Lcz's Oly Notes
Contents
Introduction
So I failed the AOIME. I don't really want to do any more AIME prep, so I have decided to go do some Oly prep :). Here's some oly notes/favorite problems. (There aren't that many examples, sorry.)
Edit (7/2/2020): This thing is getting big, woah
ALGEBRA
Algebra is cool. I'm pretty good at it (by my standards shup smh).
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Before oly...
B. Problem 1:
Find if .
B. Problem 2:
Find the sum of all such that .
B. Problem 3 (troll):
Find the sum of all in such that
B. Problem 4 (2018 I/6)
Let be the number of complex numbers with the properties that and is a real number. Find the remainder when is divided by .
B. Problem 5 (AOIME/8 sigh...)
Define a sequence recursively by and for integers . Find the least value of such that the sum of the zeros of exceeds .
B. Problem 6 (AOIME/11 bashish?)
Let , and let and be two quadratic polynomials also with the coefficient of equal to . David computes each of the three sums , , and and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If , then , where and are relatively prime positive integers. Find .
B. Problem 7 (1984/15):
Determine if
B. Problem 8 (2020 I/14):
Let be a quadratic polynomial with complex coefficients whose coefficient is Suppose the equation has four distinct solutions, Find the sum of all possible values of
B. Problem 9 (HINT: THIS IS GEOMETRY 2006 II/15):
Given that and are real numbers that satisfy:
and that where and are positive integers and is not divisible by the square of any prime, find
B. Problem 10 (2011 I/15):
For some integer , the polynomial has the three integer roots , , and . Find .
B. Problem 11 (2014 I/13):
The polynomial has complex roots of the form with and Given that where and are relatively prime positive integers, find
B. Problem 12 (2007 I/14):
A sequence is defined over nonnegative integral indexes in the following way: , .
Find the greatest integer that does not exceed
B. Problem 13 (2007 II/14):
Let be a polynomial with real coefficients such that and for all , Find
B. Problem 14 (Gotta save the easy ones for the end, 2010 I/14):
For each positive integer n, let . Find the largest value of for which .
B. Problem 15 (2021 usamo/6):
Dad and mom are playing outside while you are in a conference call discussing serious matters.
What is ?
B. Problem 16 (For difficulty of 16, 2016 II/15):
For let and . Let be positive real numbers such that and . The maximum possible value of , where and are relatively prime positive integers. Find .
B. Problem 17 (2014 usamo/1):
Let be real numbers such that and all zeros and of the polynomial are real. Find the smallest value the product can take.
B. Problem 18 (1985 usamo/1):
Determine whether or not there are any positive integral solutions of the simultaneous equations \begin{align*}x_1^2+x_2^2+\cdots+x_{1985}^2&=y^3,\\ x_1^3+x_2^3+\cdots+x_{1985}^3&=z^2\end{align*}with distinct integers , , , .
B. Problem 19 (1994 usamo/4):
Let be a sequence of positive real numbers satisfying for all . Prove that, for all
B. Problem 20 (THIS COUNTS AS ALGEBRA 1996 usamo/1):
Prove that the average of the numbers is .
Inequalities
Yay! I love inequalities. Clever algebraic manipulation+thereoms is all you need. It all comes from experience though...
Basics
Note that the set is .
. It is "symmetric".
. It is "cyclic".
AMGM says that for nonnegative reals , their Arithmetic Mean is their Geometric Mean. .
Cauchy's Inequality says that for nonnegative reals , Equality holds when is constant for all .
Titu's lemma follows directly from Cauchy, it says for the same variables,
We say a set majorizes if (a) the sum of the elements are the same. (b) the elements are in decreasing order. (c) the leftmost sums of are those of .
For example, majorizes and majorizes .
Muirhead's Inequality says that for reals and nonnegative reals , where majorizes ,
Holder's Inequality (for , you'll probably only need it for , plus this is the "basics" chapter anyway...) says that for positive reals ,
These are cool, but remember that these should only be used when the inequality is homogenized already.. These are all pretty easy to prove as well (with the help of the trivial inequality, but because it is trivial, we didn't include it here.)
Weighted Power Mean says that for positive reals and ("weights") that sum to , if , if , If , with equality if and only if all the are equal.
Example 1:
Prove that
Solution:
Expand everything. Canceling terms, you should get This is true by muirhead's on and or by AMGM
Example 2 (2001 imo/2):
Let be positive real numbers. Prove that .
Solution:
By Holder's, . Now we need to prove but this is the same as the previous question.
Example 3 (Creds. to Evan Chen):
If , prove that .
