Difference between revisions of "1968 IMO Problems/Problem 2"

Revision as of 13:05, 20 March 2012

Problem

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

Solution

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. $\blacksquare$

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 ==See Also==
 {{IMO box|year=1968|num-b=1|num-a=3}}
 * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=361673&sid=6798c42a2ab57f3ca82ffba974ed589c#p361673 Discussion on AoPS/MathLinks]
 [[Category:Olympiad Number Theory Problems]]

1968 IMO (Problems) • Resources
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Difference between revisions of "1968 IMO Problems/Problem 2"

Revision as of 13:05, 20 March 2012

Problem

Solution

See Also