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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100