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by molt_llest » Sun May 11, 2008 12:45 am
Thank you for your help.

If the number X is equal to the product of the first n positive integers {1,2,3...n}, what is the minimum value of n such that the last six digits of x are equal to 0?

a.- 20
b.- 25
c.- 30
d.- 35
e.- 40

Correct answer: b (answer in white color, drag over to see the answer)
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by molt_llest » Mon May 12, 2008 5:50 am
Anyone knows how to do it?
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by netigen » Mon May 12, 2008 9:46 am
To get the 0s in the last digit you need to multiply a number by 10

10 has two factors 2 x 5

which means you need to figure how many 5s and 2s will make up for 6 trailing zeros

Note the factor 2 can come from any even but 5 can only come from a number ending in 5 or a number ending in 0

Akshat's asnwer below is correct.
Last edited by netigen on Mon May 12, 2008 9:45 pm, edited 2 times in total.
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by akshatsingh » Mon May 12, 2008 9:30 pm
2 x 5 x 10 = 2 zeros
2 x 15 x 20 = 2 zeros
25 x 4 = 100 = 2 more zeroes


The answer should be 25. B
Aks
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by molt_llest » Tue May 13, 2008 6:48 am
Thank you for your answer, I realize now the way to get it. Thank you!
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