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.) | ||
| − | ==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>. | 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== | ||
| + | But <math>\frac{1}{3} + \frac{1}{6} = \frac{1}{2}</math>! | ||
| + | |||
| + | ==Solution 2== | ||
| + | It is well-known that <math>0.ab...k \overline{pq...t} = \frac{ab...t - ab...k}{99...900...0}</math>, where there are a number of 9s equal to the count of digits in <math>pq...t</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...t</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== | ||
Revision as of 16:45, 5 October 2014
Problem
The repeating decimal 
, where 
and 
are relatively prime integers, and there is at least one decimal before the repeating part. Show that is divisble by 2 or 5 (or both). (For example, 
, 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 
and the repeating parts of the decimal is equal to 
.
Suppose that the length of 
is 
digits. Then 
Since 
, after reducing the fraction, there MUST be either a factor of 2 or 5 remaining in the denominator. After adding the fractions 
, the simplified denominator will be 
and since 
has a factor of 
or 
, must also have a factor of 2 or 5.
Rebuttal to Solution 1
But 
!
Solution 2
It is well-known that 
, where there are a number of 9s equal to the count of digits in 
, and there are a number of 0s equal to the count of digits 
in 
. Obviously is different from (which is itself the repeating part), so the numerator cannot have 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
