Olympiad Corner
The 2000 Canadian Mathematical Olympiad
Problem 1. At 12:00 noon, Anne, Beth
and Carmen begin running laps around a circular track of length three hundred meters, all starting from the same point on the track. Each jogger maintains a constant speed in one of the two possible directions for an indefinite period of time. Show that if Anne’s speed is different from the other two speeds, then at some later time Anne will be at least one hundred meters from each of the other runners. (Here, distance is measured along the shorter of the two arcs separating two runners).
Problem 2. A permutation of the
integers 1901, 1902, …, 2000 is a sequence a , 1 a , …, 2 a100 in which each of those integers appears exactly once. Given such a permutation, we form the sequence of partial sums
1 1 a s = , s2 =a1+a2, . ..., , 100 1 2 100 3 2 1 3 a a a s a a a s = + + = + ++ (continued on page 4)
When we write down a number, it is understood that the number is written in base 10. We learn many interesting facts at a very young age. Some of these can be easily explained in terms of base 10 representation of a number. Here is an example.
Example 1. Show that a number is divisible by 9 if and only if the sum of its digits is divisible by 9. How about divisibility by 11? Solution. Let 10M =dm10m++d1 , 0 d + where d = 0, 1, 2, …, 9. The i binomial theorem tells us 10k =(9+1)k
. 1 9 + = Nk So 0 1(9 1) ) 1 9 ( N d d d M = m m+ ++ + + ). ( ) ( 9 dmNm+ +d1 + dm+ +d1+d0 =
Therefore, M is a multiple of 9 if and only if dm++d1+d0 is a multiple of 9.
Similarly, we have 10k =11Nk′ +(−1)k. So M is divisible by 11 if and only if
0 1 ) 1 (− mdm+−d +d is divisible by 11.
Remarks. In fact, we can also see that the remainder when M is divided by 9 is the same as the remainder when the sum of the digits of M is divided by 9. Recall the notation a≡b (mod c) means a and b have the same remainder when divided by c. So we have M ≡dm++d1+d0 (mod 9).
The following is an IMO problem that can be solved using the above remarks. Example 2. (1975 IMO) Let A be the sum of the decimal digits of 44444444, and B be the sum of the decimal digits of A. Find the sum of the decimal digits of B. Solution. Since 44444444<(105)4444 =
,
1022220 so A<22220×9=199980. Then 46B<1+9×5= and the sum of the decimal digits of B is at most 3+9=12. Now 74444≡ (mod 9) and 73=343
1
≡ (mod 9) imply 44443≡1 (mod 9). Then 744444444=(44443)14814444≡ (mod 9). By the remarks above, A, B and the sum of the decimal digits of B also have remainder 7 when divided by 9. So the sum of the decimal digits of B being at most 12 must be 7.
Although base 10 representations are common, numbers expressed in other bases are sometimes useful in solving problems, for example, base 2 is common. Here are a few examples using other bases.
Example 3. (A Magic Trick) A magician asks you to look at four cards. On the first card are the numbers 1, 3, 5, 7, 9, 11, 13, 15; on the second card are the numbers 2, 3, 6, 7, 10, 11, 14, 15; on the third card are the numbers 4, 5, 6, 7, 12, 13, 14, 15; on the fourth card are the numbers 8, 9, 10, 11, 12, 13, 14, 15. He then asks you to pick a number you saw in one of these cards and hand him all the cards that have that number on them. Instantly he knows the number. Why?
Solution. For n = 1, 2, 3, 4, the numbers on the n-th card have the common feature that their n-th digits from the end in base 2 representation are equal to 1. So you are handing the base 2 representation of your number to the magician. As the numbers are less than 24, he gets your number easily.
Remarks. A variation of this problem is the following. A positive integer less than
4
2 is picked at random. What is the least number of yes-no questions you can ask
Volume 6, Number 2 April 2001 – May 2001
Base n Representations
Kin Y. Li
Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK 高 子 眉 (KO Tsz-Mei)
梁 達 榮 (LEUNG Tat-Wing), Appl. Math Dept, HKPU 李 健 賢 (LI Kin-Yin), Math Dept, HKUST 吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, MATH Dept, HKUST for general assistance.
