For this problem we know that the sum of 3 consecutive numbers is one hundred five. Let's assume that the 3 numbers is: a, a+1, a+2
And we can set up the following equation:
[tex]a+(a+1)+(a+2)=105[/tex]And we can solve for a like this:
[tex]3a+3=105[/tex]We can subtract 3 in both sides and we got:
[tex]3a=102[/tex]And we can divide both sides by 3 and we got:
[tex]a=\frac{102}{3}=34[/tex]And we can conclude that the smallest number is 34
