The sum of three numbers $a, b$ and $c$ is $60$. If we decrease $a$ by $1$, we get the value $N$. If we increase $b$ by $2$, we get the value $N$. If we multiply $c$ by $3$, we also get the value $N$. What is the value of $N$?

Answer 1

We have the equations \begin{align*}
a + b + c &= 60, \\
a - 1 &= N, \\
b + 2 &= N, \\
3c &= N.
\end{align*}Subtracting these equations in pairs, we get \begin{align*}
(b+2) - (a-1) &= 3, \\
(3c) - (b+2) &= 0, \\
(3c) - (a-1) &= 3b+3.
\end{align*}The first equation tells us that $b - a = 4,$ so $b = a + 4.$ The third equation tells us that $3c - a + 1 = 3(a+4) + 3,$ so $3c - 4a + 8 = 3a + 15.$ Thus, $4a - 3c = -7.$

Substituting the equation $a+b+c = 60$ into the equation $b = a+4,$ we get $2a + 4 + c = 60,$ so $2a + c = 56.$ Therefore, the equations $4a - 3c = -7$ and $2a + c = 56$ tell us that \[4(2a+c) - 2(4a-3c) = 8a + 4c - 8a + 6c = 10c = 4(56) - 2(-7) = 232.\]Therefore, $c = \frac{232}{10} = 23.2,$ so $N = 3c = \boxed{69.6}.$