Problem:
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?
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
We are given that a<b<c and abc=100. As such, we know that a3<abc=100, which implies that a3<100.
Considering 3100​, we know that 43=64<100 and 53=125>100, and therefore, we can conclude that a≤4. As a is a positive integer, we have four possible values of a: 1,2,3,4. Since a must be a factor of 100, we can eliminate 3, leaving us with the cases a=1,2,4.
If a=1, then bc=100. Since b<c, we know b2<bc=100⟹1<b<10. Since b is a factor of 100 in this range, b can be 2,4,5. This gives the solutions:
(1,2,50),(1,4,25),(1,5,20).
If a=2, then bc=50. Since b<c, we know b2<bc=50⟹2<b<8. The factor of 50 in this range is 5, which gives:
(2,5,10).
If a=4, then bc=25. Since b<c, we know b2<bc=25⟹4<b<5, which gives no integer solution for b.