# 2005 Alabama ARML TST Problems/Problem 15

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## Problem

Virginia has a pair of fair 8 sided dice with faces numbered 1-8 on each die. Montana has a pair of fair 8 sided dice with faces numbered with positive integers in such a way that when her pair of dice is rolled, the probability of any particular sum occurring is the same as when Virginia rolls her dice. The largest number on any face of either of Montana's dice is 11. Find the sum of the numbers on the faces of Montana's die whose faces are all less than 11.

## Solution

Since the probability of any particular sum occurring on Montana's dice is the same as when Virginia rolls her dice, if we make a grid of the possible sums, then it will be identical to Virginia's grid with the exception of the entries are rearranged. So the sum of the entries in the grid is the same as on Virginia's grid. The sum of the numbers on Virginia's grid is $(2+3+\cdots +9)+(3+4+\cdots +10)+\cdots +(9+10+\cdots +16)=576$. Since the entry in an intersection of a row and column is the sum of the number of the row and the number of the column, 576 is 8 times the sum of the numbers in the row and column of the grid. We divide by 8 to get 72 is the sum of the numbers on both dice. Since 16 is the highest valued entry in the grid, 11 must be on only one of the dice, so we only need to subtract 11 once to get $\boxed{61}$.