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 1== | + | ==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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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 , 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
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.
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.