2022 AMC 8

Complete problem set with solutions and individual problem pages

Problem 3 Easy

When three positive integers a, b, and c are multiplied together, their product is 100. Suppose a < b < c. In how many ways can the numbers be chosen?

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

  • E.

    4

Answer:E

Solution 1

The positive divisors of 100 are

1,2,4,5,10,20,25,50,100.

It is clear that 10\leq c\leq50, so we apply casework to c:

- If c=10, then (a,b,c)=(2,5,10).

- If c=20, then (a,b,c)=(1,5,20).

- If c=25, then (a,b,c)=(1,4,25).

- If c=50, then (a,b,c)=(1,2,50).

Together, the numbers a,b, and c can be chosen in \boxed{\textbf{(E) } 4} ways.

 

Solution 2

The positive divisors of 100 are

1,2,4,5,10,20,25,50,100.

We apply casework to a:

If a=1, then there are 3 cases:

- b=2,c=50

- b=4,c=25

- b=5,c=20

If a=2, then there is only 1 case:

- b=5,c=10

In total, there are 3+1=\boxed{\textbf{(E) } 4} ways to choose distinct positive integer values of a,b,c.

Related Problems
Problem 1 2022 AMC 8
Problem 2 2022 AMC 8
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Problem 5 2022 AMC 8
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