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Help with this one!!! Tricky one...

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by gmat740 » Sun Apr 12, 2009 9:26 pm
Hello

10! = 10*9*8*7*6*5*4*2*1

so this has 2 zeros because: 10 and 5*2


50! = 50*49*48..................3*2*1

here 5 zeros for 10,20,30 ,40 and 50

now consider multiples of 5
5, 15 25,35,45

these when multiplied with even number produce a further 0 each.

However, 25 has 2 multiples = 5*5
so no. of zeros produced by multiples of 5 = 6


So total zeros =5 +6 = 11

can you again check if the answer is 11 or 12

Hope this helps

Karan
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by kirvar » Sun Apr 12, 2009 9:43 pm
Answer is 12.

Karan: You missed an extra 5 in 50 (5*5*2)
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by kirvar » Sun Apr 12, 2009 9:43 pm
Answer is 12.

Karan: You missed an extra 5 in 50 (5*5*2)
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by gmat740 » Sun Apr 12, 2009 9:50 pm
Opps!!
Thanks a lot Kirvar!!
I was wondering why I am not getting 12

Thanks a lot again
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by gmat740 » Sun Apr 12, 2009 9:51 pm
These silly mistakes are going to Kill me one day :evil: :twisted: :(
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by scoobydooby » Sun Apr 12, 2009 10:22 pm
another approach:

we must have a 5 and a 2 to get the units digit as 0 (from 10), so we should count the number of 2s and 5s to count the zeroes.

there are lots of 2s. the 5s are the limiting factor, so we should in effect count the number of 5s in 50!

50/5+50/(5^2) (we stop at 5^2, as 5^3>50)
=10+2
=12 5s in 50!
=> there are 12 zeroes in 50!
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Thank you!!!

by cfarrera » Mon Apr 13, 2009 1:20 pm
Thank you all for your explanations, now it is perfectly clear!!!
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Challenge

by Deznos » Mon Apr 13, 2009 3:28 pm
Good job guys,

I have a solution similar to that of scoobydooby.

We'll know the number of ending zeros of 10! by finding the largest power of 10 contained in 50!

Why? Because for example 1000= 10^3;so 1000 has 3 ending zeros.
900= 9*10^2; so 900 has 2 ending zeros.
Now to solve the problem, we first need to break down 10 into prime numbers. 10 = 2*5.

We then need to find how many 2s (call it x) and how many 5s (call it y) go into 50!
Once we do that, we'll take the lower number between x and y.
Clearly, there will be more 2s than 5s in 50!; so let's save time by focusing on y (the number of 5).

Let's do it:
Divide 50 by 5 and the resulting quotient by 5 repeatedly until the quotient of the division is less than 5, which is the divisor, and you stop.

We only write out and take the quotients in the divisions.
50/5 = 10; 10/5= 2. Stop! since 2 is less than 5.
Let's now add all the quotients: y= 10 + 2 = 12.

12 is the largest power of 5 in 50!
12 is also the largest power of 10 contained in 50!
50! has 12 ending zeros!

That's all folks!

This concept can be applied to any number including 10!, which you do not need to find the ending zeroes of 50!

Deznos.
You can tame any beast, the GMAT included.
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Re: Challenge

by quriousaddict » Sun Apr 26, 2009 7:15 pm
[quote="Deznos"]

This concept can be applied to any number including 10!, which you do not need to find the ending zeroes of 50!

Deznos.[/quote]

Deznos,

How can your method be applied to find the number of 2's in 50! or 50?

50/2 = 25
25/2=[b]12.5[/b] What do you do when there's a remainder?
Does this concept only apply to the number 5? Can you explain further?

Thanks
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