Solution:
We have never encountered something where the degrees aren't the same, or there is a condition on the variables. Put the two together! If we multiply the right hand side by , it will make both sides equal degree. Now we want to prove that . This is muirhead's on and , but what about the condition? It's gone. In fact, could be anything at this point. If we multiply and by , the inequality becomes , and we can just divide both sides by ! This is essentially the idea of [homogenizing]. Since there are now no conditions, it is a normal inequality and we are done by muirhead's.
Problem 1:
Prove that .
Problem 2 (Evan Chen?):
Find the minimum possible value of if .
Example 4 (1981 imo/1):
is a point inside a given triangle . are the feet of the perpendiculars from to the lines , respectively. Find all for which
is least.
Solution:
Note that is and is thus constant. By Cauchy,
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with equality when , or , or is the incenter.
Problem 3: (Cauchy, 2009 usamo/4):
For let , , ..., be positive real numbers such that Prove that .
Example 5 (2004 usamo/5)
Let , , and be positive real numbers. Prove that .
Solution:
1. The is cubed, so we try to use Holder's. The simplest way to do this is just to use on the LHS.
2. Now all we have to prove is that , or . Now note that if , this is true, if , this is true, and if , this is true as well, and as we have exhausted all cases, we are done.
More advanced stuff, learn some calculus
You will need to know derivatives for this part. It's actually quite simple. Derivative=Tangent.
The derivative of is
Adding and other stuff works in the same way. You also probably need to know
Product Rule:
Quotient Rule:
Summing:
The second derivative of a function is just applying the derivative twice. A function is convex on an interval if it's second derivative is always positive in that interval. A function is convex if it's second derivative is negative in that interval.
Jensen's inequality says that if is a convex function in the interval , for all in ,
The opposite holds if is concave.
Karamata's inequality says that if is convex in the interval , the sequence majorizes ,, and all are in ,
TLT (Tangent Line Trick) is basically where you either a. take the derivative, and plug in the equality cases or b. plugging in both equality cases to form a line.
Problem 6:
Show that
Problem 7: Using Jensen's, solve 2001 IMO/2:
Let be positive real numbers. Prove .
Problem 8: (2017 usamo/6)
Find the minimum possible value of given that are nonnegative real numbers such that .
Problem 9: (Japanese MO 1997/6)
Prove that
for any positive real numbers , , .
Problems
Problem 10 (2011 usamo/1):
Let , , be positive real numbers such that . Prove that
Problem 11 (1974 usamo/2):
Prove that if , , and are positive real numbers, then
Problem 12 (1995 imo/2):
Let be positive real numbers such that . Prove that
Problem 13 (2000 imo/2):
Let be positive real numbers so that . Prove that .
Problem 14 (2003 usamo/5):
Let , , be positive real numbers. Prove that
Problem 15 (2001 usamo/3):
Let and satisfy
Show that
Problem 16 (2012 usajmo/3):
Let , , be positive real numbers. Prove that
Problem 17 (2005 imo/3):
Let be three positive reals such that . Prove that
Problem 18 (2003 imo shortlist/A6. cough.):
Let be a positive integer and let , be two sequences of positive real numbers. Suppose is a sequence of positive real numbers such that for all .
Let . Prove that
Extra
Ravi Substitution for triangles. .
Problem 19 (1983 imo/6, using Ravi!):
Let , and be the lengths of the sides of a triangle. Prove that
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Determine when equality occurs.
Problem 20:
Lcz is good. Lcz is bad, contradiction. What is wrong with Lcz's proof?
Inequality Problems:
https://artofproblemsolving.com/wiki/index.php/Category:Olympiad_Inequality_Problems
https://artofproblemsolving.com/wiki/index.php/Category:Olympiad_Algebra_Problems
Function Equations
Oops...I kind of suck at these :P
~Lcz 6/9/2020 at 12:49 CST
Basics
injective: a1, b3, c2, nothing4 (There are more "things mapped to" in the range) surjective: a1, b1, b2, c (more "things mapped to" in the domain) bijective: a1, b3, c2 (11)
(I have no excersises, but let's just say this is one): A function where its domain and range are same is an involution if for every in its range/domain. Prove that this function is bijective.
Function=anything. (HINT: IT DOESNT HAVE TO BE A POLYNOMIAL)
Symmetry (switch x and y)
Plug in , , , , , , , etc.. first
Check for linear/constant solutions first. They are usually the only ones.
 trick: if , by using the inner two 's, and the outer two.
Pointwise trap!!!!!! If , this does NOT mean that for all . Remember, function=anything and you could have , for some . The most common way to get rid of this is just by doing , , and then show that this leads to a contradiction.
