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Number theory Difficulty 3.1 AMC 10/12 Find the answer

When three positive integers aa, bb, and cc are multiplied together, their product is 100100. Suppose a<b<ca < b < c. In how many ways can the numbers be chosen?

Pick one

Solution

The positive divisors of 100100 are 1,2,4,5,10,20,25,50,100.1,2,4,5,10,20,25,50,100.
It is clear that 10≤c≤50,10\leq c\leq50, so we apply casework to c:c:

If c=10,c=10, then (a,b,c)=(2,5,10).(a,b,c)=(2,5,10).
If c=20,c=20, then (a,b,c)=(1,5,20).(a,b,c)=(1,5,20).
If c=25,c=25, then (a,b,c)=(1,4,25).(a,b,c)=(1,4,25).
If c=50,c=50, then (a,b,c)=(1,2,50).(a,b,c)=(1,2,50).
Together, the numbers a,b,a,b, and cc can be chosen in (E) 4\boxed{\textbf{(E) } 4} ways.
~MRENTHUSIASM

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.