Henry Ernest Dudeney/Puzzles and Curious Problems/173 - Short Cuts/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $173$

Short Cuts
Can you multiply $993$ by $879$ mentally?
It is remarkable that any two numbers of two figures each,
where the tens are the same, and the sum of the units make ten, can always be multiplied thus:
$97 \times 93 = 9021$
Multiply the $7$ by $3$ and set it down,
then add the $1$ to the $9$ and multiply by the other $9$, $9 \times 10 = 90$.
This is very useful for squaring any number ending in $5$, as $85^2 = 7225$.
With two fractions, when we have the whole numbers the same and the sum of the fractions equal unity,
we get an easy rule for multiplying them.
Take $7 \tfrac 1 4 \times 7 \tfrac 3 4 = 56 \tfrac 3 {16}$.
Multiply the fractions $= \tfrac 3 {16}$, add $1$ to one of the $7$'s, and multiply by the other, $7 \times 8 = 56$.


Solution

To multiply $993$ by $879$, proceed as follows:

Transfer $7$ from $879$ to $993$, and we get $872$ and $1000$.
Multiply these together to make $872 \, 000$.
Subtract $872$ from $993$ to get $121$.
Multiply $121$ by the $7$ we mentioned above, and get $847$.

Add the two results together, and you get $872 \, 847$.


Proof

\(\ds \) \(\) \(\ds \paren {879 - 7} \paren {993 + 7} + \paren {993 - 872} \times 7\)
\(\ds \) \(=\) \(\ds 879 \times 993 - 7 \times 993 + 7 \times 879 - 7^2 + 7 \times 993 - 7 \times 872\)
\(\ds \) \(=\) \(\ds 879 \times 993 + 7 \times 879 - 7^2 - 7 \times 872\)
\(\ds \) \(=\) \(\ds 879 \times 993 + 7 \times 879 - 7 \times \paren {872 + 7}\)
\(\ds \) \(=\) \(\ds 879 \times 993 + 7 \times 879 - 7 \times 879\)
\(\ds \) \(=\) \(\ds 879 \times 993\)

$\blacksquare$


Sources

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