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Difference between revisions of "1988 USAMO Problems/Problem 1"
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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.
+
<math>\blacksquare</math>
  
 
==See Also==
 
==See Also==

Revision as of 17:27, 5 April 2013

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

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.

$\blacksquare$

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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