like \(1! + 2! + 3! + 4! + \text{(something ending with 0) } = 33 + \text{ (something ending with 0) } = \text{ integer ending with } 3\) .

Statement (1) by itself is sufficient. From S1 it follows that \(N\) is 4 or bigger.

Statement (2) by itself is sufficient. From S2 it follows that \(N\) is not 1 or 3. \(N\) is either 2 or larger than 3. In either case, sum \(1! + ... + N!\) ends with 3. The correct answer is D

Can some one please explain the solution for both cases?

Re: If N is a positive integer, what is the last digit of 1! + [#permalink]

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14 Mar 2011, 03:00

I think the answer would be A

Lets analyse statement 1 first.

1 != 1*1 2 != 2*1 3 !=3*2*1

3x2x1+2x1+1*1/4= 2 Hence 3! Hence Sufficient

Lets analyse statement 2 now

N/2+1/5 is a odd number 5(N+2)/2 so you can subsititude N=4, which will get you 15, its odd, so sufficient however if you subsititude N=5 which will you get you 45/2 non intenger so insufficient

Re: If N is a positive integer, what is the last digit of 1! + [#permalink]

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17 Nov 2011, 22:58

Guys, do you assume that there are no numbers between 2! and N!? I also don't understand this point: N=1 --> last digit 1. Can someone explain it to me? Please.

Re: If N is a positive integer, what is the last digit of 1! + [#permalink]

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17 Dec 2011, 20:43

MariaBez wrote:

Guys, do you assume that there are no numbers between 2! and N!? I also don't understand this point: N=1 --> last digit 1. Can someone explain it to me? Please.

I guess we assume that there are (N - 2) numbers between 2 and N. We have to find the sum of the factorials of each of these (N - 2) numbers.

But if we want to find the last digit of 1! (that is what Bunuel meant by N = 1) then the units digit will be 1. If we are asked to find the sum of factorials of the first three numbers (N = 3 in this case) then the units digit will be 9. And for all other numbers (any other value of N), the sum of the factorials of the numbers till N will have 3 as the units digit.

24 + 6 + 2 + 1 = 33 last digit is 3 If x! and x>=5 then the number will always end in zero

Essentially you will always be adding 4! + 3! + 2! + 1! and always have a value of 3 as the last digit

Generally the last digit of \(1! + 2! + ... +N!\) can take ONLY 3 values: A. N=1 --> last digit 1; B. N=3 --> last digit 9; C. N=any other value --> last digit 3 (N=2 --> 1!+2!=3 and for N=4 --> 1!+2!+3!+4!=33, \(N\geq{4}\) the terms after N=4 will end by 0 thus not affect last digit and it'll remain 3).

So basically question asks whether we can determine which of three cases we have. (1) N is divisible by 4 --> N is not 1 or 3, thus third case. Sufficient.

(2) (N^2 + 1)/5 is an odd integer --> N is not 1 or 3, thus third case. Sufficient.

Answer: D.

Hope it helps.

Could you please explain the bold face part? how is having 0 not affecting last digit and still remain 3? Wouldn't the last digit be 0?

Could you please explain the bold face part? how is having 0 not affecting last digit and still remain 3? Wouldn't the last digit be 0?

Just plug numbers to check.

If N=4 then \(1! + 2! + 3!+4!=1+2+6+24=33\); If N=5 then \(1! + 2! + 3!+4!+5!=(1+2+6+24)+120=33+120=153\); If N=6 then \(1! + 2! + 3!+4!+5!+6!=(1+2+6+24)+120+720=33+120=873\); ... As you can see the units digit if N=4 is 3, now the terms after N=4 (5!, 6!, ...) will have the units digit of 0 (since they all have the product of 2 and 5 in them), so adding those terms won't affect the units digit.

