Difference between revisions of "2013 USAJMO"
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==Day 2== | ==Day 2== | ||
===Problem 4=== | ===Problem 4=== | ||
− | Let be the number of ways to write as a sum of powers of , where we keep track of the order of the summation. For example, because can be written as , , , , , and . Find the smallest greater than for which is odd. | + | Let <math>f(n)</math> be the number of ways to write <math>n</math> as a sum of powers of <math>2</math>, where we keep track of the order of the summation. For example, <math>f(4)=6</math> because <math>4</math> can be written as <math>4</math>, <math>2+2</math>, <math>2+1+1</math>, <math>1+2+1</math>, <math>1+1+2</math>, and <math>1+1+1+1</math>. Find the smallest <math>n</math> greater than <math>2013</math> for which <math>f(n)</math> is odd. |
[[2013 USAJMO Problems/Problem 4|Solution]] | [[2013 USAJMO Problems/Problem 4|Solution]] |
Revision as of 19:19, 11 May 2013
Contents
Day 1
Problem 1
Are there integers and
such that
and
are both perfect cubes of integers?
Problem 2
Each cell of an board is filled with some nonnegative integer. Two numbers in the filling are said to be adjacent if their cells share a common side. (Note that two numbers in cells that share only a corner are not adjacent). The filling is called a garden if it satisfies the following two conditions:
(i) The difference between any two adjacent numbers is either or
.
(ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to
.
Determine the number of distinct gardens in terms of and
.
Problem 3
In triangle , points
lie on sides
respectively. Let
,
,
denote the circumcircles of triangles
,
,
, respectively. Given the fact that segment
intersects
,
,
again at
respectively, prove that
.
Day 2
Problem 4
Let be the number of ways to write
as a sum of powers of
, where we keep track of the order of the summation. For example,
because
can be written as
,
,
,
,
, and
. Find the smallest
greater than
for which
is odd.
Problem 5
Quadrilateral is inscribed in the semicircle with diameter . Segments and meet at . Point is the foot of the perpendicular from to line . Point lies on such that line is perpendicular to line . Let be the intersection of segments and . Prove that
Problem 6
Find all real numbers satisfying
See Also
2013 USAJMO (Problems • Resources) | ||
Preceded by 2012 USAJMO |
Followed by 2014 USAJMO | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |