MHB Problem #73: Finding the Minimum Value of Integers with an Average of 75

[Jameson]

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]

Congratulations to the following members for their correct solutions:

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

Solution (from Reckoner):

Spoiler

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.