2008 iTest Problems/Problem 40

Problem

Find the number of integers n that satisfy $\textit{both}$ of the following conditions: $208<n<2008$ , n has the same remainder when divided by $24$ or by $30$ .

Solution

First, find which numbers have the same remainder when divided by $24$ or by $30$ . With some experimentation, $n$ must be congruent to numbers from $0$ to $23$ modulo $120$ , so $n = 120k + m$ , where $k$ is an integer and $0 \le m \le 23$ .

When $2 \le k \le 16$ , all values of $m$ will result in an $n$ where $208 < n < 2008$ . Additionally, when checking $k = 1$ and $k = 23$ , no values of $m$ satisfies the boundaries. Thus, there are $15 \cdot 24 = \boxed{360}$ integers that satisfy both conditions.

2008 iTest (Problems)
Preceded by: Problem 39	Followed by: Problem 41
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2008 iTest Problems/Problem 40

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