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

by : $174$

 * More Curious Multiplication

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:

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 $b$.

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.

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

$n$ 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 $d \ge 3$.

Example

 * Solution 1: $987 \, 654 \, 321$, by noting that:
 * $987 \, 654 \, 321 > \dfrac 8 {81} \times 10^{11}$