请教各位一道数学题哈！

astronautking · 发表于 2011-4-24 19:56:12

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

rainny0906 · 发表于 2011-4-24 23:34:33

把990分解为9*10*11,那1*2*3*...*n应该至少包含11这个数，所以，最小的n应该为11.
如果是10，那the product就是3628800，无法被990整除。

astronautking · 发表于 2011-4-24 23:12:21

Got it! Zui xiao ke neng zhi

sdcar2010 · 发表于 2011-4-24 20:53:19

When n is 10, n! is NOT a multiple of 990.

When n is 11, n! IS a multiple of 990.

So 11 is the LEAST POSSIBLE value of n, when n! is a multiple of 990.

Reading comprehension.

astronautking · 发表于 2011-4-24 20:48:13

各位答案是B啊，求解！

lysdelyon · 发表于 2011-4-24 20:44:13

我也觉得是B。。

luoweijia0511 · 发表于 2011-4-24 20:43:21

我对这个题目的意思不是很理解，楼上的能用中文翻译一下吗？谢谢了。我的感觉好像是连续的乘积。。

astronautking · 发表于 2011-4-24 20:40:24

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!

nikejason · 发表于 2011-4-24 20:40:06

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

sdcar2010 · 发表于 2011-4-24 20:33:58

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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请教各位一道数学题哈！

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