GC Blue Combin Num Theory 2 (1)

7/28/2019 GC Blue Combin Num Theory 2 (1)

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Combinatorial Number Theory (Teachers Edition)

Gabriel Carroll

MOP 2010 (Blue)

Combinatorial number theory refers to combinatorics flavored with the rich juicy arith-metical structure of the integers. At the elementary level, like many other areas of com-binatorics, combinatorial number theory doesnt require a lot of deep theorems; insteadits a big hodgepodge of ideas and tricks.

A few notational conventions are useful, in particular in stating additive problems. If

A and B are sets of integers, we often write A + B for the set {a + b | a A, b B}. Forc a constant, we often write A + c for {a + c | a A} and cA = {ca | a A}. Also, if weare interested in sums or products of generic sets of integers, the sum of the empty set isgenerally taken to be 0, and the product of the empty set is 1.

1 Problem-solving techniques

For the most part, the ideas that are useful in solving combinatorial number theoryproblems are the same ones that are useful in other areas of combinatorics.

Use the pigeonhole principle (or probabilistic methods)

Use induction

Use greedy algorithms

Look at prime factorizations and the divisibility lattice

Look at largest or smallest elements

Think about orders of magnitude

Count things in two ways

Look at things mod n, for conveniently chosen n

Transform things to make them convenient to work with

Dont be afraid of case analysis and brute force

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Use generating functions or similar algebraic techniques

Translate the problem into graph theory

Use actual number theory

2 Problems

1. Determine whether or not there exists an increasing sequence a1, a2, . . . of positiveintegers with the following property: for any integer k, only finitely many of thenumbers a1 + k, a2 + k , . . . are prime.

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2. [BMC, 1999] The set of positive integers is partitioned into finitely many subsets.Show that some subset S has the following property: for every positive integer n, Scontains infinitely many multiples of n.

3. Given is a list of n positive integers whose sum is less than 2n. Prove that, for anypositive integer m not exceeding the sum of these integers, one can choose a sublistof the integers whose sum is m.

greedy

4. Let S be an infinite set of integers, such that every finite subset of S has a commondivisor greater than 1. Show that all the elements of S have a common divisorgreater than 1.

5. Show that any positive integer can be expressed as a sum of terms of the form 2a3b,

where a, b are nonnegative integers, and no term is divisible by any other.

take powers of 3 greedily; if the number is even, divide by 2 and use induction

6. [Canada, 2000] Given are 2000 integers, each one having absolute value at most1000, and such that their sum equals 1. Prove that we can choose some of theintegers so that their sum equals 0.

order them so that the sum of each sublist is in [2000, 1999], then pigeonhole

7. [Paul Erdos] Show that if n + 1 numbers are chosen from the set {1, 2, . . . , 2n},then one of these numbers divides another.

8. [BAMO, 2009] A set S of positive integers is magic if for any two distinct membersi, j S, (i + j)/ gcd(i, j) is also in S. Find all finite magic sets.

cant have two coprime numbers, else we generate infinitely many numbers. let a, bbe the smallest two numbers. so (a + b)/(a, b)


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then b, d give e = a2(a+2)/2. now keep averaging b, e, etc. and get a contradiction.so only such sets are {a, a2 a}.

9. [IMO, 1991] Let n > 6 be an integer with the following property: all the integersin {1, 2, . . . , n 1} that are relatively prime to n form an arithmetic progression.

Prove that n is either prime or a power of 2.let d be the difference of the progression. if d 3 then 3 | n, so 3 | d, but thend + 1 or 2d + 1 is divisible by 3, contradicting coprimality. so d = 1 (n prime) ord = 2 (n a power of 2).

10. [USAMO, 1998] Prove that, for each integer n 2, there is a set S of n integerssuch that ab is divisible by (a b)2 for all distinct a, b S.

11. [Reid Barton] Let a1 < a2 < be an increasing sequence of positive integers, suchthat an+1 an < 1000000 for all n. Prove that there exist indices i < j such thataj is divisible by ai.

12. [China, 2009] Find all pairs of distinct nonzero integers (a, b) such that there existsa set S of integers with the following property: for any integer n, exactly one ofn, n + a, n + b is in S.

answer: (kc,kd) where c, d 1, 2 mod 3 in some order. we can reduce to the casea, b coprime. if theyre 1, 2 mod 3 then just take the set of numbers that are 0mod 3. lets show this is necessary. for x S we have x + (b a), x + b / S sox+(2ba) S, likewise x+(2ab) S. if gcd(2ab, 2ba) = 1 then everythingsin S, which is bad. but the gcd is at most 3, possible only if a, b are 1, 2 mod 3 insome order.

