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chetan2u

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 ?
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chetan2u

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
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25! = 25*24*23*...*14!...
2*14!

When 25*24, or 21*20, or 20*19, or 16*15, these are the checkpoint and the last digit always comes to zero

After that, no matter what numbers multiply it, it should be zero too
0 * n = 0

When zero divided by 2, it becomes 5...

In this case, why the ans is not 5???

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25! = 25*24*23*...*14!...
2*14!

When 25*24, or 21*20, or 20*19, or 16*15, these are the checkpoint and the last digit always comes to zero

After that, no matter what numbers multiply it, it should be zero too
0 * n = 0

When zero divided by 2, it becomes 5...

In this case, why the ans is not 5???

Posted from my mobile device

Posted from my mobile device




Hi,

It is not the case always..
It depends on the non zero digit next to zero..
1) if it's odd, then answer is 5.... 10/2=5 or 30/2=15
2) if it's even, it will depend on the digit... 20/2=10 or 40/2=20 and so on..

Hope it helps
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Hi Chetan
Could you please elaborate your strategy in detail..I still did not get it. I have the exam day after tomorrow.. kind request
Thank you

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Also, if possible give some example of similar question and how would you solve them

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Also, if possible give some example of similar question and how would you solve them

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Hey, I know that I'm very late for this but here it is for others who might need some more details. This is a self-made question. The strategy is simple.

Check the number of 5s and 2s. The following 3 cases are possible:

1. If there are more 5s than 2s, the Answer is 5 as an Odd number when multiplied by 5 give the unit digit 5.

2. If there are equal number of 5s and 2s, the answer is Odd but not 5.

3. If there are more 2s, which is the case here, the answer has to be Even. This eliminates 2 options.

Now, remove all the 5s and equal number of 2s and multiply for the Unit Digit.
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I did it in 5 minutes is there a shorter way to do it
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What is the last non-zero digit of expression

\(\frac{(25!)}{(2 * 14!)}\)

Where a! = a * (a-1) * (a-2) * (a-3) * ... * 1

A. 2
B. 3
C. 4
D. 5
E. 6

bond001 and for others looking for a quicker way...
The answer depends on the multiplication of units digit, so we have to find a way that eases the calculations..

If pure guess, there are more number of 2s than the number of 5s, so in xyz000, z will be even. Eliminate odd choices, and you are left with 3 options. Now, to be sure of your answer, you will have to multiply all units digit after removing all 5s and an equal number of 2s.

\(\frac{(25!)}{(2 * 14!)}=\frac{25*24*...*20*....*16*15}{2}=25*24*23*22*21*10*19*18*17*16*15\)
Remove the 5s by multiplying with the same number of 2s => 25*4*6*23*22*21*10*19*18*17*8*2*15=100*3*22*21*10*19*18*17*8*3*10
Take only the units digits now...
3*2*1*9*8*7*8*3
Remove 3*3*9 as we will get 1 as the units digit => 2*8*7*8
It's easy to see now that we are getting 8 from 2*8*8, and finally 6 from 8*7

there are various means to cancel out terms to get to singe units digit. you can multiply 3*7 or 9*9 to cancel out those terms.
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