Difference between revisions of "1988 USAMO Problems"
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+ | Problems from the '''1988 [[USAMO]].''' | ||
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==Problem 1== | ==Problem 1== | ||
− | 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 | + | 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 divisible 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.) |
[[1988 USAMO Problems/Problem 1|Solution]] | [[1988 USAMO Problems/Problem 1|Solution]] | ||
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== See Also == | == See Also == | ||
{{USAMO box|year=1988|before=[[1987 USAMO]]|after=[[1989 USAMO]]}} | {{USAMO box|year=1988|before=[[1987 USAMO]]|after=[[1989 USAMO]]}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:33, 5 February 2018
Problems from the 1988 USAMO.
Problem 1
The repeating decimal , where
and
are relatively prime integers, and there is at least one decimal before the repeating part. Show that
is divisible by 2 or 5 (or both). (For example,
, and 88 is divisible by 2.)
Problem 2
The cubic polynomial has real coefficients and three real roots
. Show that
and that
.
Problem 3
Let be the set
and let
be the set of all 9-element subsets of
. Show that for any map
we can find a 10-element subset
of
, such that
for any
in
.
Problem 4
is a triangle with incenter
. Show that the circumcenters of
,
, and
lie on a circle whose center is the circumcenter of
.
Problem 5
Let be the polynomial
, where
are integers. When expanded in powers of
, the coefficient of
is
and the coefficients of
,
, ...,
are all zero. Find
.
See Also
1988 USAMO (Problems • Resources) | ||
Preceded by 1987 USAMO |
Followed by 1989 USAMO | |
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.