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

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.

@@ Line 3: / Line 3: @@
 ==Solution==
+===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>.
@@ Line 8: / Line 9: @@
 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>
+===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.
 ==See Also==
 {{USAMO box|year=1988|before=First Question|num-a=2}}
+{{MAA Notice}}
 [[Category:Olympiad Number Theory Problems]]

1988 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
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