Be aware of your domain/range. ((Probably not)). For example, "Cauchy Functional Equation" Changes a LOT when we go from to .
Ok these things are sort of cool
Example 1 (2002 usamo/4):
Let be the set of real numbers. Determine all functions such that
for all pairs of real numbers and .
Solution:
I claim that the sole solution is for some constant .
First note that setting and as zero respectively yields and . Letting in the second equation means that is odd.
Now, we can substitute into the original equation our findings:
where .
Finally,
. Setting , we get the desired.
Problem 1 (2009 imo/5):
Determine all functions from the set of positive integers to the set of positive integers such that, for all positive integers and , there exists a nondegenerate triangle with sides of lengths
and .
This question...is...cool...?
Example 2 (ISL/2015):
Determine all functions
with the property that
holds for all .
BOGUS Solution:
The answer is or . These clearly work.
Claim: there exists a such that .
Proof: set .
Then (*) where . Now for some constant , or .
The former yields , the latter works and yields .
Why this is wrong: You can't just assume because or . Remember, a function is anything. Maybe this function is surjective.
Solution (after that step):
Now we plug in (because we know that the will turn into , and therefore will be proving that is linear).
(from (*) on where is actually .
(follows directly from (*)).
Therefore, this equation is linear.
Now, letting ,
From (*), we get
If , then
So . Otherwise, is constant and plugging back into (*), we get the only solution as
Ay, ok. Heres a semirecent one.
Problem 2 (2016 usamo/4):
Find all functions such that for all real numbers and ,
Problem 3(One of my favorites, found it in Inter. Alg I think; 1981 imo/6)
The function satisfies
(1)
(2)
(3)
for all nonnegative integers . Determine .
Problem 4(cute, 1983 imo/1):
Find all functions defined on the set of positive reals which take positive real values and satisfy the conditions:
(i) for all ;
(ii) as .
Problem 5 (1986 imo/5, easy but lots of pitfalls oops):
Find all (if any) functions taking the nonnegative reals onto the nonnegative reals, such that
(a) for all nonnegative , ;
(b) ;
(c) for every .
Problems
Problem 6 (1993 usamo/3):
Consider functions which satisfy
(i) for all in , (ii) , (iii) whenever , , and are all in . Find, with proof, the smallest constant such that
for every function satisfying (i)(iii) and every in .
Problem 7 (2015 usajmo/4):
Find all functions such thatfor all rational numbers that form an arithmetic progression. ( is the set of all rational numbers.)
Problem 8 (2015 imo/5):
Let be the set of real numbers. Determine all functions : satisfying the equation
for all real numbers and .
Problem 9 (POINTWISE TRAP!!! 2014 usamo/2):
Let be the set of integers. Find all functions such thatfor all with .
Problem 10 (woah casework, 2012 usamo/4):
Find all functions (where is the set of positive integers) such that for all positive integers and such that divides for all distinct positive integers , .
Problem 11 (2012 imo/4, misplaced i think):
Find all functions such that, for all integers and that satisfy , the following equality holds:
Problem 12 (2011 imo/3):
Let be a realvalued function defined on the set of real numbers that satisfiesfor all real numbers and . Prove that for all .
Problem 13 (This was the first oly FE question I solved. That's why it's last :P 2010 imo/1):
Find all function such that for all the following equality holds
where is greatest integer not greater than
Extra
Nothin much
https://artofproblemsolving.com/wiki/index.php/Category:Functional_Equation_Problems
COMBINATORICS
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This will be a big chapter because I have no idea what the categories of combo are.
Yo I'm trash at this stuff. I'm trash at everything tbh. Let's give it a try, shall we?
Before oly
B. Problem 1
How many ways can people stand in a line?
B. Problem 2
How many ways can people stand in a circle?
B. Problem 3
How many ways can you choose indistinguishable cupcakes from of them?
B. Problem 4
How many ways can you choose distinguishable cupcakes from of them?
B. Problem 5 (2015 I/5):
In a drawer Sandy has pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the socks in the drawer. On Tuesday Sandy selects of the remaining socks at random and on Wednesday two of the remaining socks at random. The probability that Wednesday is the first day Sandy selects matching socks is , where and are relatively prime positive integers, Find
B. Problem 6 (2015 10A/22):
Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
B. Problem 7 (2013 10A/24):
Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?
B. Problem 8 (1988/10):
A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?
