#math/mental For **two-by-one multiplication** the trend continues. We go left to right and solve the problem sequentially. This time the loop is multiply add multiply add etc. Consider $42\times 7$ $ \begin{align} 42\times 7 & \text{ is} \\ 280+14 & \text{ is} \\ 294 \end{align} $ The $2\times 7=14$ is calculated in the same breath. Eventually $42\times7$ will be as well when we do larger problems. The last digit multiplication only has two possibilities. One, it results in a single digit like $74\times 2=14\textcolor{#1f9d57}{8}$. Second, it results in two digits like $ \begin{align} 76\times 2 =1\textcolor{#1f9d57}{4} & 0 \\ +\textcolor{#1f9d57}{1} & 2 \\ \hline 15 & 2 \end{align} $ . While saying "one-hundred" we should be looking ahead and seeing if we are case one or two. That way we quickly say "forty" or "fifty" in that same breath. --- For **two-by-two multiplication** you can either round and add or subtract the multiplied difference or you can find a factor to simplify the problem. Consider the following two approaches for $46\times 42$ both with the factoring method. Approach 1 $ \begin{array}{ccc} \begin{array}{r} 46 \\ \times\ 42 \\ \hline \phantom{} \end{array} & \implies \begin{array}{r} 42 \\ \times\ (7\times \textcolor{#1f9d57}{6}) \\ \hline \phantom{} \end{array} &\implies \begin{array}{r} 276 \\ \times\ 7 \\ \hline 1932 \end{array} & \qquad \phantom{000} \\[25pt] \end{array} $ Approach 2 $ \begin{array}{ccc} \begin{array}{r} 46 \\ \times\ 42 \\ \hline \phantom{}\end{array} & \implies \begin{array}{r} 46 \\ \times\ (\textcolor{#1f9d57}{7}\times 6) \\ \hline\phantom{} \end{array} &\implies \begin{array}{r} 322 \\ \times\ 6 \\ \hline 1932\phantom{}\end{array} & \qquad \phantom{000} \\[25pt] \end{array} $ The final multiplication in approach 2 is easier. Generally, reserving the smaller factor for the three-by-one multiplication is easier. Alternatively the rounding method could be applied but typically requires more working memory. $ \begin{array}{ccc} \begin{array}{r} 46 \\ \times\ 42 \\ \hline \phantom{}\end{array} & \implies \begin{array}{r} 46 \\ \times\ (40+ 2) \\ \hline \phantom{}\end{array} &\implies \begin{array}{r} 1840 \\ +\ (46 \times 2) \\ \hline 1932\end{array} & \qquad \\[25pt] \end{array} $ _The factoring method is greatly enhanced with immediate recall of [[6 - Primitives Pt 2|friendly products]]._ --- For **two-by-eleven multiplication** add the two digits together and place that sum between them. Consider $42\times 11$ $ \begin{align} 42\times 11 & \text{ is} \\ 4\ [4{+}2]\ 2 & \text{ is} \\ 462 \end{align} $ When the inner sum reaches two digits we carry the one to the left. Consider $76\times 11$ $ \begin{align} 76\times 11 & \text{ is} \\ 7\ (7{+}6)\ 6 & \text{ is} \\ 7\ (1\textcolor{#1f9d57}{3})\ 6 & \text{ is} \\ 8\textcolor{#1f9d57}{3}6 \end{align} $ --- For **three-by-eleven multiplication** the same idea extends. Writing the number as $\text{ABC}$ the answer reads $A\ (A{+}B)\ (B{+}C)\ C$ — the outer digits stay and each neighboring pair is summed inward. Consider $314\times 11$ where $A=3,\ B=1,\ C=4$ $ \begin{align} \text{ABC}\times 11 & = A\ (A{+}B)\ (B{+}C)\ C \\ 314\times 11 & = 3\ (3{+}1)\ (1{+}4)\ 4 \\ & = 3454 \end{align} $ Carries propagate from right to left exactly as before. Consider $768\times 11$ where $A=7,\ B=6,\ C=8$ $ \begin{align} 768\times 11 & = 7\ (7{+}6)\ (6{+}8)\ 8 \\ & = 7\ (1\textcolor{#1f9d57}{3})\ (1\textcolor{#1f9d57}{4})\ 8 \\ & = 8448 \end{align} $ --- For **two-by-two multiplication** the same problem can be solved several ways. To see this, apply all three methods from this chapter to $73\times 49$; each one lands on $3577$. Begin with the addition method, splitting $73$ into $70+3$ $ \begin{array}{ccc} \begin{array}{r} 73 \\ \times\ 49 \\ \hline \phantom{}\end{array} & \implies \begin{array}{r} (70+3) \\ \times\ 49 \\ \hline \phantom{}\end{array} &\implies \begin{array}{r} 3430 \\ +\ 147 \\ \hline 3577\end{array} & \qquad \\[25pt] \end{array} $ where $3430=70\times 49$ and $147=3\times 49$. Next the subtraction method, splitting $49$ into $50-1$ $ \begin{array}{ccc} \begin{array}{r} 73 \\ \times\ 49 \\ \hline \phantom{}\end{array} & \implies \begin{array}{r} 73 \\ \times\ (50-1) \\ \hline \phantom{}\end{array} &\implies \begin{array}{r} 3650 \\ -\ 73 \\ \hline 3577\end{array} & \qquad \\[25pt] \end{array} $ where $3650=50\times 73$ and the subtracted $73$ is $1\times 73$. The trailing $77$ also comes out as $50+27$ — fifty plus the complement of $73$ — or as the complement of $73-50=23$. Finally the factoring method, writing $49$ as $7\times 7$ $ \begin{array}{ccc} \begin{array}{r} 73 \\ \times\ 49 \\ \hline \phantom{}\end{array} & \implies \begin{array}{r} 73 \\ \times\ (7\times \textcolor{#1f9d57}{7}) \\ \hline \phantom{}\end{array} &\implies \begin{array}{r} 511 \\ \times\ 7 \\ \hline 3577\end{array} & \qquad \\[25pt] \end{array} $ where $511=73\times 7$.