If a number must have the last 6 digits as 0, then that number must have 10^6=2^6 x 5^6 as factor.figs wrote: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
OA:after some explanation
I meant Choose C 30! Didn't consider 25! because it has only one more 5 than 20! and is still less than 6dtweah wrote:If a number must have the last 6 digits as 0, then that number must have 10^6=2^6 x 5^6 as factor.figs wrote: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
OA:after some explanation
So let n!= G x 10^6. G contains factors of 2 and 5 of the form 2^m-6 and 5^n-6, where m is number of 2 and n number of 5 in the expression. So G will contain the leftover 2's and 5's, since we are only looking for the minimum.
20! has 10 even numbers so clearly m>= 10 but n is < 6. So NO.
30! has 15 even numbers and m=6, ( 30, 25, 20, 15, 10, 5), So it is the smallest that meets the parameters above.
Choose B.
25! has two more 5's than 20!, not one, since 25 = 5^2.dtweah wrote: I meant Choose C 30! Didn't consider 25! because it has only one more 5 than 20! and is still less than 6
Yes, Ian one of those costly errors!!Ian Stewart wrote:25! has two more 5's than 20!, not one, since 25 = 5^2.dtweah wrote: I meant Choose C 30! Didn't consider 25! because it has only one more 5 than 20! and is still less than 6
Vemuri has the right answer above; in the prime factorization of 25!, we have six 5's -- two from the 25, and one each from 20, 15, 10 and 5 -- and lots of 2's, so 25! is divisible by 10^6, and therefore ends in six zeros.
