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.
60!
This topic has expert replies
gnenerally... 2 * 5 = 10 gives u a 0mbadreams wrote:Can somebody please explain this?
Thanks
(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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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)
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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- Newbie | Next Rank: 10 Posts
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wow! gmatplayer - you really nailed this; i missed the trap part.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)
mbadreams - can you share the official explanation please.
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.
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.