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No. of Zeros in 60!

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anurag_7 Just gettin' started! Default Avatar
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No. of Zeros in 60! Post Thu Jul 17, 2014 9:42 am
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    If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
    6
    12
    14
    42
    56

    I can only come up with 12. Help!!

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    Post Thu Jul 17, 2014 9:51 am
    Number of zeros are representation of number of pairs of (2x5) because 2x5=10 which makes one zero

    But 60! on factorizing will have higher power of 2 than the power of 5 because higher the prime number, lower the power available in any factorial

    Power of 5 in 60! = [60/5] + [60/5^2] + [60/5^3]
    Power of 2 in 60! = [60/2] + [60/2^2] + [60/2^3] + [60/2^4] + [60/2^5] + [60/2^6] = 30+15+7+3+1+0 = 56

    Where [60/5] refers the Greatest Integer Function
    i.e. [60/5]= 12
    and [60/5^2] = [60/25] = 2
    and [60/5^3] = [60/125] = 0

    Power of 5 in 60! = 12 + 2 +0 = 14

    which means, 60!=(2^56) x (5^14) ...and other factors = 10^14

    Therefore Number of Zeros = 14

    Answer: Option C

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    Post Thu Jul 17, 2014 9:56 am
    Using other basic method...

    Power of 5 will be available in the following numbers in 60!...

    5 x 10 x 15 x 20 x 25 x 30 x 35 x 40 x 45 x 50 x 55 x 60
    Powers of 5 in respective numbers
    5 --->1
    10 --->1
    15 --->1
    20 --->1
    25 --->2
    30 --->1
    35 --->1
    40 --->1
    45 --->1
    50 --->2
    55 --->1
    60 --->1

    Total powers of 5 in 60! = 1+1+1+1+2+1+1+1+1+2+1+1 = 14

    Answer: Option C

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    Post Thu Jul 17, 2014 10:00 am
    You must have missed the additional power of 5 available in 25 and 50 which if counted will correct your answer.

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    anurag_7 Just gettin' started! Default Avatar
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    Post Thu Jul 17, 2014 10:59 am
    Bhoopendra Sir,
    First of all thanks for responding. I calculated by counting the nos which will have 0 as the last digit i.e 10,20,30,40,50 and 60. Then by making the pairs which will make multiples of 10 i.e 2*5, 12*15, ... 52*55. By using these two logic I was able to get only 12. Sorry to say but still unable to figure out where I went wrong?

    Post Thu Jul 17, 2014 11:15 am
    Quote:
    If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
    6
    12
    14
    42
    56
    60! = 60*59*58*....*3*2*1.

    Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 60! will yield a 0 at the end of the integer representation of 60!.
    The prime-factorization of 60! is composed of FAR MORE 2'S than 5's.
    Thus, the number of 0's depends on the NUMBER OF 5's contained within 60!.

    To count the number of 5's, simply divide increasing POWERS OF 5 into 60.

    Every multiple of 5 within 60! provides at least one 5:
    60/5 = 12 --> twelves 5's.
    Every multiple of 5² provides a SECOND 5:
    60/5² = 2 --> two more 5's.
    Thus, the total number of 5's contained within 60! = 12+2 = 14.

    The correct answer is C.

    Another example:

    If 200! is written out as an integer, with how many consecutive 0’s will that integer end?

    Every multiple of 5 within 200! provides at least one 5:
    200/5 = 40 --> forty 5's.
    Every multiple of 5² provides a SECOND 5:
    200/5² = 8 --> eight more 5's.
    Every multiple of 5³ provides a THIRD 5:
    200/5³ = 1 --> one more 5.
    Thus, the total number of 5's contained within 200! = 40+8+1 = 49.

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    Post Thu Jul 17, 2014 12:56 pm
    Quote:
    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
    60! = (1)(2)(3)(4)(5)(6) . . . (57)(58)(59)(60)
    For every pair of one 2 and one 5, we get a product of 10, which accounts for one zero at the end of the integer.
    So, the question is "How many pairs of one 2 and one 5 are "hiding" in the product?"

    Well, there is no shortage of 2's hiding in the product. In fact, there are FAR MORE 2's than 5's. So, all we need to do is determine how many 5's are hiding in the product.
    60! = (1)(2)(3)(4)(5)(6) . . . (57)(58)(59)(60)
    = (1)(2)(3)(4)(5)(6)(7)(8)(9)(2)(5)(11)...(3)(5)...(4)(5)...(5)(5)...(6)(5)...(7)(5)...(8)(5)...(9)(5)...(2)(5)(5)...(11)(5)...(56)(57)(58)(59)(12)(5)
    In total, there are 14 5's hiding in the product.
    And there are MORE THAN 14 2's hiding in the product.

    So, there are 14 pairs of 2's and 5'2, which means the integer ends with 14 zeros

    Answer: C

    If you'd like additional practice, here's a similar question: http://www.beatthegmat.com/integers-t270158.html

    Cheers,
    Brent

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    Post Thu Jul 17, 2014 5:56 pm
    Hi Anurag,

    Please check the answer in your own way however

    10 - 1 zero
    20 - 1 zero
    30 - 1 Zero
    40 - 1 zero
    50x2 - 2 zero
    60 - 1 Zero

    But now
    5x2 =10 -- 1 Zero
    15x8 =120 -- 1 Zero
    25x4 =100 -- 2 Zero
    35x6 =210 -- 1 Zero
    45x12 =540 -- 1 Zero
    55*14 = 770 -- 1 Zero

    Total Zeros = 14

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    Post Thu Jul 17, 2014 6:01 pm
    Suggestion: looking at the numbers from the perspective of finding the total power of 5 available in 60! to find out number of zero would be the best approach.
    The logic is every zero is representation of every multiple of 10 but every 10 is obtained by making one pair of one 2 and one 5 [because 2 x 5 = 10]

    The power of any prime number in any factorial function will be higher if it's smaller Prime number and the power of any prime number that is higher will have lower power in comparison to the other prime numbers available in the number

    Therefore power of 2 available in the number will be higher than the power of 5 available in the number.

    But the numbers of pair that can be made by using one power of 2 and one power of 5 will be restricted by the power of prime number which is lower than the other

    Since the power of 5 is lower than the power of 2 in 60! therefore we only need to calculate the power of 5 which (as calculated in the second post on this thread) becomes 14

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