Difference between revisions of "1988 USAMO Problems/Problem 1"
(Created page with "==Problem== 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 ...") |
(Removed extraneous rebuttal) |
||
(6 intermediate revisions by 3 users not shown) | |||
Line 3: | Line 3: | ||
==Solution== | ==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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | <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== | ==See Also== | ||
{{USAMO box|year=1988|before=First Question|num-a=2}} | {{USAMO box|year=1988|before=First Question|num-a=2}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] |
Latest revision as of 18:58, 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.