B. Problem 9 (1997/10):
Every card in a deck has a picture of one shape  circle, square, or triangle, which is painted in one of the three colors  red, blue, or green. Furthermore, each color is applied in one of three shades  light, medium, or dark. The deck has 27 cards, with every shapecolorshade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:
i. Either each of the three cards has a different shape or all three of the card have the same shape.
ii. Either each of the three cards has a different color or all three of the cards have the same color.
iii. Either each of the three cards has a different shade or all three of the cards have the same shade.
How many different complementary threecard sets are there?
B. Problem 10 (2008 II/10):
In a graph of points each a unit away from it's nearest neighbor, define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the sequence is strictly increasing. Let be the maximum possible number of points in a growing path, and let be the number of growing paths consisting of exactly points. Find .
B. Problem 11 (2020 I/9):
Let be the set of positive integer divisors of Three numbers are chosen independently and at random with replacement from the set and labeled and in the order they are chosen. The probability that both divides and divides is where and are relatively prime positive integers. Find
B. Problem 12 (2011 II/12):
Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be , where and are relatively prime positive integers. Find .
B. Problem 13 (2020 I/7):
A club consisting of men and women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as member or as many as members. Let be the number of such committees that can be formed. Find the sum of the prime numbers that divide
B. Problem 14 (this question sucks, that's why it's here. AOIME/9):
While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next to each other after the break. Find the number of possible seating orders they could have chosen after the break.
B. Problem 15 (2012 II/14):
In a group of nine people each person shakes hands with exactly two of the other people from the group. Let be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when is divided by .
B. Problem 16 (All combo is easy, sorry. 2011 II/14):
There are permutations of such that for , divides for all integers with . Find the remainder when is divided by .
B. Problem 17 (2017 I/7, the only combo question I've ever liked on the AIME):
For nonnegative integers and with , let . Let denote the sum of all , where and are nonnegative integers with . Find the remainder when is divided by .
B. CHALLENGE PROBLEM 18 (2021 usajmo/6):
Find
B. Problem 19 (I guess this is combo??? ISL 2012/C1):
Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers and such that and is to the left of , and replaces the pair by either or . Prove that she can perform only finitely many such iterations.
B. Problem 20 (Is this combo though? usojmo 1):
Let be an integer. Carl has books arranged on a bookshelf. Each book has a height and a width. No two books have the same height, and no two books have the same width. Initially, the books are arranged in increasing order of height from left to right. In a move, Carl picks any two adjacent books where the left book is wider and shorter than the right book, and swaps their locations. Carl does this repeatedly until no further moves are possible. Prove that regardless of how Carl makes his moves, he must stop after a finite number of moves, and when he does stop, the books are sorted in increasing order of width from left to right.
Basics
We're pretty much only going to be working with Expected value and pigeonhole principle here, which corresponds to about 25%(a lot less) of olympiad combo problems.
First of all, we will always denote expected value as . Denote several events and their probabilities as . Then
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Next, there's this cool (I think so) property of the expected value.
Say the average number of cookies someone in this world has ate is . Then there exists a person that has ate at least cookies, and also a person who ate at most cookies. It seems simple, but it's pretty powerful.
Pigeonhole principle says that if you have pigeonholes and pigeons, there must exist a pigeonhole with at least ceiling() pigeons.
This follows directly from our previous claim.
Combo identities: You should probably know all of these from AIME...
Ok. More advanced stuff, a.k.a. "Greedy" Algorithm:
Repeat until you can't anymore
(Well, that's basically it. Just do a thing like "If this thing is , make it into a and a . Repeat until you can't anymore.)
You'll actually need Jensen's Inequality a lot here, watch out!
Counting in two ways. Prove that they are equal. Solve the problem . . . ?!?!?
(That's pretty much all of olympiad combinatorics, for real. A bunch of random strategies! Innovative.)
This stuff is pretty easy
Example 1 (AoPS):
In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is where and are relatively prime positive integers, find
Solution:
The four aces divide the deck into parts, and the expected number of cards in each of them is . Therefore the answer is
Example 2 (2014 AIME I/15):
For all positive integers , let and define a sequence as follows: and for all positive integers . Let be the smallest such that . (For example, and .) Let be the number of positive integers such that . Find the sum of the distinct prime factors of .
Solution (Ripoff of AoPS Wiki, ik ik):
This is equivalent of doing two operations, and times, and seeing how many integers we can reach in this way. Note the first operation must be a , so now we have operations, starting at .
The first operations: there are in this case, since we can not have all 's. Since we did at least in the first operations, it follows that we are now above , so we do not need to worry about going back to anymore.
The last operations: . So far we have sequences.
However, there is still more! If we have or more 's in a row, then the number was once divisible by , but we didn't use any 's, contradiction. There are places to put our , and ways to label the other places. Thus our total is , where is prime. .
