Problem 237

Solution 1:

We assign each of $r$, $s$ and $t$ a different number from the list $3$, $4$, $5$, and then determine the value of $r\times s+t$ for each possibility.

If $r=3$, $s=4$ and $t=5$, then $r\times s+t=3\times 4+5=17$, which is
odd.

If $r=3$, $s=5$ and $t=4$, then $r\times s+t=3\times 5+4=19$, which is
odd.

If $r=4$, $s=3$ and $t=5$, then $r\times s+t=4\times 3+5=17$, which is
odd.

If $r=4$, $s=5$ and $t=3$, then $r\times s+t=4\times 5+3=23$, which is
odd.

If $r=5$, $s=3$ and $t=4$, then $r\times s+t=5\times 3+4=19$, which is
odd.

If $r=5$, $s=4$ and $t=3$, then $r\times s+t=5\times 4+3=23$, which is
odd.

We have considered all possible ways to assign the values $3$, $4$
and $5$, and so there are $6$ such ways for which $r\times s+t$ is odd.

Solution 2:

First, recognize that both $3$ and
$5$ are odd, and $4$ is even.

When each of $r$, $s$ and $t$ is assigned a different number from the
list $3$, $4$, $5$, then either $r$ and $s$ are both odd, or exactly one of $r$ and $s$ is odd.

If $r$ and $s$ are both odd, then $r\times s$ is odd and $t=4$, and so $r\times s+t$ is odd.

If exactly one of $r$ and $s$ is odd, then $r\times s$ is even and $t$ is odd, and so $r\times s+t$ is odd.

Thus, the value of $r\times s+t$ is
always odd.

When assigning the values $3$, $4$, $5$, there are $3$ choices for the value of $r$, followed by $2$ choices for the value of $s$, and then $1$ choice for the value of $t$.

Since $r\times s+t$ is odd for all
such choices, then there are $3\times2\times1=6$ such ways to assign the
numbers.