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.

See Also

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