Example 3 (1987 imo/1):
Let be the number of permutations of the set which have exactly fixed points. Prove that .
Solution:
Look at the summation. It's actually just the expected number of fixed points, multiplied by the number of permutations, or .
Because every integer has a chance of being a fixed point, the expected number of fixed points is , and multiplying by , we see that the requested sum is indeed .
Problem 1 (Easy Pigeonhole! 1972 imo/1):
Prove that from a set of ten distinct twodigit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum.
Problem 2 (another easy Pigeonhole, 1976 usamo/1):
(a) Suppose that each square of a chessboard is colored either black or white. Prove that with any such coloring, the board must contain a rectangle (formed by the horizontal and vertical lines of the board such as the one outlined in the figure) whose four distinct unit corner squares are all of the same color.
(b) Exhibit a blackwhite coloring of a board in which the four corner squares of every rectangle, as described above, are not all of the same color.
Problem 3 (Pigeonhole mastery lol, 2012 usamo/2):
A circle is divided into congruent arcs by points. The points are colored in four colors such that some points are colored Red, some points are colored Green, some points are colored Blue, and the remaining points are colored Yellow. Prove that one can choose three points of each color in such a way that the four triangles formed by the chosen points of the same color are congruent.
Problem 4 (Get out your combinatorics identities, 1981 imo/2):
Let and consider all subsets of elements of the set . Each of these subsets has a smallest member. Let denote the arithmetic mean of these smallest numbers; prove that
Problem 5 (Well, it's trivial. Prove it! 2003 imo/1):
is the set . Show that for any subset of with elements we can find distinct elements of , such that the sets are all pairwise disjoint.
Problem 6 (1998 imo/2):
In a competition, there are contestants and judges, where is an odd integer. Each judge rates each contestant as either “pass” or “fail”. Suppose is a number such that, for any two judges, their ratings coincide for at most contestants. Prove that
Problemsoof
Problem 7 (Construction! 2014 imo/5):
For each positive integer , the Bank of Cape Town issues coins of denomination . Given a finite collection of such coins (of not necessarily different denominations) with total value at most , prove that it is possible to split this collection into or fewer groups, such that each group has total value at most .
Problem 8 (cough. 2016 imo/2):
Find all integers for which each cell of table can be filled with one of the letters and in such a way that:
in each row and each column, one third of the entries are , one third are and one third are ; and
in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are , one third are and one third are .
Problem 9 (I don't feel like putting this farther up. 2017 IMO/1):
For each integer , define the sequence for as Determine all values of such that there exists a number such that for infinitely many values of .
Problem 10 (Cool!!! 2013 usajmo/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 11 (This is literally trivial :P 2005 imo/2):
Let be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer the numbers leave different remainders upon division by .
Prove that every integer occurs exactly once in the sequence .
Problem 12 (Induction. 2010 usamo/2):
There are students standing in a circle, one behind the other. The students have heights . If a student with height is standing directly behind a student with height or less, the two students are permitted to switch places. Prove that it is not possible to make more than such switches before reaching a position in which no further switches are possible.
Problem 13 (Not even that hard, seriously. 2014 imo/6):
A set of lines in the plane is in if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite are; we call these its . Prove that for all sufficiently large , in any set of lines in general position it is possible to colour at least of the lines blue in such a way that none of its finite regions has a completely blue boundary.
Problem 14 (2020 usojmo/1. trivial):
Let be an integer. Carl has books arranged on a bookshelf. Each book has a height and a width. No two books have the same height, and no two books have the same width. Initially, the books are arranged in increasing order of height from left to right. In a move, Carl picks any two adjacent books where the left book is wider and shorter than the right book, and swaps their locations. Carl does this repeatedly until no further moves are possible. Prove that regardless of how Carl makes his moves, he must stop after a finite number of moves, and when he does stop, the books are sorted in increasing order of width from left to right.
Problem 15 (2020 usojmo/3):
An empty cube is given, and a grid of square unit cells is drawn on each of its six faces. A [i]beam[/i] is a rectangular prism. Several beams are placed inside the cube subject to the following conditions:
 The two faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are possible positions for a beam.)
 No two beams have intersecting interiors.
 The interiors of each of the four faces of each beam touch either a face of the cube or the interior of the face of another beam.
What is the smallest positive number of beams that can be placed to satisfy these conditions?
Problem 16 (2020 usojmo/5):
Suppose that are distinct ordered pairs of nonnegative integers. Let denote the number of pairs of integers satisfying and . Determine the largest possible value of over all possible choices of the ordered pairs.
