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Difference between revisions of "1968 IMO Problems/Problem 2"
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Find all natural numbers <math>x</math> such that the product of their digits (in decimal notation) is equal to <math>x^2 - 10x - 22</math>.
 
Find all natural numbers <math>x</math> such that the product of their digits (in decimal notation) is equal to <math>x^2 - 10x - 22</math>.
  
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==Solution==
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==Solution 1==
  
 
Let the decimal expansion of <math>x</math> be <math>\overline{d_1d_2d_3\dots d_n}</math>, where <math>d_i</math> are base-10 digits. We then have that <math>x\geq d_1\cdot 10^{n-1}</math>. However, the product of the digits of <math>x</math> is <math>d_1d_2d_3\dots d_n\leq d_1\cdot 10\cdot 10\dots 10=d_1\cdot 10^{n-1}</math>, with equality only when <math>x</math> is a one-digit integer. Therefore the product of the digits of <math>x</math> is always at most <math>x</math>, with equality only when <math>x</math> is a base-10 digit. This implies that <math>x^2-10x-22\leq x</math>, so <math>x^2-11x-22\leq 0</math>. 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 <math>x^2-10x-22<0</math> for those values. However, <math>12^2-10\cdot 12-22=2</math>, which is the product of the digits of 12. Therefore <math>\boxed{12}</math> is the only natural number with the desired properties. <math>\blacksquare</math>
 
Let the decimal expansion of <math>x</math> be <math>\overline{d_1d_2d_3\dots d_n}</math>, where <math>d_i</math> are base-10 digits. We then have that <math>x\geq d_1\cdot 10^{n-1}</math>. However, the product of the digits of <math>x</math> is <math>d_1d_2d_3\dots d_n\leq d_1\cdot 10\cdot 10\dots 10=d_1\cdot 10^{n-1}</math>, with equality only when <math>x</math> is a one-digit integer. Therefore the product of the digits of <math>x</math> is always at most <math>x</math>, with equality only when <math>x</math> is a base-10 digit. This implies that <math>x^2-10x-22\leq x</math>, so <math>x^2-11x-22\leq 0</math>. 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 <math>x^2-10x-22<0</math> for those values. However, <math>12^2-10\cdot 12-22=2</math>, which is the product of the digits of 12. Therefore <math>\boxed{12}</math> is the only natural number with the desired properties. <math>\blacksquare</math>
  
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==Solution 2==
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Let,
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<math>x^2-10x-22=y</math>
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<math>\implies x^2-10+25-47=y</math>
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<math>\implies (x-5)^2=47+y</math>
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Now note that, if <math>p</math> is a prime such that <math>p|y</math> then <math>7\geq p</math>.
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That means, <math>y=2^a*3^b*5^c*7^d</math>
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But, <math>a^2 \not\equiv 2 (mod3), a^2 \not\equiv 2 (mod5), a^2 \not\equiv 5 (mod7)</math> which means <math>3,5,7</math> don't divivde <math>(x-5)^2-47=y.</math>
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So, <math>y=2^a</math> and <math>y+17=2^a+47=(x-5)^2</math>
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It is easy to see that <math>a</math> has one solution  and that is <math>2.</math>( Prove it by contradiction)
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So, <math>(x-5)^2=47+2=49</math>
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<math>\implies x=12</math>
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<cmath> A.W.D</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 03:30, 26 April 2021

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

$\implies x=12$

\[A.W.D\]

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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