Olympiad Maths Prep

Track / Stage 3 / 132 of 260 #132 of 2000

Problem 132

AMC 10/12, early questions
Number theory Difficulty 3.6 Find the answer

The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
(A) 11(B) 12(C) 13(D) 14(E) 15\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15

Official solution

As the unique mode is 88, there are at least two 88s.
As the range is 88 and one of the numbers is 88, the largest one can be at most 1616.
If the largest one is 1616, then the smallest one is 88, and thus the mean is strictly larger than 88, which is a contradiction.
If we have 2 8's we can add find the numbers 4, 6, 7, 8, 8, 9, 10, 12.
This is a possible solution but has not reached the maximum.
If we have 4 8's we can find the numbers 6, 6, 6, 8, 8, 8, 8, 14.
We can also see that they satisfy the need for the mode, median, and range to be 8. This means that the answer will be
(D) 14\boxed{\text{(D)}\ 14 }. ~By QWERTYUIOPASDFGHJKLZXCVBNM

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.