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

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The repeating decimal <math>0.ab\cdots k\overline{pq\cdots u}=\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime integers, and there is at least one decimal before the repeating part. Show that <math>n</math> is divisble by 2 or 5 (or both). (For example, <math>0.011\overline{36}=0.01136363636\cdots=\frac 1{88}</math>, and 88 is divisible by 2.)
 
The repeating decimal <math>0.ab\cdots k\overline{pq\cdots u}=\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime integers, and there is at least one decimal before the repeating part. Show that <math>n</math> is divisble by 2 or 5 (or both). (For example, <math>0.011\overline{36}=0.01136363636\cdots=\frac 1{88}</math>, and 88 is divisible by 2.)
  
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==Solution 1==
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==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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<math>\blacksquare</math>
 
<math>\blacksquare</math>
  
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==Rebuttal to Solution 1==
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===Solution 2===
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But <math>\frac{1}{3} + \frac{1}{6} = \frac{1}{2}</math>!
 
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==Solution 2==
 
 
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.
 
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.
  

Latest revision as of 17:58, 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 (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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