Find the integer $n,$ $0 \le n \le 180,$ such that $\cos n^\circ = \cos 259^\circ.$
Since the cosine function has period $360^\circ,$
\[\cos 259^\circ = \cos (259^\circ - 360^\circ) = \cos (-101^\circ).\]And since the cosine function is even, $\cos (-101^\circ) = \cos 101^\circ,$ so $n = \boxed{101}.$
101
Level 1
Precalculus
Problem: Find the integer $n,$ $0 \le n \le 180,$ such that $\cos n^\circ = \cos 259^\circ.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for the integer \( n \) such that \( \cos n^\circ = \cos 259^\circ \), we utilize the property of the cosine function, which states that \( \cos \theta = \cos (360^\circ - \theta) \) and \( \cos \theta = \cos (360^\circ k \pm \theta) \) for any integer \( k \).

First, we find the reference angle for \( 259^\circ \):
1. Calculate \( 360^\circ - 259^\circ = 101^\circ \).
2. This means \( \cos 259^\circ = \cos 101^\circ \).

Next, we consider the general solutions for \( \cos n^\circ = \cos 101^\circ \):
1. The cosine function is periodic with a period of \( 360^\circ \), so the solutions can be expressed as:
   - \( n = 101^\circ + 360^\circ k \) for any integer \( k \)
   - \( n = -101^\circ + 360^\circ k \)

Now, we need to find valid integer values of \( n \) within the range \( 0 \le n \le 180 \):
1. For \( k = 0 \):
   - \( n = 101^\circ \) (valid since \( 0 \le 101 \le 180 \))
   - \( n = -101^\circ \) (not valid since it is negative)

2. For \( k = 1 \):
   - \( n = 101^\circ + 360^\circ = 461^\circ \) (not valid since it exceeds 180)
   - \( n = -101^\circ + 360^\circ = 259^\circ \) (not valid since it exceeds 180)

3. For \( k = -1 \):
   - \( n = 101^\circ - 360^\circ = -259^\circ \) (not valid since it is negative)
   - \( n = -101^\circ - 360^\circ = -461^\circ \) (not valid since it is negative)

Since the only valid solution within the specified range is \( n = 101^\circ \), we conclude that the final answer is \( 101 \).<|END_OF_LATENT|> The final answer is $101$.