The sum of three consecutive numbers is 42. What is the smallest of these numbers?

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Answer 1 · 2016-12-16T14:45:31

The smallest of the three consecutive integers summing to 42 is 13.

Let's call the smallest of the three consecutive numbers $s$ .

The next two consecutive integers, by definition of consecutive and the fact they are integers as: $s + 1$ and $s + 2$

We know there sum is 42 so we can add our three numbers and solve for $s$ :

$s + (s + 1) + (s + 2) = 42$

$s + s + 1 + s + 2 = 42$

$3s + 3 = 42$

$3s + 3 - 3 = 42 - 3$

$3s + 0 = 39$

$3s = 39$

$(3s)/3 = 39/3$

$s = 13$

Checking the solution:

The three consecutive integers would be:

$13$

$13 + 1 = 14$

$13 + 2 = 15$

Adding the three integers gives:

$13 + 14 + 15 = 27 + 15 = 42$