24 + 6 + 2 + 1 = 33 last digit is 3 If x! and x>=5 then the number will always end in zero

Essentially you will always be adding 4! + 3! + 2! + 1! and always have a value of 3 as the last digit

Generally the last digit of \(1! + 2! + ... +N!\) can take ONLY 3 values: A. N=1 --> last digit 1; B. N=3 --> last digit 9; C. N=any other value --> last digit 3 (N=2 --> 1!+2!=3 and for N=4 --> 1!+2!+3!+4!=33, \(N\geq{4}\) the terms after N=4 will end by 0 thus not affect last digit and it'll remain 3).

So basically question asks whether we can determine which of three cases we have. (1) N is divisible by 4 --> N is not 1 or 3, thus third case. Sufficient.

(2) (N^2 + 1)/5 is an odd integer --> N is not 1 or 3, thus third case. Sufficient.

Answer: D.

Hope it helps.

BB i am confused here...

A: N = 4 , u said the 4! = 4*3*2*1 = 24, 24+6+2+1 = 33 N= 8, tthe value will be someting else, but the last digit is 0 so the last digit is always 3....

B: N = 2, then the integer from B is 5/5 = 1, then unit digit is ( 2!+1!) 3, N = 3, does not get odd integer N = 4, does not get integer N = 5, does not get integer N = 6, does not get integer N = 7, does not get odd integer N = 8, does not get integer.... here, since N>4, last digit is 0, the value is 3...

very tricky.. i gave it for B, but the anser is D
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Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

A: N = 4 , u said the 4! = 4*3*2*1 = 24, 24+6+2+1 = 33 N= 8, tthe value will be someting else, but the last digit is 0 so it would be two cases

B: N = 2, then the integer from B is 5/5 = 1, then unit digit is 2, N = 3, does not get odd integer N = 4, does not get integer N = 5, does not get integer N = 6, does not get integer N = 7, does not get odd integer N = 8, does not get integer.... here, since N>4, last digit is 0, the value is fixed and hence answer should be B...

but your explanation says it is D.... Can you plese explain thoroughly...i could not understand your earlier explanation... sorry..

Again, the units digit of \(1! + 2! + ... +N!\) can take ONLY 3 values 1, 9, or 3: A. If N=1 then the units digit is 1; B. If N=3 then the units digit is 9; C. If N=any other value then the units digit is 3.

So basically question asks whether we can determine which of three cases we have.

(1) N is divisible by 4 --> N is not 1 or 3,because neither 1 nor 3 is divisible by 4, hence we have case C. For case C the units digit of \(1! + 2! + ... +N!\) is 3. Sufficient.

(2) (N^2 + 1)/5 is an odd integer --> again N is not 1 or 3, because neither 1 nor 3 gives odd value for (N^2 + 1)/5, hence we have case C. For case C the units digit of \(1! + 2! + ... +N!\) is 3. Sufficient.

Answer: D.

In either of statement it doesn't matter what the actual value of N is. Since we can get that it's not 1 or 3 then in all other cases the units digit of \(1! + 2! + ... +N!\) will be 3.
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Re: If N is a positive integer, what is the last digit of 1! + [#permalink]

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12 Apr 2012, 08:50

Bunuel wrote:

catty2004 wrote:

Could you please explain the bold face part? how is having 0 not affecting last digit and still remain 3? Wouldn't the last digit be 0?

Just plug numbers to check.

If N=4 then \(1! + 2! + 3!+4!=1+2+6+24=33\); If N=5 then \(1! + 2! + 3!+4!+5!=(1+2+6+24)+120=33+120=153\); If N=6 then \(1! + 2! + 3!+4!+5!+6!=(1+2+6+24)+120+720=33+120=873\); ... As you can see the units digit if N=4 is 3, now the terms after N=4 (5!, 6!, ...) will have the units digit of 0 (since they all have the product of 2 and 5 in them), so adding those terms won't affect the units digit.

Hope it's clear.

Perfectly clear and i can't thank you enough Bunuel!!!!

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