How many factors of 2020 have more than 3 factors? (As an example, 12 has 6 factors, namely 1, 2, 3, 4, 6, and 12.)
A. 6
B. 7
C. 8
D. 9
E. 10

Work through it with AI →

It won't just hand you the answer — it asks where you're stuck.How the checker works

Worked solution

1. 2020 AMC 8 官方解析

The number 2020 has 12 factors.
2020=1⋅20202020 = 1 \cdot 2020
=2⋅1010= 2 \cdot 1010
=4⋅505= 4 \cdot 505
=5⋅404= 5 \cdot 404
=10⋅202= 10 \cdot 202
=20⋅101= 20 \cdot 101

These factors may be classified as follows:
• The number 1 has exactly 1 factor.
• The numbers 2, 5, and 101 are primes, so each has exactly 2 factors.
• The number 4 has exactly 3 factors, namely 1, 2, and 4.
• The remaining 7 numbers, 10, 20, 202, 404, 505, 1010, and 2020, each have more than 3 factors. Thus 7 of the factors of 2020 have more than 3 factors.

2020 有 12 个约数 .
2020=1⋅20202020 = 1 \cdot 2020
=2⋅1010= 2 \cdot 1010
=4⋅505= 4 \cdot 505
=5⋅404= 5 \cdot 404
=10⋅202= 10 \cdot 202
=20⋅101= 20 \cdot 101

这些约数可以按照如下分类:
• 数字1有且只有1个约数 .
• 数字2 , 5和101都是质数,所有每个数都有且只有2个约数 .
• 数字4有且只有3个约数,分别是1 , 2和4 .
• 其余7个数字,10 , 20 , 202 , 404 , 505 , 1010 , 和2020,每个数都有3个以上的约数 .
因此2020的约数中有7个数具有3个以上的约数 .

In order to determine the number of factors of 2020, note that the prime factorization of 2020 is 2020=22⋅51⋅10112020 = 2^2 \cdot 5^1 \cdot 101^1.
This means that each factor of 2020 has the form 2a⋅5b⋅101c2^a \cdot 5^b \cdot 101^c, where a=0,1a = 0, 1, or 22; b=0b = 0 or 11; and c=0c = 0 or 11. Thus the number of factors of 2020 is 3⋅2⋅2=123 \cdot 2 \cdot 2 = 12.
By similar reasoning, any positive integer with at least 2 distinct prime factors, say pp and qq, has at least 4 factors, namely 11, pp, qq, and pqpq.
The factors of 2020 that have no more than 1 distinct prime factor are 1, 2, 4, 5, and 101. Each of these has at most 3 factors. The remaining 7 factors, 10, 20, 202, 404, 505, 1010, and 2020, each have at least 2 distinct prime factors. Therefore 7 of the factors of 2020 have more than 3 factors.

为了确定2020的所有约数的数量,可以将2020进行素因数分解如下
2020=22⋅51⋅10112020 = 2^2 \cdot 5^1 \cdot 101^1
这意味着2020的每个约数都可以表示为2a⋅5b⋅101c2^a \cdot 5^b \cdot 101^c的形式,其中a=0,1a = 0, 1或22;b=0b = 0或11;并且c=0c = 0或11. 因此,2020的约数数量为3⋅2⋅2=123 \cdot 2 \cdot 2 = 12.
类似地,任何具有至少2个不同的素数因子(例如pp和qq)的正整数都具有至少4个约数,即11,pp,qq和pqpq. 2020的约数中具有不超过1个素数因子的是1,2,4,51, 2, 4, 5和101101. 其中每个数最多具有3个约数 . 其余的7个因子,即10,20,202,404,505,101010, 20, 202, 404, 505, 1010和20202020,每个数至少具有2个素数因子 . 因此,在2020的所有约数中有7个数有3个以上的约数 .

Answer

B. 7

AI analysis — may contain mistakes.

Worked this one out on paper? Snap your steps and Gigle's AI math checker will circle the step where it went wrong.

More questions on this topic

  1. 1任何质数加1都成为偶数。( )
  2. 2一个5位数,十位上的数字是最小的质数,个位上的数字是最小的自然数,百位上的数字是最小的合数,千位上的数是最小的偶数,万位上最小的奇数,这个5位数是___.
  3. 3ξ={integers x:2<x≤12}\xi = \{\text{integers } x: 2 < x \le 12\} A={prime numbers}A = \{\text{prime numbers}\} B={factors of 12}B = \{\text{factors of 12}\} C={greater than 100}C = \{\text{greater than } \sqrt{100}\}…
  4. 4The smallest of a set of five number is an even number and it is prime. The range and the mean is 7. The median is also…
  5. 51995的约数共有( ) A. 16 B. 17 C. 18 D. 19