1968 IMO Problems/Problem 2

Problem

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

Video Solution (in Chinese)

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 $(x - 5)^2 - 47 = y$ ".

(When writing $a^2$ the author means the square of an arbitrary natural number, not the square of the number $a$ used in the line above this statement.) Neither the modulo statements, nor the conclusion are obvious; proofs should be given. The second unproven statement 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)

Let the decimal expansion of $x$ be $\overline{d_1d_2d_3\dots d_n}$ , where $d_i$ are base-10 digits. Let us prove first that $n \le 2$ .

Using $x^2 - 10x -22 = (x - 5)^2 - 47$ and the fact that this expression equals the product $d_1d_2d_3 \dots d_n$ we have that $(x - 5)^2 - 47 \le 9^n$ . Since $x$ has $n$ digits, we also have $(x - 5)^2 - 47 \ge (10^{n - 1} - 5)^2 - 47$ . Thus, we have $9^n \ge (10^{n - 1} - 5)^2 - 47$ . We will show that this can not be true for $n \ge 3$ .

Divide by $9^{n - 1}$ , rearrange factors and terms, and get $\left[\left(\frac{10}{3}\right)^{n - 1} - \frac{5}{3^{n - 1}}\right]^2 \le 9 + \frac{47}{9^{n - 1}} < 10$ . Again using $n \ge 3$ , we get $\left(\frac{10}{3}\right)^{n - 1} < \sqrt{10} + \frac{5}{3^{n - 1}} < 5$ . This is false for $n \ge 3$ , so $n =$ the number of digits of $x$ can not be more than $2$ .

Let then $x = 10a + b$ , with $0 \le a, b \le 9$ the digits of $x$ . We have $x^2 - 10x - 22 = ab \le 81$ . By solving the quadratic equation $x^2 - 10x - 103 = 0$ we see that the inequality is true when $0 \le x \le 16$ .

At this point, we could simply check which of the numbers $x = 0, 1, \dots, 16$ has the product of its digits equal to $x^2 - 10x -22$ . It would turn out that $x = 12$ is the only one. But we can take a short cut by writing $(10a + b)^2 - 10(10a + b) - 22 = ab$ . and using that $a = 0$ or $a = 1$ . In the case $a = 0$ we get the equation $b^2 - 10b - 22 = 0$ in $b$ which has no integer solutions. In the case $a = 1$ we get the equation $b^2 + 9b - 22 = 0$ in $b$ , which has solutions $-11$ and $2$ . Since $b$ is a digit, $b = 2$ is the only acceptable solution, so $x = 12$ .

Solution 3 (corrected and complete)

Let $y = x^2 - 10x - 22$ . Since $y$ is a product of digits, the only prime factors of $y$ can be $2, 3, 5, 7$ . Write $y = (x - 5)^2 - 47$ .

Note that if $A$ is a natural number then $A^2 \equiv 0$ or $1\ (mod\ 3)$ . Indeed, $A = 3k$ or $A = 3k + 1$ or $A = 3k + 2$ for some $k$ . By writing out $A^2$ the statement follows immediately. Then $(x - 5)^2 - 47 \equiv 1$ or $2\ (mod\ 3)$ , so $3$ can not be a factor of $y$ .

Similarly, if $A$ is a natural number then $A^2 \equiv 0, 1$ or $4\ (mod\ 5)$ . Then $(x - 5)^2 - 47 \equiv 3, 4$ or $2\ (mod\ 5)$ , so $5$ can not be a factor of $y$ .

And finally if $A$ is a natural number then $A^2 \equiv 0, 1, 2$ or $4\ (mod\ 7)$ . Then $(x - 5)^2 - 47 \equiv 2, 3, 4$ or $6\ (mod\ 7)$ , so $7$ can not be a factor of $y$ .

The only possible prime factor of $y$ is $2$ . So $2^z = (x - 5)^2 - 47$ for some $z \ge 0$ . Write this as $2^z -2 = (x - 5)^2 - 49$ , or $2(2^{z - 1} - 1) = (x - 12)(x + 2)$ .

Clearly $z = 0$ is not acceptable because the equation in $x$ has no integer solutions.

If $z > 1$ the left hand side is divisible by $2$ , but not by $4$ . The right hand side is not divisible by $2$ if $x$ is odd, and is divisible by $4$ if $x$ is even. So $z > 1$ yields no solutions.

If $z = 1$ , the equation becomes $x^2 - 10x - 24 = 0$ which has solutions $-2, 12$ . The only acceptable solution is $x = 12$ .

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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