# Problem of the Week Problem E and Solution Just an Average Sum

## Problem

Faisal chooses four numbers. When each number is added to the mean (average) of the other three, the following sums are obtained: $$25$$, $$37$$, $$43$$, and $$51$$.

Determine the mean of the four numbers Faisal chose.

Extra Problem: Can you interpret the following picture puzzle? You may need to research the meanings of some mathematical symbols used in the puzzle.

## Solution

Let $$a,b,c,$$ and $$d$$ represent the four numbers. It is possible to precisely determine the four numbers, but the problem asks for only their average, which is $$\frac{a+b+c+d}{4}$$.

When the first number is added to the average of the other three numbers, the result is $$25$$. Thus,

$a+\frac{b+c+d}{3}=25$

which can be rewritten as

$3a+b+c+d=75 \tag{1}$

When the second number is added to the average of the other three numbers, the result is $$37$$. Thus,

$b+\frac{a+c+d}{3}=37$

which can be rewritten as

$a+3b+c+d=111 \tag{2}$

When the third number is added to the average of the other three numbers, the result is $$43$$. Thus,

$c+\frac{a+b+d}{3}=43$

which can be rewritten as

$a+b+3c+d=129 \tag{3}$

When the fourth number is added to the average of the other three numbers, the result is $$51$$. Thus,

$d+\frac{a+b+c}{3}=51$

which can be rewritten as

$a+b+c+3d=153 \tag{4}$

Adding equations $$(1)$$, $$(2)$$, $$(3)$$, and $$(4)$$, we obtain $$6a+6b+6c+6d=468$$. Dividing this equation by $$6$$ gives $$a+b+c+d=78$$. It follows that $$\dfrac{a+b+c+d}{4}=19.5$$.

Therefore, the average of the four numbers is $$19.5$$.

Although it is not required, we could solve the system of equations to determine that the numbers are: $$-1.5$$, $$16.5$$, $$25.5$$, and $$37.5$$.

Extra Problem Solution:
The notation $$\dfrac{1}{n} \displaystyle \sum_{i=1}^{n} x_i$$ is a mathematical short form which represents the average of the $$n$$ numbers $$x_1,x_2,\ldots,x_n$$. So the picture puzzle can be interpreted as “Be greater than average”.