请教各位一道数学题哈！

astronautking

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

luoweijia0511

(n-1)n(n+1)=n的三次方-N

所以当N为10的时候，是990

luoweijia0511

答案是A？

luoweijia0511

写错了，，N-1乘以N乘以N+1等于990

得出N的三次方-10等于990

所以等于1000-10=990

所以为A

astronautking

。。。我也选的A，但答案是B啊！郁闷！

luoweijia0511

有详细解答么？

astronautking

今天PREP模拟遇到的，没有详解啊！

sdcar2010

You guys missed the point of this question.

If n, then the product of of all the integers from 1 to n, inclusive, is 1*2*3*. . .*(n-1)*n. Let's say this product is A.

Then A is a MULTIPLE of 990. So A = m*990 = m* 3*3*2*5*11. So A has a factor  of 11. Then n has to be equal to or great than 11 in order to contain a factor of 11.

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nikejason

你这样看，要是990的倍数，那这个1*...n的积就必须包含9,10,11，因为990=9*10*11

astronautking

You guys missed the point of this question.

If n, then the product of of all the integers from 1 to n, inclusive, is 1*2*3*. . .*(n-1)*n. Let's say this product is A.

Then A is a MULTIPLE of 990. So A = m*990 = m* 3*3*2*5*11. So A has a factor  of 11. Then n has to be equal to or great than 11 in order to contain a factor of 11.

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-- by 会员 sdcar2010 (2011/4/24 20:33:58)

Yep, I also conduct that n has to be equal to or greater than 11, so 10 is the least possible value because if n is 10, A will not contain 11. Could find where my thought is wrong? Thanks a lot!