Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication/Solution 1

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Puzzles and Curious Problems by Henry Ernest Dudeney: $174$

More Curious Multiplication
What number is it that, when multiplied by $18$, $27$, $36$, $45$, $54$, $63$, $72$, $81$ or $99$,
gives a product in which the first and last figures are the same as those in the multiplier,
but which when multiplied by $90$ gives a product in which the last two figures are the same as those in the multiplier?


Solution

$987 \, 654 \, 321$


Proof

We have:

\(\ds 987 \, 654 \, 321 \times 18\) \(=\) \(\ds 17 \, 777 \, 777 \, 778\) beginning with $1$ and ending in $8$
\(\ds 987 \, 654 \, 321 \times 27\) \(=\) \(\ds 26 \, 666 \, 666 \, 667\) beginning with $2$ and ending in $7$
\(\ds 987 \, 654 \, 321 \times 36\) \(=\) \(\ds 35 \, 555 \, 555 \, 556\) beginning with $3$ and ending in $6$
\(\ds 987 \, 654 \, 321 \times 45\) \(=\) \(\ds 44 \, 444 \, 444 \, 445\) beginning with $4$ and ending in $5$
\(\ds 987 \, 654 \, 321 \times 54\) \(=\) \(\ds 53 \, 333 \, 333 \, 334\) beginning with $5$ and ending in $4$
\(\ds 987 \, 654 \, 321 \times 63\) \(=\) \(\ds 62 \, 222 \, 222 \, 223\) beginning with $6$ and ending in $3$
\(\ds 987 \, 654 \, 321 \times 72\) \(=\) \(\ds 71 \, 111 \, 111 \, 112\) beginning with $7$ and ending in $2$
\(\ds 987 \, 654 \, 321 \times 81\) \(=\) \(\ds 80 \, 000 \, 000 \, 001\) beginning with $8$ and ending in $1$
\(\ds 987 \, 654 \, 321 \times 90\) \(=\) \(\ds 88 \, 888 \, 888 \, 890\) ending in $90$

$\blacksquare$


Sources

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