The problem states that a set of 4 consecutive integers has a sum of 42. We need to find the least of the 4 integers.

1. Problem Description

The problem states that a set of 4 consecutive integers has a sum of

4

2. We need to find the least of the 4 integers.

2. Solution Steps

Let $x$ be the smallest of the four consecutive integers. Then the four consecutive integers are $x$, $x+1$, $x+2$, and $x+3$.

The sum of these integers is given as

4

2. So, we have the equation:

$x + (x+1) + (x+2) + (x+3) = 42$

Combining the terms, we get:

$4x + 6 = 42$

Subtracting 6 from both sides:

$4x = 42 - 6$

$4x = 36$

Dividing by 4:

$x = \frac{36}{4}$

$x = 9$

Thus, the four consecutive integers are 9, 10, 11, and

1

2. The least of these integers is

9.

3. Final Answer

9