Difference between revisions of "1988 USAMO Problems/Problem 1"

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==Solution==
 
==Solution==
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===Solution 1===
  
 
First, split up the nonrepeating parts and the repeating parts of the decimal, so that the nonrepeating parts equal to <math>\frac{a}{b}</math> and the repeating parts of the decimal is equal to <math>\frac{c}{d}</math>.
 
First, split up the nonrepeating parts and the repeating parts of the decimal, so that the nonrepeating parts equal to <math>\frac{a}{b}</math> and the repeating parts of the decimal is equal to <math>\frac{c}{d}</math>.
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Suppose that the length of  <math>0.ab\cdots k</math> is <math>p</math> digits. Then <math>\frac{a}{b} = \frac{0.ab\cdots k}{10^{p+1}}</math> Since <math>0.ab\cdots k < 10^{p+1}</math>, after reducing the fraction, there MUST be either a factor of 2 or 5 remaining in the denominator. After adding the fractions <math>\frac{a}{b}+\frac{c}{d}</math>, the simplified denominator <math>n</math> will be <math>\\lcm(b,d)</math> and since <math>b</math> has a factor of <math>2</math> or <math>5</math>,  <math>n</math> must also have a factor of 2 or 5.
 
Suppose that the length of  <math>0.ab\cdots k</math> is <math>p</math> digits. Then <math>\frac{a}{b} = \frac{0.ab\cdots k}{10^{p+1}}</math> Since <math>0.ab\cdots k < 10^{p+1}</math>, after reducing the fraction, there MUST be either a factor of 2 or 5 remaining in the denominator. After adding the fractions <math>\frac{a}{b}+\frac{c}{d}</math>, the simplified denominator <math>n</math> will be <math>\\lcm(b,d)</math> and since <math>b</math> has a factor of <math>2</math> or <math>5</math>,  <math>n</math> must also have a factor of 2 or 5.
  
Q.E.D.
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<math>\blacksquare</math>
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===Solution 2===
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It is well-known that <math>0.ab...k \overline{pq...u} = \frac{ab...u - ab...k}{99...900...0}</math>, where there are a number of 9s equal to the count of digits in <math>pq...u</math>, and there are a number of 0s equal to the count of digits <math>c</math> in <math>ab...k</math>. Obviously <math>ab...k</math> is different from <math>pq...u</math> (which is itself the repeating part), so the numerator cannot have <math>c</math> consecutive terminating zeros, and hence the denominator still possesses a factor of 2 or 5 not canceled out. This completes the proof.
  
 
==See Also==
 
==See Also==
 
{{USAMO box|year=1988|before=First Question|num-a=2}}
 
{{USAMO box|year=1988|before=First Question|num-a=2}}
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{{MAA Notice}}
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

Latest revision as of 18:58, 18 July 2016

Problem

The repeating decimal $0.ab\cdots k\overline{pq\cdots u}=\frac mn$, where $m$ and $n$ are relatively prime integers, and there is at least one decimal before the repeating part. Show that $n$ is divisble by 2 or 5 (or both). (For example, $0.011\overline{36}=0.01136363636\cdots=\frac 1{88}$, and 88 is divisible by 2.)

Solution

Solution 1

First, split up the nonrepeating parts and the repeating parts of the decimal, so that the nonrepeating parts equal to $\frac{a}{b}$ and the repeating parts of the decimal is equal to $\frac{c}{d}$.

Suppose that the length of $0.ab\cdots k$ is $p$ digits. Then $\frac{a}{b} = \frac{0.ab\cdots k}{10^{p+1}}$ Since $0.ab\cdots k < 10^{p+1}$, after reducing the fraction, there MUST be either a factor of 2 or 5 remaining in the denominator. After adding the fractions $\frac{a}{b}+\frac{c}{d}$, the simplified denominator $n$ will be $\\lcm(b,d)$ and since $b$ has a factor of $2$ or $5$, $n$ must also have a factor of 2 or 5.

$\blacksquare$

Solution 2

It is well-known that $0.ab...k \overline{pq...u} = \frac{ab...u - ab...k}{99...900...0}$, where there are a number of 9s equal to the count of digits in $pq...u$, and there are a number of 0s equal to the count of digits $c$ in $ab...k$. Obviously $ab...k$ is different from $pq...u$ (which is itself the repeating part), so the numerator cannot have $c$ consecutive terminating zeros, and hence the denominator still possesses a factor of 2 or 5 not canceled out. This completes the proof.

See Also

1988 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

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