1988 USAMO Problems/Problem 1

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 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.

$\blacksquare$

Rebuttal to Solution 1

But $\frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ !

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.

1988 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5
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1988 USAMO Problems/Problem 1

Contents

Problem

Solution 1

Rebuttal to Solution 1

Solution 2

See Also