Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication/Solution 3

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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

All $d$-digit numbers $n$ in the form of:

$n = \sqbrk {9 a_1 a_2 \dots a_{d - 2} 1}$

where $n \ge \dfrac 8 {81} \times 10^{d + 1}$ and $d \ge 3$.

The smallest solution of this form is $991$.


Proof

Let $\sqbrk {a b}$ be a $2$-digit integer that is not $91$.

We will show that $n$ satisfies more restrictions than those stated in the problem.


For the case $b \ne 0$, we will show that $n \sqbrk {a b}$ starts with $a$ and ends with $b$.


First we show that $\dfrac {\sqbrk {a b} } {\sqbrk {a 0} } \ge \dfrac {81}{80}$.

This is equivalent to showing $\dfrac b {10 a} \ge \dfrac 1 {80}$,

which in turn is equivalent to showing $\dfrac b a \ge \dfrac 1 8$.

For $a \le 8$:

$\dfrac b a \ge \dfrac 1 a \ge \dfrac 1 8$

For $a = 9$, we have $b \ge 2$.

Thus:

$\dfrac b a \ge \dfrac 2 9 > \dfrac 1 8$

and hence our inequality is proved.


Now we have:

\(\ds \paren {a + 1} \times 10^{d + 1}\) \(\ge\) \(\ds \sqbrk {a b} \times 10^d\)
\(\ds \) \(\ge\) \(\ds \sqbrk {a b} \times n\) $n$ has $d$ digits
\(\ds \) \(\ge\) \(\ds \sqbrk {a b} \times \frac 8 {81} \times 10^{d + 1}\)
\(\ds \) \(=\) \(\ds a \times 10 \times \frac {\sqbrk {a b} } {\sqbrk {a 0} } \times \frac 8 {81} \times 10^{d + 1}\)
\(\ds \) \(\ge\) \(\ds a \times 10 \times \frac {81} {80} \times \frac 8 {81} \times 10^{d + 1}\) from above
\(\ds \) \(=\) \(\ds a \times 10^{d + 1}\)

therefore $n \sqbrk {a b}$ starts with $a$.


We also have $n \equiv 1 \pmod {10}$ and $\sqbrk {a b} \equiv b \pmod {10}$.

Thus $n \sqbrk {a b}$ ends with $b$.

$\Box$


For $b = 0$, we will show that $n \sqbrk {a 0}$ ends with $\sqbrk {a 0}$.

Both $\sqbrk {a 0}$ and $n \sqbrk {a 0}$ are multiples of $10$, so they end with $0$.

The second-last digit of $n \sqbrk {a 0}$ is the last digit of $n \sqbrk {a 0} \div 10 = n a$.

Since $n \equiv 1 \pmod {10}$, $n a \equiv a \pmod {10}$.

This implies that $n a$ ends with $a$, thus proving our claim.

$\Box$


Finally we show that $n$ starts with a $9$ and $d \ge 3$.

Aiming for a contradiction, suppose $n$ does not start with a $9$.

Then $n < 9 \times 10^{d - 1}$.

Thus:

$9 \times 10^{d - 1} > \dfrac 8 {81} \times 10^{d + 1}$
$9 > \dfrac {800} {81} \approx 9.87654 \ldots$

which is a contradiction.

Therefore $n$ must start with a $9$ and contain at least $2$ digits.

If $d = 2$, $n = 91$.

However $91 < \dfrac 8 {81} \times 10^3 \approx 98.7654 \ldots$.

Thus $d \ge 3$.

$\blacksquare$


Example

$987 \, 654 \, 321 > \dfrac 8 {81} \times 10^{11}$
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