# Problem of the week #73 - August 19th, 2013

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#### Jameson

Staff member
Here's a problem geared towards those beginning high school. For those students from the US, welcome back to school!
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Problem: The average (arithmetic mean) of 4 different integers is 75. If the largest integer is 90, what is the least possible value of the smallest integer?
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#### Jameson

Staff member
Congratulations to the following members for their correct solutions:

1) MarkFL
3) anemone
4) Reckoner
5) Deveno

Solution (from Reckoner):
Call the three smaller integers $x_1,$ $x_2,$ and $x_3,$ with $x_1 < x_2 < x_3 < 90.$ Then we have
\begin{gather*}
\frac{x_1 + x_2 + x_3 + 90}4 = 75\$5pt] \Rightarrow x_1 = 210 - (x_2 + x_3). \end{gather*} Since x_2 < x_3 < 90, \[ x_2 + x_3 \leq 88 + 89 = 177$
so that
$x_1 \geq 210 - 177 = 33.$
Therefore, $33$ is the least possible value of the smallest integer.

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