# 1976 AHSME Problems/Problem 12

## Problem 12

A supermarket has $128$ crates of apples. Each crate contains at least $120$ apples and at most $144$ apples. What is the largest integer $n$ such that there must be at least $n$ crates containing the same number of apples? $\textbf{(A) }4\qquad \textbf{(B) }5\qquad \textbf{(C) }6\qquad \textbf{(D) }24\qquad \textbf{(E) }25$

## Solution

To find the largest number of "repeated" crates necessary, we must account for all the possibilities of the number of apples in each crate. Since each crate contains a minimum of $120$ apples and a maximum of $144$ apples, there are $144 - 120 + 1 = 25$ different amounts possible for the number of apples per crate.

Now, we have to count for the worst case scenario: the $25$ amounts are repeated as many times as possible. $25$ can go into $128$ exactly $5$ times because $5 \cdot 25 = 125$, which is less than $128$. This leaves a remainder of $3$ crates.

The worst case scenario would be that these $3$ crates have a different number of apples each. It doesn't actually matter how many apples are in these $3$ crates because any of the $25$ values would be repeated again anyway. So, the answer is $5 + 1 = \boxed{\textbf{(C) }6}$ ~jiang147369

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