Extra
Nothin much
https://artofproblemsolving.com/wiki/index.php/Category:Olympiad_Combinatorics_Problems
Number Theory
Ok, I'm not sure of my skill level here... Number Theory is pretty easy, right?
https://s3.amazonaws.com/aopscdn.artofproblemsolving.com/resources/articles/olympiadnumbertheory.pdf (Highly recommended)
Before Oly
B. Problem 1:
What is mod ?
B. Problem 2:
Chicken nuggets at McDonald's come in packs of and . What is the largest number of chicken nuggets you can not buy?
B. Problem 3:
Find the last two digits of .
B. Problem 4 (Classic):
Find mod .
B. Problem 5 (2020 AOIME/1):
Find the number of ordered pairs of positive integers such that .
B. Problem 6 (2020 AOIME/5):
For each positive integer , let be the sum of the digits in the basefour representation of and let be the sum of the digits in the baseeight representation of . For example, , and . Let be the least value of such that the basesixteen representation of cannot be expressed using only the digits through . Find the remainder when is divided by .
B. Problem 7 (2020 AOIME/6):
Define a sequence recursively by , , andfor all . Then can be written as , where and are relatively prime positive integers. Find .
B. Problem 8 (2020 AOIME/10):
Find the sum of all positive integers such that when is divided by , the remainder is .
B. Problem 9 (2018 10B/19):
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is year older than Chloe, and Zoe is exactly year old today. Today is the first of the birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
B. Problem 10 (2018 10A/25):
For a positive integer and nonzero digits , , and , let be the digit integer each of whose digits is equal to ; let be the digit integer each of whose digits is equal to , and let be the digit (not digit) integer each of whose digits is equal to . What is the greatest possible value of for which there are at least two values of such that ?
B. Problem 11 (2016 II/11):
For positive integers and , define to be nice if there exists a positive integer such that has exactly positive divisors. Find the number of positive integers less than that are neither nice nor nice.
B. Problem 12 (2014 I/8):
The positive integers and both end in the same sequence of four digits when written in base , where digit is not zero. Find the threedigit number .
B. Problem 13 (2011 I/11):
Let be the set of all possible remainders when a number of the form , a nonnegative integer, is divided by . Let be the sum of the elements in . Find the remainder when is divided by .
B. Problem 14 (2010 I/12):
Let be an integer and let . Find the smallest value of such that for every partition of into two subsets, at least one of the subsets contains integers , , and (not necessarily distinct) such that .
Note: a partition of is a pair of sets , such that , .
B. Problem 15 (2006 I/13):
For each even positive integer , let denote the greatest power of 2 that divides For example, and For each positive integer let Find the greatest integer less than 1000 such that is a perfect square.
B. Problem 16 (2006 II/14):
Let be the sum of the reciprocals of the nonzero digits of the integers from to inclusive. Find the smallest positive integer for which is an integer.
B. Problem 17 (2008 II/15):
Find the largest integer satisfying the following conditions:
(i) can be expressed as the difference of two consecutive cubes
(ii) is a perfect square.
B. Problem 18 (2021 usajmo/3):
What is mod ? Specify how you got your answer. Write at least essays on all aspects of the Measles morbillivirus as well.
B. Problem 19 (1959 imo/1):
Prove that the fraction is irreducible for every natural number .
B. Problem 20 (1975 imo/4):
When is written in decimal notation, the sum of its digits is . Let be the sum of the digits of . Find the sum of the digits of . ( and are written in decimal notation.)
Exponents
Basics of NT
If , for some integer , and positive integer , then we can say . We can add, subtract, and multiply in the mods.
Perfect squares can never be . corresponds to . Notice how they repeat every , as all are , so since we can multiply, they will all be . Similarly, all of will be , so their squares will be .
(Exercise): Prove that no squares are congruent to or .
Chinese remainder theorem (CRT) states that for relatively prime , the congruences
...
has congruence for , and it is .
Orders
(totient of n) is the number positive integers less than that are relatively prime to .
Immediately from PIE,
if where the are distinct primes,
Euler's Totient Theorem states that if and are relatively prime positive integers, then is .
A direct result is Fermat's Little Theorem, which states that if is prime, is .
The order of an integer modulo a prime is the smallest positive integer such that is .
Primitive Roots: Let be a prime. Then there exists a residue such that has order .
Example 1 (2016 HMMT Feb. Team/2):
For positive integers , let be the smallest positive integer for which is divisible by , if such a positive integer exists, and otherwise. What is ?
Solution:
is nonzero if and only if is relatively prime to , so we'll only consider those numbers.
