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How many 10s we can get out of 30! ?

Expert replies
by jerryragland » Wed Apr 07, 2010 8:30 am
This is from one of the questions from yesterday's post. I could not understand the logic properly.

How to find how many 10s we can get out of 30! ?
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Source: — Problem Solving |

by thephoenix » Wed Apr 07, 2010 9:18 am
jerryragland wrote:This is from one of the questions from yesterday's post. I could not understand the logic properly.

How to find how many 10s we can get out of 30! ?
10=2*5 so it depends on how many 5's do we have
short cut
divide the # i.e 30 by 5 note the quotient 6 in this case
now divide the quotient 6 ahain by 5 and note down the quotient i.e 1
beyond this repaeting the same thing we get 0 as the quotient
so total is 6+1=7
hence 7
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by jerryragland » Wed Apr 07, 2010 9:49 am
thephoenix wrote:
jerryragland wrote:This is from one of the questions from yesterday's post. I could not understand the logic properly.

How to find how many 10s we can get out of 30! ?
10=2*5 so it depends on how many 5's do we have
short cut
divide the # i.e 30 by 5 note the quotient 6 in this case
now divide the quotient 6 ahain by 5 and note down the quotient i.e 1
beyond this repaeting the same thing we get 0 as the quotient
so total is 6+1=7
hence 7
Why did not we pick 2? why 5? I know I am asking something basic here but could not figure out what.
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by thephoenix » Wed Apr 07, 2010 9:56 am
becoz the # of 2's wont be able to tell u the # of 10's
for eg in a smaller no such as 5!
there 3 two's but there is only one 10 becoz there is only one 5
thats y we decide on the # of five's
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by harshavardhanc » Wed Apr 07, 2010 10:34 am
jerryragland wrote:
Why did not we pick 2? why 5? I know I am asking something basic here but could not figure out what.
Because, twos will be always more than fives and for making a ten, you need a two and a five . So, obviously it is the number of fives that you've to concentrate on (you've twos in surplus).

here is one more link in which I've shown a short cut to tackle this kind of problem :

https://www.beatthegmat.com/post236803.html#236803

and here the problem boils down to : "finding power of five that divides 30!."
Regards,
Harsha
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by jerryragland » Wed Apr 07, 2010 1:38 pm
Thank you everyone.. I get it now..
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by supritajoshi » Wed Apr 07, 2010 7:45 pm
Can you please explain what will be answer if we are asked "find out how many 20 are there in 30!" ? I mean in this case also we have to calculate number of 5s in 30 ! . Thanks.
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by lkm » Wed Apr 07, 2010 9:12 pm
supritajoshi wrote:Can you please explain what will be answer if we are asked "find out how many 20 are there in 30!" ? I mean in this case also we have to calculate number of 5s in 30 ! . Thanks.
Dude!

Here is the simple logic for finding all the numbers:-

Let's say number to find = N

Then factorize the number N with all possible prime number. Let's say n1, n2, n2.... nn are the prime factors of number N.

Now find the possible number of combination of these prime factors in the given factorial number.

The number will be our answer.

Shortcut: Choose the prime factors whose chances is least to appear in that factorial.

For example:

Let's say N = 20

On factorizing 20, we get 2 x 2 x 5 or 4 x 5.

So, now in 30! we need to find the possible combination of 4 x 5.

Now 4 is the factor of following numbers in 30! :

4, 8, 12, 16, 20, 24, 28 (Total 7)

And 5 is the factor of following numbers in 30! :

5, 10, 15, 20, 25 (5*5), 30 (Total 7)

So possible number of combination = 7

Hence, there SEVEN 20s in 30!
The RULE Breaker
----------------------
I am born to fly, not to live in boundaries
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by supritajoshi » Thu Apr 08, 2010 12:41 am
Thanks...I got it now
lkm wrote:
supritajoshi wrote:Can you please explain what will be answer if we are asked "find out how many 20 are there in 30!" ? I mean in this case also we have to calculate number of 5s in 30 ! . Thanks.
Dude!

Here is the simple logic for finding all the numbers:-

Let's say number to find = N

Then factorize the number N with all possible prime number. Let's say n1, n2, n2.... nn are the prime factors of number N.

Now find the possible number of combination of these prime factors in the given factorial number.

The number will be our answer.

Shortcut: Choose the prime factors whose chances is least to appear in that factorial.

For example:

Let's say N = 20

On factorizing 20, we get 2 x 2 x 5 or 4 x 5.

So, now in 30! we need to find the possible combination of 4 x 5.

Now 4 is the factor of following numbers in 30! :

4, 8, 12, 16, 20, 24, 28 (Total 7)

And 5 is the factor of following numbers in 30! :

5, 10, 15, 20, 25 (5*5), 30 (Total 7)

So possible number of combination = 7

Hence, there SEVEN 20s in 30!
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by sunchopper » Thu Apr 08, 2010 3:27 pm
lkm wrote:
supritajoshi wrote:Can you please explain what will be answer if we are asked "find out how many 20 are there in 30!" ? I mean in this case also we have to calculate number of 5s in 30 ! . Thanks.
Dude!

Here is the simple logic for finding all the numbers:-

Let's say number to find = N

Then factorize the number N with all possible prime number. Let's say n1, n2, n2.... nn are the prime factors of number N.

Now find the possible number of combination of these prime factors in the given factorial number.

The number will be our answer.

Shortcut: Choose the prime factors whose chances is least to appear in that factorial.

For example:

Let's say N = 20

On factorizing 20, we get 2 x 2 x 5 or 4 x 5.

So, now in 30! we need to find the possible combination of 4 x 5.

Now 4 is the factor of following numbers in 30! :

4, 8, 12, 16, 20, 24, 28 (Total 7)

And 5 is the factor of following numbers in 30! :

5, 10, 15, 20, 25 (5*5), 30 (Total 7)

So possible number of combination = 7

Hence, there SEVEN 20s in 30!
So 30s in 50! would be??? I'm trying to use your shortcut and I don't get it.
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