On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is June 30, 2001.
For individual subscription for the next five issues for the 01-02 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin Li Department of Mathematics Hong Kong University of Science and Technology
Clear Water Bay, Kowloon, Hong Kong Fax: 2358-1643
that always allow you to know the number? Four questions are enough as you can ask if each of the four digits of the number in base 2 is 1 or not. Three questions are not enough as there are 15 numbers and three questions can only provide 23=8 different yes-no combinations.
Example 4. (Bachet’s Weight Problem) Give a set of distinct integral weights that allowed you to measure any object having weight 40n=1,2,3, , on a balance. Can you do it with a set of no more than four distinct integral weights?
Solution. Since the numbers 1 to 40 in base 2 have at most 6 digits, we can do it with the set 1, 2, 4, 8, 16, 32. To get a set with fewer weights, we observe that we can put weights from this set on both sides of the balance! Consider the set of weights 1, 3, 9, 27. For example to determine an object with weight 2, we can put it with a weight of 1 on one side to balance a weight of 3 on the other side. Note the sum of 1, 3, 9, 27 is 40. For any integer n between 1 and 40, we can write it in base 3. If the digit 2 appears, change it to 13− so that n can be written as a unique sum and difference of 1, 3, 9, 27. For example, 22 = 2⋅9 + 3 + 1 = (3 – 1)9 + 3 + 1 = 27 – 9 + 3 + 1 suggests we put the weights of 22 with 9 on one side and the weights of 27, 3, 1 on the other side. Example 5. (1983 IMO) Can you choose 1983 pairwise distinct nonnegative integers less than 10 such that no three 5 are in arithmetic progression?
Solution. Start with 0, 1 and at each step add the smallest integer which is not in arithmetic progression with any two preceding terms. We get 0, 1, 3, 4, 9, 10, 12, 13, 27, 18, … . In base 3, this sequence is ... , 1001 , 1000 , 111 , 110 , 101 , 100 , 11 , 10 , 1 , 0
(Note this sequence is the nonnegative integers in base 2.) Since 1982 in base 2 is 11110111110, so switching this from base 3 to base 10, we get the 1983th term of the sequence is 87843<105. To see this sequence works, suppose x, y, z with x <y < z are three terms of the sequence in arithmetic progression. Consider the
rightmost digit in base 3 where x differs from y, then that digit for z is a 2, a contradiction.
Example 6. Let ][r be the greatest integer less than or equal to r. Solve the equation ] 8 [ ] 4 [ ] 2 [ ] [x + x + x + x . 12345 ] 32 [ ] 16 [ + = + x x
Solution. If x is a solution, then since , ] [ 1 r r r− < ≤ we have 63x – 6 < 12345 . 63x ≤ It follows that 195<x<196. Now write the number x in base 2 as
, .
11000011abcde where the digits , , , , ,b c d e a are 0 or 1. Substituting this into the equation, we will get 12285 + 31a + 15b + 7c + 3d + e = 12345. Then 31a + 15b + 7c + 3d + e = 60, which is impossible as the left side is at most 31 + 15 + 7 + 3 + 1 = 57. Therefore, the equation has no solution.
Example 7. (Proposed by Romania for 1985 IMO) Show that the sequence {an} defined by an =[n 2] for n = 1, 2, 3, … (where the brackets denote the greatest integer function) contains an infinite number of integral powers of 2.
