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1968 IMO Problems/Problem 2

Revision as of 22:51, 6 December 2022 by Mathboy100 (talk | contribs)

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

$\implies x=12$

$\blacksquare$

See Also

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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