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 \( \boxed{14} \).
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