SOL:
Well first we will need to calculate how many zeroes are there in 20!. Since a zero in a factorial is always obtained when a 2 is multiplied with a 5, we need to find out how many powers of 5 are there in 20!. We know that in any factorial, powers of 2 will always be more than the powers of 5.

Note: Find the method for finding out the no of powers of a prime number in a factorial at the following link: factor-of-p-82381.html?view-post=617879#p617879)

Powers of 5 in 20!
20/5 + 20/5^2
= 4 + 0
= 4

Using the above method express 20! as follows:
20! = 2^18 * 3^8 * 5^4 * 7^2 * 11 * 13 * 17 * 19

Now from the above remove 4 2's and 4 5's as they are responsible for zeroes at the end. The remaining factors will decide the last non-zero digit.
=> 2^14 * 3^8 * 7^2 * 11 * 13 * 17 * 19

Last digit of this product is decided by the individual last digits of each number in the product.
=> 4 * 1 * 9 * 1 * 3 * 7 * 9
=> 36 * 21 * 9
=> 6 * 1 * 9
=> 54
=> 4

You sure the options are correct?
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2x
3x
2^2 x
2x3 x
2^3 x
3^3(p) x
2^2x3 x
2^4 x
2x3^2 x
2^3x3 x
3^3 x
2^2x3^2 x
2^4x3 x
2x3^3 x
2^3x3^2 x
2^2x3^3 x
2^4x3^2 x
2^3x3^3 x
2^4x3^3

3^3(p) x For 9, you have written 3^3 instead of 3^2. But thts all I could find out.
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Can you pls elaborate:
|||ly, the number of powers of 2 with:
One power of 3 i.e. 3^1 = 2^10
Two powers of 3 i.e. 3^2 = 2^10
Three powers of 3 i.e. 3^3 = 2^10

Couldn't get the idea, because we are concerned with the product of factors.

Thats the best approach. I was also considering using combination approach.

Although you did not mention combination in your solution, your approach is similar to that.

Good job...




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this looks more simpler. i tried with 36 too. more simpler. this must be 700+ level question(?)
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432 = 2^4 + 3^3
so no of factors = 5*4 =20

to find the product .... just divide total no. of factors by 2 so = 10
ans is

432^10 = 2^40 * 3 ^30

There is a brute force way to solve this. The principle is the same, you have to remove all powers of 5 and a corresponding number of powers of 2 from the product. The product of the rest will give you the non-zero digit.

First of all what I'd do is split the numbers out into four sets. Set 1 is {2,12,22, ... ,82}. Set 2 is {5,15,25,...,85}. Set 3 is {10,20,30,...90} and Set 4 is everything else {1,3,4,6,7,8,9, 11,13,14,16,17,18,19, .... ,81,83,84,86,87,88,89}

Set 1,2,3 have 9 elements each and have all the powers of 5 within them. Set 4 has the remaining 7x9=63 elements

Lets process Set 3 first, I can take out all powers of 10, to get {1,2,3,4,5,6,7,8,9}
Then I take out the 5 and a corresponding 2 to get {1,1,3,4,1,6,7,8,9}

Now lets process Set 1 & 2, For every 5 I take out of set 2, I will take out a 2 from set 1 :
So set 1 becomes : {1,6,11,16,....,36,41} and set 2 : {1,3,5,7,9,11,13,15,17}

We are almost done now, the only powers of 5 left are in set 2 which I take out and against that I take out 4 in set 3 to leave me with with :

Set 1 : {1,6,.....,36,41} : Last digit is 6
Set 2 : {1,3,1,7,9,11,13,3,17} : Last digit is 7
Set 3 : {1,1,3,1,1,6,7,8,9} : Last digit 2
Set 4 : Which has a pattern to it and easy to show last digit is 8 (It has subsets ending in 1,3,4,6,7,8,9 each of which has last digit 8 and 9 such subsets. The "cyclicity" of 8 in a product is 8,4,2,6,repeat so with 9 subsets the answer will be 8)

Multiplying those out, the last digit of 90! is 2



The approach is brute force, but once you get a hang of it, it doesnt take very long to calculate it out. I would be very surprised if this is a GMAT level question, unless of course there is a trick to doing this which has completely evaded me
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What is the answer for 20!'s last digit?
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doh ! my bad ... edited post
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Can you please explain how you calculating the last digit for Set 4 ?