Verified Answer :

TaylorBayley

\(n+(n+2)+(n+4)=153\) -- combine like terms
\(3n+6=153\) -- subtract 6 from both sides
\(3n=147\) -- divide by 3
\(n=49\)

So the numbers are 49, (49+2) and (49+4)
I.e. 49, 51, and 53
But the question only wants the smallest (i.e. the first of the consecutive odd numbers).

The smallest of the 3 consecutive odd numbers that sum to 153 is 49

mariamikayla

\(a+a+2+a+4=153 \\ \\ 3a+6=153 \\ \\ 3a=153-6 \\ \\ 3a=147 \\ \\ \boxed{a=\frac{147}{3}=49- \ the \ smallest} \\ \\ a+2=49+2=51 \\ a+4=49+4=53\)

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