The sum of three consecutive integers is 123. What would be the three integers?

2 Answers
Jul 23, 2017

The 3 integers are 40, 41, and 42.

Explanation:

Let n = the first integer, then n + 1, is second, and n + 2, is the third:

#n + n + 1 + n + 2 = 123#

#3n + 3 = 123#

#3n = 120#

#n = 40#

#n + 1 = 41#

#n + 2 = 42#

Jul 24, 2017

The three integers are #40#, #41#, and #42#.

Explanation:

There are two ways to solve this problem.

A] Since we are looking for three consecutive integers that add up to #123#, if we divide the total by #3#, we get the average, which is also the middle integer.

It follows that the first and third integers will be that integer minus and plus one respectively.

#123/3=41#

#41# is the middle integer.

#41-1=40#

#40# is the first integer.

#41+1=42#

#42# is the third integer.

Hence, the three integers are #40#, #41#, and #42#.


B] The other way to solve this problem involves very basic algebra (even though this question is in Prealgebra), in which we represent the first integer as #x#, in which case the second and third integers will be #x+1# and #x+2#.

Since we know the sum of these three integers is #123#, we can write:

#x+x+1+x+2=123#

Adding the #x# separately and the numbers separately, we get:

#3x+3=123#

Subtract #3# from each side.

#3x+3-3=123-3#

#3x+cancel3-cancel3=120#

#3x=120#

Divide each side by #3#.

#(3x)/3=120/3#

#(cancel3x)/(cancel3)=40#

#x=40#

#:.x+1=40+1=41# and #x+2=40+2=42#

Hence the three consecutive integers are #40#, #41#, and #42#.