You can represent an odd integer with the expression 2n+1, where n is any integer. Write and solve an equation to find three consecutive odd integers that have a sum of 63

[tex](2n+1)+((2n+1)+2)+((2n+1)+4)=63[/tex]
[tex]6n+9=63[/tex]
[tex]6n=54[/tex]
[tex]n=9[/tex]

The numbers are:
[tex]2n+1=2(9)+1=19[/tex]
[tex]2n+3=21[/tex]
and
[tex]2n+5=23[/tex]

To check
19+21+23=63
63=63-->Check TRUE

Answer Link

facundo3141592 facundo3141592 · Answer 2 · 2021-08-12T19:15:09+02:00

The 3 consecutive odd numbers are 19, 21, and 23.

We know that we can write an odd number as (2*n + 1) where n is an integer.

We want to find 3 consecutive odd numbers that have a sum of 63.

First, notice that if (2*n + 1) is an odd number, the next odd number will be:

(2*(n + 1) + 1) = (2*n + 3)

and the next one is:

(2*(n + 1) + 3) = (2*n + 5)

Then 3 consecutive odd numbers can be written as:

(2*n + 1), (2*n + 3), and (2*n + 5)

Now the sum of these numbers must be equal to 63, then we have:

(2*n + 1) + (2*n + 3) + (2*n + 5) = 63

Now we can solve this for n.

6*n + 1 + 3 + 5 = 63

6*n = 63 - 1 - 3 - 5 = 54

n = 54/6 = 9

Now that we found the value of n, we can find the 3 odd numbers as:

(2*n + 1) = (2*9 + 1) = 19

(2*n + 3) = (2*9 + 3) = 21

(2*n + 5) = (2*9 + 5) = 23

The 3 consecutive odd numbers are 19, 21, and 23.

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