If n is a positive integer and the product of all the integers from 1 to n,inclusive,is a multiple of 990,what is the least possible value of n
A 10
B 11
C 12
D 13
E 14
A 10
B 11
C 12
D 13
E 14
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Hi,winniethepooh wrote:Hey Frankenstine, where do 7 and 8 divide 990 with no remainder?
And if this is not the question wants then why haven't u selected 10 as the least possible value?
Hi Frankenstein,Frankenstein wrote:Hi,winniethepooh wrote:Hey Frankenstine, where do 7 and 8 divide 990 with no remainder?
And if this is not the question wants then why haven't u selected 10 as the least possible value?
I think you have misinterpreted the question. What the question says is:
Find the least value of n such that 990 is a factor of n!.
990 = 2*3^2*5*11
Now for 990 to be a factor of n!, 11 should be a factor of n!. The least value of n that satisfies this condition is 11.
Consider 10! = 1.2.3...10
11 is not a factor of this right?
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