, 2024

Solution 1:

In this solution, we work backward from each of the given choices.
Since we are asked to find the smallest possible integer in the list, we
begin with the smallest of the five choices, $39$.

If the smallest integer in the list is $39$, then the largest integer in the list
is $39+14=53$ (since the three
integers have a range of $14$).

If the three integers have a mean of $50$, then they have a sum of $50\times3=150$. Two of the integers are
$39$ and $53$, and so the third (the middle) integer
is $150-39-53=58$.

Since $58$ is greater than $53$, this is not possible (the range of
these three integers is $58-39=19$,
not $14$).

If the smallest integer in the list is $40$ (the next smallest answer given), then
the largest integer in the list is $40+14=54$.

If two of the integers are $40$ and
$54$, then the third (the middle)
integer is $150-40-54=56$.

Since $56$ is greater than $54$, this is not possible (the range of
these three integers is $56-40=16$,
not $14$).

If the smallest integer in the list is $41$ (the next smallest answer given), then
the largest integer in the list is $41+14=55$.

If two of the integers are $41$ and
$55$, then the third (the middle)
integer is $150-41-55=54$.

We may confirm that the three integers $41$, $54$, $55$ indeed have a range of $14$ and a mean of $50$.

We have shown that $41$ is the
smallest of the five choices to satisfy the given conditions, and so
$41$ is the smallest possible integer
in the list.

Solution 2:

Assume that the list of $3$
integers, ordered from smallest to largest, is $a,b,c$.

Since $a$ is the smallest integer in
the list and $c$ is the largest, and
the list has a range of $14$, then
$c$ is $14$ more than $a$ or $c=a+14$.

The three integers have a mean of $50$, and so $\frac{a+b+c}{3}=50$ or $a+b+c=150$.

Substituting $c=a+14$, the previous
equation becomes $a+b+a+14=150$ or
$2a+b=136$.

To find the smallest possible value of $a$, we determine the largest possible
value of $b$, recalling that $a<b$ and $b<c$ and so $b<a+14$.

Since $2a$ is even for all possible
values of $a$, and $136$ is even, then $b$ must be even (since $2a+b=136$).

We begin by choosing an arbitrary value of $b$ and using this value to determine $a$ and $c$.

If $b=52$, then $2a=136-52=84$ and so $a=42$ and $c=42+14=56$.

In this case, the three integers are $42$, $52$, $56$.

We continue to increase the value of $b$ in order to determine the smallest
possible value for $a$.

If $b=54$, then $2a=136-54=82$ and so $a=41$ and $c=41+14=55$.

In this case, the three integers are $41,54,55$.

If $b=56$, then $2a=136-56=80$ and so $a=40$ and $c=40+14=54$.

In this case, the three integers are $40,56,54$, which is not possible since
$56>54$.

Continuing to increase the value of $b$ will continue to give values of $c$ that are less than $b$.

Decreasing the value of $b$ will give
values of $a$ that are greater than
$41$.

Therefore, the smallest possible integer in the list is $41$.