Matth971 wrote:

chetan2u wrote:

Hi..

\(\frac{(25!)}{(2 * 14!)}=\frac{25*24*23*22*21*20*19*18*17*16*15}{2}\)..

25*15*16=\(5^3*2^4*3\).. last non-0 digit is 6..

Let's check with last non-0 digit of remaining terms..

4*3*2*1*20*9*8*7/2=240*9*56=4*9*6=36*6=6*6=36..

So last digit =6..

E..

Otherwise if you know product of all digits, it will help

Hello,

Can you explain more please ?

This is too quick to be understandable by the beginners :/

I tried :

25!

--- = 25x24x23x21x20x19x18x17x8x15 The 8 is coming from 16/2 =8

(2x14!)

With that the last digit will be = 5x4x3x2x1x0X9x8x78x5 = 0

So now i have to do all the 25x24x23x21x20x19x18x17x8x15 to find something ?

Where am I wrong ?

Hi,

I'll take your method ahead..

Remove EQUAL number of 5s and 2s as combined they will give you 0....

1) if there are more 5s than 2s, straight way the last digit will be 5.

Here 25 has 2*5s and 20 and 15 have one 5 each...

So 4*5s remove 4*2s also..

So 25*24*23*22*21*20*19*18*17*8*15 will become 24*23*22*21*4*19*18*17*8*3/(2*2*2*2)=3*23*22*21*2*19*18*17*8*3..

Now check for last digit..

4*3*2*1*4*9*8*7*8*3 =(4*3)*2*(9*8)*(4*3)*(8*7)...

Last digit in brackets are (2)*2*(2)*(2)*(6)=16*6..

Last digit is 6..

Hope it helps

_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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