Question

Three consecutive integers have a sum of 45. What is the least of these integers?

Answer 1

Let the three consecutive integers be represented as \( x \), \( x+1 \), and \( x+2 \).

We can express the sum of these three integers as follows:
\[
x + (x + 1) + (x + 2) = 45
\]

Now, we simplify the left-hand side:
\[
x + x + 1 + x + 2 = 3x + 3
\]

Setting this equal to 45 gives us the equation:
\[
3x + 3 = 45
\]

To isolate \( x \), we first subtract 3 from both sides:
\[
3x = 45 - 3
\]
\[
3x = 42
\]

Next, we divide both sides by 3:
\[
x = \frac{42}{3} = 14
\]

Now we have found that \( x = 14 \). Therefore, the three consecutive integers are:
- The first integer (least) is \( 14 \),
- The second integer is \( 15 \),
- The third integer is \( 16 \).

Thus, the least of these integers is:
\[
\boxed{14}
\]

To verify, we can check the sum:
\[
14 + 15 + 16 = 45
\]
Since the sum is indeed 45, our solution is confirmed. The least of the three integers is \( \boxed{14} \).