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(SFFT)

It is pretty obvious that $x$ cannot be three digits or more, because then $x^2 - 10x - 22$ is way too big.

Write $x = 10a + b$ where $a$ and $b$ are digits satisfying $0 \leq a, b < 10$ . Then, we can use SFFT: $(10a + b)^2 - 10(10a + b) - 22 = ab$ $(10a + b)^2 - 10(10a + b) - 24 = ab - 2$ $(10a + b + 2)(10a + b - 12) = ab - 2.$ We have $(10a + b + 2)(10a + b - 12) \geq (10a + 2)(10a - 12) = 100a^2 - 100a + 24 = 100(a^2 - a) + 24.$ It is therefore clear that $a$ must be either $0$ or $1$ . We can then split into two cases:

$\mathbf{a = 0:}$

We have $(b + 2)(b - 12) = -2$ or $b^2 - 10b - 22 = 0$ , which is only satisfied when $b = -2$ or $12$ .

$\mathbf{a = 1:}$

We have $(b + 12)(b - 2) = b - 2$ . This is only satisfied when $b = 2$ , or $b + 12 = 0$ . Therefore, $b = 2$ , and so $x = \boxed{12}.\square$

~mathboy100

Solution 3

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

$\implies x=12$

$\blacksquare$

Remarks (added by pf02, August 2024)

Solutions 2 and 3 are not satisfactory. In fact, they can not be called solutions, since they make statements which are not proven. Specifically:

In Solution 2, the author writes "It is pretty obvious that $x$ cannot be three digits or more, because then $x^2 - 10x - 22$ is way too big." This is intuitively true, but not obvious at all. As a crucial step in the solution, it should be proven. Later, the author states

" $(10a + b + 2)(10a + b - 12) \ge \cdots = 100(a^2 - a) + 24$ . It is therefore clear that $a$ must be either $0$ or $1$ ".

First, the last term should be $-24$ instead of $24$ . Either way, the conclusion about $a$ is not clear at all. As a second crucial step in the solution, it should be proven.

In Solution 3, the notation and writing are very confusing. However, a diligent reader can make sense of them. But in this solution as well, there are statements which beg for a proof. The first such statement is

" $a^2 \not\equiv 2 (mod\ 3), a^2 \not\equiv 2 (mod\ 5), a^2 \not\equiv 5 (mod\ 7)$ which means $3, 5, 7$ don't divide $(x - 5)^2 - 47 = y$ ".

Neither the modulo statements, nor the conclusion are obvious; proofs should be given. The second one is

" $2^a + 47 = (x - 5)^2$ . It is easy to see that $a$ has one solution and that is $2$ . (Prove it by contradiction.)"

(The author means "the equation has a unique solution for $a$ ".) The conclusion about the uniqueness of $a$ is not easy to see, and as a crucial step in the solution, it should be proven.

Below, I will give corrected, complete, and somewhat simplified versions of these two solutions.

Solution 2 (corrected and complete)

TO BE CONTINUED. I AM SAVING UNFINISHED TEXT SO I DON'T LOSE WORK DONE SO FAR.

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

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