13. [APMC, 1990] Let a1, . . . , ar be integers such that

iIai = 0 for every nonemptyset I {1, . . . , r}. Prove that the positive integers can be partitioned into a finitenumber of classes so that, whenever n1, . . . , nr are integers from the same class,a1n1 + + arnr = 0.

let p be a prime not dividing any partial sum; class them according to their lastnonzero digit in base p

14. [IMO, 2003] Let S = {1, 2, . . . , 106}. Prove that for any A S with 101 elements,we can find B S with 100 elements such that the sums a + b, for a A and b B,are all different.

as long as |B| < 100 we can find another element to put in B without creating newcollisions. proof: only 9999 sums exist so far, and each could create a collision forat most 100 of the values of b not already used.

15. [Russia, 1998] A sequence a1, a2, . . . of positive integers contains each positive in-teger exactly once. Moreover, for every pair of distinct positive integers m and

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n,1

1998 ab+a+b and (a+k)(bj1) < ab. subtractinggives k j > 1, but then (a + k)(b j) = ab + b(k j) + (b a k)j ab + 2b isalready greater than ab + a + b.

18. [IMO Shortlist, 1990] The set of positive integers is partitioned into finitely many

subsets. Prove that there exists some subset, say Ai, and some integer m with thefollowing property: for any k, there exist numbers a1 < a2 < < ak in Ai, withaj+1 aj m for each j.

let A1, . . . , An be the subsets. if none has the desired property, show by inductionthat Ai An contains arbitrarily long sequences of consecutive numbers.

19. [St. Petersburg, 1996] The numbers 1, 2, . . . , 2n are divided into 2 sets ofn numbers.For each set, we consider all n2 possible sums a + b, where a, b are in that set (andmay be equal). Each sum is reduced mod 2n. Show that the n2 remainders fromone set are equal, in some order, to the n2 remainders from the other set.

generating functions: A2 B2 divisible by x2n 1

20. [IMO Shortlist, 1999] Let A be a set ofN residues mod N2. Prove that there existsa set B of N residues mod N2 such that the set A + B contains at least half of allresidues mod N2.

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successively choose elements of B, we can always cover at least N/2 new sums bypigeonhole (each of the N2/2 new sums can be covered by one of N differentchoices of new element, and fewer than N2 choices of new element are available)

21. [IMO shortlist, 1999] Let x and y be odd integers with |x| = |y|. Suppose that the

positive integers have been colored in four different colors. Show that there existtwo different numbers of the same color whose difference is equal to x, y, x + y, orx y.

22. [Erdos-Selfridge] For any set A of positive integers, let A(n) be the number ofways of writing n as a sum of two distinct members of A. If two different sets Aand B have the property that A(n) = B(n) for all positive integers n, prove thatthe number of elements in each set is a power of 2.

let a, b be the generating functions for the two sets and f their difference. we getf(x2) = f(x)g(x) for some poly g. factor out some power of x 1, and were leftwith h(x2)(x + 1)n = h(x)g(x) where h(1) = 0. plug in x = 1 and divide through

to get g(1) = 2n, which gives what we needed.

23. [Bulgaria, 2000] Let p 3 be a prime number, and a1, . . . , ap2 a sequence ofintegers such that, for each i, neither ai nor a

ii 1 is a multiple of p. Prove that

there exists some collection of distinct terms whose product is congruent to 2 modp.

actually every product is achievable. proof: let Sk be the set of all products ofsubsets of the first k terms, mod p. check that |Sk| > k by induction each timewe include a new term, its order isnt a factor of k, so if we had exactly k beforethen we cant keep the same set.

24. [IMO, 2009] Let a1, a2, . . . , an be distinct positive integers and let M be a set ofn 1 positive integers not containing s = a1 + a2 + + an. A grasshopper is to

jump along the real axis, starting at the point 0 and making n jumps to the rightwith lengths a1, a2, . . . , an in some order. Prove that the order can be chosen in sucha way that the grasshopper never lands on any point in M.

induction. let m be the largest element of M and a1 < < an. if s an Mand less than m, theres some i such that s ai and s ai an are both not inM, so apply the induction hypothesis to all the elements except ai, an, then jumpby an and then ai. otherwise, use the induction hypothesis on a1, . . . , an1 to avoidlanding at any element of M except possibly m. if we never land at m were home

free. otherwise, take the preceding hop, replace it with an, and then fill in theremaining hops.

25. [Van der Waerdens Theorem] For any positive integers k and m, there exists Nwith the following property: if the integers 1, 2, . . . , N are colored in k colors, thereexists an arithmetic progression of length m, all of whose members are the samecolor.

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multidimensional grid proof induction on length of progresssions, proving for allvalues of k simultaneously

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GC Blue Combin Num Theory 2 (1)