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If n is a positive integer, what is the units digit of the sum of the following series:1! + 2! + ... + n!? (The series includes every integer between 1 and n, inclusive)

(1) n is divisible by 4. (2) n^2 + 1 is an odd integer.

Re: If n is a positive integer, what is the units digit of the sum of the [#permalink]

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08 Jul 2015, 03:37

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Bunuel wrote:

If n is a positive integer, what is the units digit of the sum of the following series:1! + 2! + ... + n!? (The series includes every integer between 1 and n, inclusive)

(1) n is divisible by 4. (2) n^2 + 1 is an odd integer.

Kudos for a correct solution.

We know 1!=1, 2!=2, 3!=6, 4!=24, 5!=120 etc with emphasis on 5! and greater have 0 as the unit's digit.

1!+2!+3!+4! = 33 with unit's digit =3.

1. Per statement 1, n=4p.

If n=4, unit digit =3

If n =8 or 12 or 16 or 20 , unit digit =3 again as all factorials from >=5 have 0 as the unit's digit (as mentioned above). Thus this statement is sufficient.

2. Per statement 2, n^2+1=odd --> n^2=even --> n=even integer

n=2, unit digit =3 n=4 . unit digit =3 and finally n=6 and above , unit digit is 0

Thus for this case as well unit digit = 3 and is thus sufficient.

Ans is D, both statements are sufficient to answer this question.

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08 Jul 2015, 04:31

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1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 From 5! onwards, all the factorials will end in 0, as they include a factor of 10. (5! = 5*4*2*1)

1!+2! = 1+ 2 = 3 1! + 2!+3! = 1+2+6=9 1! + 2!+3! + 4! = 1+2+6+24=33 = ...3 1! + 2!+3! + 4! + 5! = 1+2+6+24+120 =153 = ...3 Now since all the higher factors contain 0 as its units digit, the unit digit of their sum will not be affected by further addition. So, the threshold figure is 4.

Statement (1) n is divisible by 4. From the explanation above, the unit digit will always be 3. Sufficient.

Statement (2) \(n^2 + 1\) is an odd integer. \(n^2 + 1\)= Odd \(n^2\) = Even \(n\)= Even So, \(n\)= 2, 4, 6.. \(n\)=2; 1!+2! = 3 And from 4 onwards, the unit digit will always be 3. In both the cases (\(n\)=2, even integer >2), the unit digit is 3. Sufficient.

Answer (D).
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Re: If n is a positive integer, what is the units digit of the sum of the [#permalink]

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08 Jul 2015, 17:50

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Bunuel wrote:

If n is a positive integer, what is the units digit of the sum of the following series:1! + 2! + ... + n!? (The series includes every integer between 1 and n, inclusive)

(1) n is divisible by 4. (2) n^2 + 1 is an odd integer.

Kudos for a correct solution.

To find the Unit digit of 1! + 2! + ... + n! we need to find the unit digit of each component.

Unit digit of n! where n > 4 will be equal to Zero (since it will have component of 5 and 2). So any addition to series above 4 (n>4) won't impact unit digit.

Unit digit of series will be some thing like below 1! = 1 1! + 2! = 3 1! + 2! + 3! = 9 1! + 2! + 3! + 4! = 3 1! + 2! + 3! + 4! + 5! = 3 (Here on unit digit will be 3 only)

So we need to know, if n = 1,2,3 or n >= 4

Statement 1: n is divisible by 4. So lowest possible value of n = 4. So unit digit will be 3 (See above explanation)

Hence Sufficient

Statement 2 : n^2 + 1 is an odd integer

n^2 - is even integer n is even integer, So possible value of n = 2,4,6,8 So in each case unit digit is 3

If n is a positive integer, what is the units digit of the sum of the following series:1! + 2! + ... + n!? (The series includes every integer between 1 and n, inclusive)

(1) n is divisible by 4. (2) n^2 + 1 is an odd integer.

Kudos for a correct solution.

1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 and so on

While we look at the series we understand that Factorial of every value greater than 4 has unit digit 0

i.e. 1! + 2! + ... + n! will not given any different value of Unit Digit for any value of n>4 for

Statement 1: n is divisible by 4

i.e. n can be any one of 4, 8, 12, 16,... etc.

The Unit digit of given expression will always be 1+2+6+4 = 3 (Unit digit) because every Factorial value greater than 4! will have unit digit 0 and won't influence the result SUFFICIENT

Statement 2: n^2 + 1 is an odd integer i.e. n^2 is Even i.e. n may be 2 or 4 or 6 etc. @n=2, the unit digit of 1!+2! = 3 @n=4, the unit digit of 1!+2!+3!+4! = 3 and every Factorial value greater than 4! will have unit digit 0 and won't influence the result SUFFICIENT

Answer: Option D
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If n is a positive integer, what is the units digit of the sum of the following series:1! + 2! + ... + n!? (The series includes every integer between 1 and n, inclusive)

(1) n is divisible by 4. (2) n^2 + 1 is an odd integer.

Kudos for a correct solution.

800score Official Solution:

In this question, we need to look for a pattern in the units digit of the sum of factorials for positive integers. Start with 1! + 2! = 3. Then since 3! = 6 add this to the previous sum and get 9. 4! = 24. Add this to the previous sum: 33. 5! = 120. Adding this: 153. 6! = 720…

Let’s stop here, because the pattern should now be clear. Every additional number that will be added to the series will have a units digit of zero (because it has 2 and 5 as factors and therefore 10).

So for all n > 3, the units digit of the sum will be 3. Now, let's look at the statements.

Statement (1) tells us that n is a multiple of 4. Since n must be greater than 3, we know the units digit of the sum of the series will always be 3, so Statement (1) is sufficient.

From Statement (2), we can determine that n cannot be either 1 or 3. It is possible that n = 2, in which case: 1! + 2! = 3. Any other number that satisfies the statement is greater than 4, in which case the sum 1! + ... + n! will have a units digit of 3 as well. So Statement (2) is also sufficient because the units digit is always 3.

Since both statements are sufficient individually, the correct answer is choice (D).
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If n is a positive integer, what is the units digit of the sum of the [#permalink]

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20 Jul 2015, 08:40

The sum always ends in 3.

Before even looking at (1) or (2).... 1*2=2 *3=6 *4=24 *5... focus only on the units digit from now on... 4*5 = 20 .... 20*6 = 120 all factorials beyond 4 will always end in 0, because 0 times anything is always 0. Summing up the factorials before the 0 units digit factorials we get:

1+2+6+24 = 33. Adding 33 to any factorial where n = 5 or greater will result in a units digit of 3, because 3+0+0+0+0+0+0+0+... will always be 3.

Therefore, we know that all units digits for the sum of n>=4 will always be 3.

(1) n is a multiple of 4, so n = 4k. N = 4, 8, 12, 16, etc. Since n >= 4, n must have a units of 3. Sufficient.

(2) (2) n^2 + 1 is an odd integer. This tells us that n is even. All positive, even integers equal to 4 or greater must have a units digit of 3. How about 2? The sum of factorials where n = 2 is also 3, since 2! + 1! = 3. Therefore, all even values of n will have a units digit of 3 when we sum up the given factorial series. Sufficient.

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