The orders modulo are , respectively. Now we compute the LCM, picking one from each of them (noting that by Chinese Remainder Theorem, they each generate a unique number). For and , we make a table.
Adding them all up, we get . This is for orders of for each of modulo and . If we have a order of for modulo , in the first row we change the to a , adding , so the total is .
Problem 1:
Find all positive integers such that divides .
Problem 2 (ISL 2000 A1):
Determine all positive integers that satisfy the following condition: for all and relatively prime to we have
Problem 3 (Fermat's Christmas Theorem):
Let and be positive integers, and let be a prime. Suppose divides . Prove that divides both and .
LTE
Define for prime and positive integer to be the highest power of dividing . For example, .
LTE lemma (Lifting The Exponent) says that if and are both , , If , . If and is odd, . If is even, .
The proofs are left as an exercise. Just expand .
Zsigmondy theorem: (i) For any sequence (, and are relatively prime), there exists a prime such that is not a factor of any other terms of the terms before in the sequence with the exception of , , and is a power of . (ii) For any sequence , the same holds but the exceptions are .
Example 1 (2018 I/11 showcasing the power):
Find the least positive integer such that when is written in base , its two rightmost digits in base are .
Solution:
By inspection, all are . Now, all are . Using LTE, . .
Problem 4 (2020 I/12, similar flavor...):
Let be the least positive integer for which is divisible by Find the number of positive integer divisors of
Problem 5 (2008 tst/4):
Prove that for no integer is a perfect square.
Example 2 (1991 isl/18?):
Find the highest degree of for which divides the number
Solution:


Since is divisible by but not , by LTE (On and )
Problem 6 (2016 usamo/2):
Prove that for any positive integer ,is an integer.
Problem 7 (1990 imo/3):
Determine all integers such that is an integer.
Problem 8 (2000 imo/5):
Does there exist a positive integer such that has exactly prime divisors and divides ?
Problem 9 (ISL 2014/N5):
Find all triples consisting of a prime number and two positive integers and such that and are both powers of .
Other NT/Problems
(Exercise, Wilson's): Prove that for integer , then is divisible by if and only if is prime.
Chicken McNugget Theorem states that for any two relatively prime positive integers , the greatest integer that cannot be written in the form for nonnegative integers is . Furthermore, positive integers which cannot be expressed in the form .
Example 1 (ISL 2007/N5):
Find all surjective functions such that for every and every prime the number is divisible by if and only if is divisible by .
Solution:
Assume . Then, if , by the given. By induction, this holds for all , but is surjective, contradiction . (1)
Assume and . By subjectivity, there exists a such that . Thus as well, so and . Now we have , so . Therefore . (2)
Assume . Choose a prime . Since , (2) means , contradiction. Thus is injective (3)
Assume for some prime . By (1), (2), , contradiction (4).
By (1), . By (4), . Since is actually bijective by (3), this means that (works)is the only solution.
Example 2 (usajmo 2011/1):
Find all positive integers for which is a perfect square.
Solution
If is even, taking modulo results in the expression being , contradiction. Thus is odd.
If is odd, then the expression is , contradiction.
If , the expression is , so the only solution is .
Problem 10 (Justin Stevens):
Prove that for positive integers we cannot have .
Problem 11 (1975 usamo/1):
For every natural number , evaluate the sum(The symbol denotes the greatest integer not exceeding .)
Problem 12 (1992 imo/1):
Find all integers , , satisfying such that is a divisor of .
Problem 13 (1973 usamo/5):
Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.
Problem 14 (2007 usamo/5 AHAHAHAHA CAN YOU FIND THE DIFFERENCE OF SQUARES???):
Prove that for every nonnegative integer , the number is the product of at least (not necessarily distinct) primes.
Problem 15 (1991 usamo/3):
Show that, for any fixed integer the sequenceis eventually constant.
Problem 16 (1968 imo/6):
For every natural number , evaluate the sum(The symbol denotes the greatest integer not exceeding .)
Problem 17 (1983 imo/3; 3var Chicken McNuggets!):
Let and be positive integers, no two of which have a common divisor greater than . Show that is the largest integer which cannot be expressed in the form , where are nonnegative integers.
Problem 18 (Wolstenholme's Theorem Theorem):
Prove that for a prime , ifis expressed as a fraction, then the numerator is divisible by .
Problem 19 (2005 imo/4):
Determine all positive integers relatively prime to all the terms of the infinite sequence
Problem 20 (1989 imo/5):
Prove that for each positive integer there exist consecutive positive integers none of which is an integral power of a prime number.
