# if the sum of 3 consecutive numbers is 123, what are the 3 numbers?

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• John P
Lv 7
4 weeks ago

If you think about it you can do it in your head and on a scrap of paper. 3 consecutive numbers, so each must be about one third of 120. That's 40. Add on a bit, one more each time, and you are there.  No real need for algebra.

• 1 month ago

n + (n + 1) + (n + 2) = 123

3n + 3 = 123

3n = 123 - 3

3n = 120

n = 40

n + 1 = 40 + 1 = 41

n + 2 = 40 + 2 = 42

Therefore the 3 numbers are: 40,41,42..

• 1 month ago

x + (x+1) + (x+2) = 123

3x + 3 = 123

3x = 120

x = 40

4 numbers are 40, 40+1 and 40+2

• 1 month ago

40 , 41 , 42 ..

• Phil
Lv 6
1 month ago

The three consecutive numbers are 40, 41 and 42.

• 1 month ago

Hi

Marianna

Let,

The sum of 3 consecutive numbers is 123

numbers are (n-1), n, (n+1)

sum = 3n

so,

n = 123/3

= 41

Hence,the final answer is  40,41, 42

• 1 month ago

numbers are (n-1), n, (n+1) sum = 3n so n = 123/3 = 41

hence 40,41, 42

• 1 month ago

There's no such thing as "consecutive numbers."

Did you mean consecutive integers? Consecutive odd numbers? Consecutive even numbers? Consecutive perfect squares? Please clarify.

• 1 month ago

The 3 consecutive numbers are 40, 41, and 42.

• 1 month ago

A shortcut to figuring these out is to realize the middle number will be the average.

Divide by 3 and that's the middle number.

123/3 = 41

40, 41, 42

You can see why this works algebraically if you let x represent the *middle* integer. The prior integer will be x-1 and the subsequent integer will be x+1.

x-1 + x + x+1 = 123

3x = 123

x = 123/3

x = 41 <-- middle integer

40, 41, 42

A more "typical" way to do this, though, is to let x represent the smallest integer.

Let x be the smallest integer

Let x+1 be the middle integer

Let x+2 be the largest integer

x + x+1 + x+2 = 123

3x + 3 = 123

3x = 120

x = 120/3

x = 40 <-- smallest integer