1968 IMO Problems/Problem 2

Problem

Find all natural numbers $x$ such that the product of their digits (in decimal notation) is equal to $x^2 - 10x - 22$ .

Solution 1

Let the decimal expansion of $x$ be $\overline{d_1d_2d_3\dots d_n}$ , where $d_i$ are base-10 digits. We then have that $x\geq d_1\cdot 10^{n-1}$ . However, the product of the digits of $x$ is $d_1d_2d_3\dots d_n\leq d_1\cdot 10\cdot 10\dots 10=d_1\cdot 10^{n-1}$ , with equality only when $x$ is a one-digit integer. Therefore the product of the digits of $x$ is always at most $x$ , with equality only when $x$ is a base-10 digit. This implies that $x^2-10x-22\leq x$ , so $x^2-11x-22\leq 0$ . Every natural number from 1 to 12 satisfies this inequality, so we only need to check these possibilities. It is easy to rule out 1 through 11, since $x^2-10x-22<0$ for those values. However, $12^2-10\cdot 12-22=2$ , which is the product of the digits of 12. Therefore $\boxed{12}$ is the only natural number with the desired properties. $\blacksquare$

Solution 2

Let, $x^2-10x-22=y$

$\implies x^2-10+25-47=y$

$\implies (x-5)^2=47+y$

Now note that, if $p$ is a prime such that $p|y$ then $7\geq p$ .

That means, $y=2^a*3^b*5^c*7^d$

But, $a^2 \not\equiv 2 (mod3), a^2 \not\equiv 2 (mod5), a^2 \not\equiv 5 (mod7)$ which means $3,5,7$ don't divivde $(x-5)^2-47=y.$

So, $y=2^a$ and $y+17=2^a+47=(x-5)^2$

It is easy to see that $a$ has one solution and that is $2.$ ( Prove it by contradiction)

So, $(x-5)^2=47+2=49$

$\implies x=12$

$\blacksquare$

1968 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
All IMO Problems and Solutions

Page

Toolbox

Search

1968 IMO Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

See Also