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60!

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60!

by mbadreams » Sat Oct 03, 2009 8:38 pm
If 60! is written out as an integer, with how many consecutive 0's will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

IMO C

The value of 60! = 8.32098711 × 10^81 as per google. So that would mean that there are 73 0's following the integer 832098711. So how is the answer correct given correct?

Source: Manhattan gmat.
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by mbadreams » Sun Oct 04, 2009 7:29 pm
Can somebody please explain this?

Thanks
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its very easy easy easy n intersting.......

by ionmat » Mon Oct 05, 2009 2:33 am
mbadreams wrote:Can somebody please explain this?

Thanks
gnenerally... 2 * 5 = 10 gives u a 0
(eg:6! contains 1 * 2 * 3 * 4 * 5 * 6 = 720)

so u ve to find the no of (2 * 5) pairs in 60!.
To find them ...first find the no of numbers dvisible by 5 in 60! ie.60/5 = 12 (this is because 60! = 1 * 2 * 3 *4 *5 * 6 .....10 * .....15....*20..60 so there are nos divisible by 5 in every placeholders divisibel by 5) and they are 5,10,15,20,25,30...60 . so in each number except 5 there are more than one 5 hidden.but in every nos except first digit 5 the no of 5s present in the preceeding no is present in its succeding digits...eg(5 has only 5 but 10=5*2 which means 2 5s out of which one five is carried from its previous digit 5 ..this is just to make u aware of the duplicate repitions present in every nos except fist digit....so the answer is 12.don bother about the presence of 2 cos there are more no of 2s in 60! than those of 5s. but those extra 2s doest make sense without there 5 pairs...
Last edited by ionmat on Mon Oct 05, 2009 4:37 am, edited 1 time in total.
thanks
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by gmatplayer » Mon Oct 05, 2009 3:01 am
All you need to know is how the product of any two number results in zero at unit place.
In this case

10 x some intereger
and
5 x 2

so we need to find out how many 10s and 5s and 2s are there in 60!
Number of 10s = 6
Number of 5s & 2s = 6 (remember that 5 x 2 gives 10, so this pair is considered as one)

So total 12, which is one of the answer. Now here is the official trap comes:
Within the 60! there are two important terms 25 and 50.
where 25 =5x5
so 25 x 2 is another pair
Also, 50 has one 10 in it (10x5), so that is another pair

So in total there are 6 + 6 + 2 =14 zeros

IMO (C)
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by bikerguy.gmat » Mon Oct 05, 2009 5:13 am
gmatplayer wrote:All you need to know is how the product of any two number results in zero at unit place.
In this case

10 x some intereger
and
5 x 2

so we need to find out how many 10s and 5s and 2s are there in 60!
Number of 10s = 6
Number of 5s & 2s = 6 (remember that 5 x 2 gives 10, so this pair is considered as one)

So total 12, which is one of the answer. Now here is the official trap comes:
Within the 60! there are two important terms 25 and 50.
where 25 =5x5
so 25 x 2 is another pair
Also, 50 has one 10 in it (10x5), so that is another pair

So in total there are 6 + 6 + 2 =14 zeros

IMO (C)
wow! gmatplayer - you really nailed this; i missed the trap part.

mbadreams - can you share the official explanation please.
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by mbadreams » Mon Oct 05, 2009 7:13 am
Thanks guys for the reply...
The explanation given is as follows and it is from manhattan gmat cat. I am still wondering why google gives such as huge answer for 60!.

If an integer ends in 0, then that integer is divisible by 10; when that integer is divided by 10, the final 0 disappears. Therefore, the number of 0's at the end of an integer is exactly the number of times that integer is divisible by 10.
Since 10 = 2 × 5, this number (the number of times the integer is divisible by 10) is the smaller of the powers to which 2 and 5 are raised in the integer's prime factorization. For instance, an integer whose prime factorization contains 28 × 57 and an integer whose prime factorization contains 27 × 58 will both end with seven 0's.

In a factorial, such as 60!, there are many more 2's than 5's in the prime factorization: the factorial product has more numbers contributing 2's than 5's, and more of those numbers contribute repeated factors. Therefore, the question can be rephrased: How many 5's are in the prime factorization of (60!) ?

60! is the product of all the integers from 1 to 60, inclusive. Each of the twelve multiples of 5 in this range - 5, 10, 15, ..., 60 - contributes a 5 to the prime factorization. Additionally, a second 5 is contributed by each of the two multiples of 25 (25 and 50). Therefore, the prime factorization of 60! contains fourteen 5's, and so 60! will end with fourteen 0's.
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