Problem 21 (1971 imo/3):
Prove that we can find an infinite set of positive integers of the from (where is a positive integer) every pair of which are relatively prime.
Problem 22 (2013 usajmo/1):
Are there integers and such that and are both perfect cubes of integers?
Problem 23 (2005 usamo/2):
Prove that the system
has no solutions in integers , , and .
Bonus! "A bit more advanced bit of theory"...
Pell Equations!
We will be working in , that is, for rational , and positive . Define the norm of .
(Exercise): Prove the norm is multiplicative. That is, . (+)
Let's say we were trying to find solutions to .
1. Guess and check a solution. Make sure it is a "smallest" solution. In this case, it is , that is, has norm .
2. We can generate all solutions, which are for all , since these all have norm by (+).
(Exercise): Now find all solutions so .
Quadratic ResiduesLet and be positive integers. is a quadratic residue modulo if for some .
Determining whether is a quadratic residue modulo is easiest if is a prime. In this case we write
The symbol is called the Legendre symbol.
follows directly.
(Things?)
(i)
(ii)
Jacobi Symbols=9.5/10 on coolness scale. Legendre=0.5/10 on coolness scale. Therefore, use Jacobi. :P. If follows basically all of the above properties, but I feel like all this stuff is best seen as an example.
Quadratic Reciprocity:
(Note that this is always integral as the bottom number has to be odd.)
Example 1:
Is a quadratic residue modulo ?
Solution:
No. .
Example 2:
Is a quadratic residue modulo ?
Solution:
from Quadratic Reciprocity :)
from .
Once again from QR.
From mods. Since is a Quadratic Residue modulo and modulo , it is a Quadratic Residue modulo , and since is prime, the answer is Yes!
Vieta Jumping! :) Assume that there is a pair such that the “given” is minimal. Rearrange the equation to get an equation in . From this equation, we get another that satisfies it. Call this . Show that this contradicts the minimality argument, so that you’re done!
Example 3 (1988 imo/6, yes it is actually trivial not sure why it's so famous):
Let and be positive integers such that divides . Show that is the square of an integer.
Solution
Let . WLOG . Then, we write a quadratic in :
, and since is an integer, so is !
Thus we can keep "decreasing" the sequence in this way. Since , we can not have any negative integers in this chain, and therefore we must hit at some time, so as desired. (This was a slight twist!)
Problems
Problem 24 (2007 imo/5):
Let and be positive integers. Show that if divides , then .
Problem 25 (Brilliant.org):
If and are positive integers such that is an integer, show that .
Problem 26 (2009 usa tst/5):
Find all pairs of positive integers such that divides .
Problem 27 (AsianPacific Olympiad 1997?):
Find an integer such that divides .
Problem 28 (ISL 2017/N6):
Find the smallest positive integer or show no such exists, with the following property: there are infinitely many distinct tuples of positive rational numbers such that both are integers.
Problem 29 (2017 usa tst/6):
Prove that there are infinitely many triples of positive integers with prime, , and , such that is a multiple of .
Appendix A. Induction
Induction:
Show that an event is true for the [base case] (most likely when ).
Assume that is true for (This is called the [inductive hypothesis]). Then, show that it is true for (This is called the [inductive step]). Since it was true for , it is also true for , , . . . and so we are done.
Strong induction:
Again show that an event is true for the base case.
Then, assume that is true for . Then show that it is true for . Since it was true for , it is true for , (because and are true), and so on, so we are done again.
Example 1:
Prove that for all .
Solution (Induction):
(Base Case):
(Inductive Hypothesis): Now assume for some .
(Inductive Step): Then, by the inductive hypothesis, .
(This might be optional, still do it of course): By the principle of mathematical induction, we are done.
Example 2 (Fundamental Theorem of Arithmetic):
Every integer can be written uniquely as the product of prime numbers.
Solution (Strong Induction):
(Base Case): This is clearly true for .
(Inductive Hypothesis): Now assume that it is true for .
(Inductive Step): If is prime, we are done. If is not prime, it has a smallest prime factor, which we will denote as . Then for some . Then, since , it can be expressed as the product of prime numbers and therefore can also be expressed as the product of prime numbers, and we are done by Strong Induction.
Problem 1:
Prove that .
Problem 2 (Evan Chen, Strong Induction):
Prove that has no prime factors .
Appendix B. Hints
All the pro people do this, so lets do it
~Lcz 6/24/2020, 4:26 PM CST
Wait actually this is going to be hard. Maybe later
~Lcz 7/3/2020, 9:07 AM CST
Hint #1: Think harder! I'm prepared to bet that you haven't spent a combined total of more than 24 hours on this problem.