What three consecutive integers sum to 42?

2 Answers
F
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#

smendyka
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

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