What three consecutive integers sum to 42?

2 Answers
Nov 6, 2017

13, 14, 15

Explanation:

Let:
x = the first integer
x+1 = the second integer
x+2 = the third integer

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

Simplifying further,
3x+3=42
3x=42-3
3x=39

x = 13
x+1 = 14
x+2 = 15

Nov 6, 2017

See a solution process below:

Explanation:

Let's call the first integer: n

Then the next two consecutive integers are: n + 1 and n + 2

From the problem we know:

n + (n + 1) + (n + 2) = 42

We can now solve for n:

n + n + 1 + n + 2 = 42

n + n + n + 1 + 2 = 42

1n + 1n + 1n + 1 + 2 = 42

(1 + 1 + 1)n + 3 = 42

3n + 3 = 42

3n + 3 - color(red)(3) = 42 - color(red)(3)

3n + 0 = 39

3n = 39

(3n)/color(red)(3) = 39/color(red)(3)

(color(red)(cancel(color(black)(3)))n)/cancel(color(red)(3)) = 13

n = 13

The first integer is: 13

The next two integers are:

n + 1 = 13 + 1 = 14

n + 2 = 13 + 2 = 15

Another process for solving the Three Consecutive Interger Problem is to divide the number the e integers sum up to and add 1 and subtract 1 to get the three integers:

42 -: 3 = 14

14 - 1 = 13

14 + 1 = 15

The three integers are the same as above: 13, 14, 15