Consider two positive even integers less than 15 (not necessarily distinct). When the sum of these two numbers is added to their product, how many different possible values may result?
Consider two positive even integers less than 15 (not necessarily distinct).
When the sum of these two numbers is added to their product, how many different possible values may result?
\(\text{Let $n_1\lt 15$ is an positive even integer.} \\ \text{Let $n_2\lt 15$ is an positive even integer.} \\ \text{The sum of these two numbers is added to their product $n_1+n_2+n_1\cdot n_2 $.} \)
\(\text{Because $n_1$ is even, we can set: $n_1 = 2i \Rightarrow i_{\text{min}} = 1 $ and $i_{\text{max}} = 7 $ } \\ \text{Because $n_2$ is even, we can set: $n_2 = 2j \Rightarrow j_{\text{min}} = 1 $ and $j_{\text{max}} = 7 $ } \)
\(\begin{array}{|rcll|} \hline && n_1+n_2+n_1\cdot n_2 \\ &=& 2i + 2j + 2i2j \\ &=& 2(i+j+2ij) \quad & 1 \leq i \leq 7 \qquad 1 \leq j \leq 7 \\ \hline \end{array} \)
\(2(i+j+2ij) \quad 1 \leq i \leq 7 \qquad 1 \leq j \leq 7 \\ \begin{array}{|c|r r r r r r r|} \hline (i,j) & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 8 & 14 & 20 & 26 & 32 & 38 & 44 \\ 2 & & 24 & 34 & \color{red}44 & 54 & 64 & 74 \\ 3 & & & 48 & 62 & 76 & 90 & 104 \\ 4 & & & & 80 & 98 & 116 & 134 \\ 5 & & & & & 120 & 142 & 164 \\ 6 & & & & & & 168 & 194 \\ 7 & & & & & & & 224 \\ \hline \end{array}\)
Only 44 is double.
There may result 27 different possible values