2019 AMC 10B 第 19 题

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19.

SS100,000100,000 的所有正整数因数组成的集合。有多少个数可以表示为 SS 中两个不同元素的乘积?

Let SS be the set of all positive integer divisors of 100,000.100,000. How many numbers are the product of two distinct elements of S?S?

9898

100100

117117

119119

121121

答案:C
知识点:质因数分解因数个数
难度评级:2010
解答:

首先注意到 100,000=2555100,000=2^5\cdot5^5

因此,SS 中任一元素都形如 2a5b2^a5^b,其中 0a,b50 \leq a,b \leq 5

设不同的 x,ySx,y \in S,并且 则 因此 表面上共有 种指数对。不过有些只能在 x=yx=y 时出现,即 (a,b)=(c,d)(a,b)=(c,d) x=2a5b,x = 2^a5^b, y=2c5d.y=2^c5^d. xy=2a+c5b+d.xy = 2^{a+c}5^{b+d}. 0a+c,b+d10.0 \leq a+c,b+d \leq 10. (10+1)(10+1)=121(10+1)(10+1)=121

a+c=0a+c=0,则必须有 a=0,c=0a=0,c=0

a+c=10a+c=10,则必须有 a=5,c=5a=5,c=5

a+ca+c 取其他值时,可以让 aca \neq c

b+db+d 也有类似结构。因此若 则 从而 x=yx=ya+c,b+d{0,10},a+c,b+d \in \{0,10\}, (a,b)=(c,d),(a,b)=(c,d),

因此必须排除 44 种选择,剩下 1214=117.121-4=117.

所以答案是 C

First, note that 100,000=2555.100,000=2^5\cdot5^5.

Therefore, any element of SS must be of the form 2a5b2^a5^b with 0a,b5.0 \leq a,b \leq 5.

Suppose I have distinct x,ySx,y \in S with x=2a5b,x = 2^a5^b,y=2c5d.y=2^c5^d. Then, xy=2a+c5b+d.xy = 2^{a+c}5^{b+d}. Thus, 0a+c,b+d10.0 \leq a+c,b+d \leq 10. This means that there are (10+1)(10+1)=121(10+1)(10+1)=121 possible exponent pairs for a product. However, some products can arise only when x=yx=y, namely when (a,b)=(c,d)(a,b)=(c,d).

If a+c=0,a+c=0, then a=0,c=0a=0,c=0 must be true.

If a+c=10,a+c=10, then a=5,c=5a=5,c=5 must be true.

With any other value of a+c,a+c, we can have ac.a \neq c.

Similar structure holds for b+d.b+d. Thus, if a+c,b+d{0,10},a+c,b+d \in \{0,10\}, then (a,b)=(c,d),(a,b)=(c,d), thus making x=y.x=y.

This means we have to eliminate 44 choices, leaving 1214=117.121-4=117.

Thus, the answer is C .

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