# Checking Cases

###### This wiki is incomplete.

PS: Here are examples of great wiki pages — Fractions, Chain Rule, and Vieta Root Jumping

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) \(\ \ 20\)

Correct Answer: D

Solution:If \( a = 1 \), then we can have \( b = 1, 2 \ldots 7,\) which gives us 7 solutions.

If \( a = 2 \), then we can have \( b = 1, 2,\) and \(3,\) which gives us 3 solutions.

If \( a = 3 \), then we can have \(b =1\) and \(2,\) which gives us 2 solutions.

If \( a =4, 5, 6\) or \(7,\) we can only have \(b = 1 \). This gives us 4 solutions.Hence, in total, there are \( 7 + 3 + 2 + 4 = 16 \) solutions.

Thus, the answer is (D).

Incorrect Choices:

(A),(B),(C), and(E)

These answers are just offered to confuse you.