What are 3 consecutive numbers whose sum is 54?

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Answer 1 · 2016-05-04T18:10:55

Another approach

The three number are: 17, 18 and 19

Let the first number be $n$ $n$
then we have:
$n + (n + 1) + (n + 2) = 54$ $n+(n+1)+(n+2)=54$

$\Rightarrow 3 n + 3 = 54$ $=> 3n+3=54$

Subtract $3$ $color(red)(3)$ from both sides

$\Rightarrow 3 n + 3 - 3 = 54 - 3$ $=>color(green)(3n+3color(red)(-3)=54color(red)(-3)$

but +3-3=0

$\Rightarrow 3 n + 0 = 54 - 3$ $=>3n+0=54-3$

Notice that the 3 has changed sides and as a consequence we now have 3n on its own

$\Rightarrow 3 n = 54 - 3$ $=> 3n=54-3$

Divide both sides by $3$ $color(red)(3)$ to get n on its own

$\Rightarrow \frac{3}{3} n = \frac{54}{3} - \frac{3}{3}$ $=> color(green)(3/(color(red)(3))n=54/(color(red)(3))-3/(color(red)(3)))$

But $\frac{3}{3} = 1 so \frac{3}{3} n$ $3/3=1 " so "3/3n$ becomes just $n$ $n$

$\Rightarrow n = \frac{51}{3} = 17$ $=>n=51/3=17$

So we have : $17 + 18 + 19 = 54$ $17+18+19=54$