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# Checking Cases

Problems which look complicated can often be broken down into simplier scenarios, where we have more information. This makes the individual case easier to consider, which allows us to solve the entire problem.

How many triples of positive integers are there such that $$a \times b + c = 8$$?

A) 4
B) 8
C) 12
D) 16
E) 20

Solution: If $$a = 1$$, then we can have $$b = 1$$ to 7, which gives us 7 solutions.
If $$a = 2$$, then we can have $$b = 1$$ to 3, which gives us 3 solutions.
If $$a = 3$$, then we can have $$b =1$$ to 2, which gives us 2 solutions.
If $$a =4$$ to 7, we can only have $$b = 1$$. This gives us 4 solutions.
Hence, in total, there are $$7 + 3 + 2 + 4 = 16$$ solutions.

Note by Arron Kau
2 years, 5 months ago