Solution. Write 2 in base 2 as ,
.1 2 3 0bb b
b where each bi =0 or 1. Since 2 is irrational, there are infinitely many .bk =1 If bk =1, then in base 2,
k k k 2 b b .b 2 −1 = 0 −1 . Let m = ], 2 2 [ k−1 then . 2 1 2 2 ] 2 2 [ 1 2 2k−1 − < k−1 =m< k−1 − Multiplying by 2 and adding 2, we
get . 2 2 2 2 ) 1 ( 2k < m+ < k + Then . 2 ] 2 ) 1 [(m+ = k
Example 8. (American Mathematical Monthly, Problem 2486) Let p be an odd prime number. For any positive integer k, show that there exists a positive integer m such that the rightmost k digits of m2,
when expressed in the base p, are all 1's. Solution. We prove by induction on k. For ,k=1 take m=1. Next, suppose
2
m in base p, ends in k 1's, i.e. 1 2=1+p+ +pk−
m +(apk +).
This implies m is not divisible by p. Let gcd stand for greatest common divisor (or highest common factor). Then gcd(m, p) = 1. Now k k k m mcp c p cp m )2 2 2 2 2 ( + = + + . ) 2 ( 1+ ++ 1+ + + = p pk− a mc pk
Since gcd(2m,p)=1, there is a positive integer c such that (2m)c≡1−a (mod p). This implies a+2mc is of the form
Np +
1 and so (m+cpk)2 will end in at least (k+1) 1's as required.
Example 9. Determine which binomial coefficients )! ( ! ! r n r n Crn − = are odd. Solution. We remark that modulo arithmetic may be extended to polynomials with integer coefficients. For example, (1+x)2=1+2x+x2≡ 1+x2 (mod 2). If n=am++a1, where the
i
a ’s are distinct powers of 2. We have
k k x x)2 1 2 1 ( + ≡ + (mod 2) by induction on k and so ) 1 ( ) 1 ( ) 1 ( +x n ≡ +xam +xa1 (mod 2). The binomial coefficient C is odd if rn and only if the coefficient of
x
r in) 1 ( ) 1 ( +xam +xa1 is 1, which is equivalent to r being 0 or a sum of one or more of the a ’s. For example, if i
, 1 4 16 21= + + =
n then C is odd for r rn = 0, 1, 4, 5, 16, 17, 20, 21 only.
Example 10. (1996 USAMO) Determine (with proof) whether there is a subset X of the integers with the following property: for any integer n there is exactly one solution of a+2b=n with a,b∈X. This is a difficult problem. Here we will try to lead the reader to a solution. For a problem that we cannot solve, we can try to change it to an easier problem. How about changing the problem to positive integers, instead of integers? At least we do not have to worry about negative integers. That is still not too obvious how to proceed. So can we change it to an even simpler problem? How about changing 2 to 10?
(continued on page 4)
Problem Corner
We welcome readers to submit their solutions to the problems posed below for publication consideration. Solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, Hong Kong University of Science & Technology, Clear Water Bay, Kowloon. The deadline for submitting solutions is June 30, 2001.
Problem 126. Prove that every integer
can be expressed in the form 2
2 2 y 5z
x + − , where x, y, z are integers.
Problem 127. For positive real numbers
a, b, c with a + b + c = abc, show that 2 3 1 1 1 1 1 1 2 2 2 + + + + ≤ +a b c , and determine when equality occurs. (Source: 1998 South Korean Math Olympiad)
Problem 128. Let M be a point on
segment AB. Let AMCD, BEHM be squares on the same side of AB. Let the circumcircles of these squares intersect at M and N. Show that B, N, C are collinear and H is the orthocenter of
. ABC
∆ (Source: 1979 Henan Province Math Competition)
Problem 129. If f(x) is a polynomial of
degree 2m+1 with integral coefficients for which there are 2m+1 integers
1 2 2
1 ,k , ,k m+
k such that f(ki)=1 for , 1 2 , , 2 , 1 + = m
i prove that f(x) is not the product of two nonconstant polynomials with integral coefficients.
Problem 130. Prove that for each
positive integer n, there exists a circle in the xy-plane which contains exactly n lattice points in its interior, where a lattice point is a point with integral coordinates. (Source: H. Steinhaus, Zadanie 498, Matematyka 10 (1957), p. 58)
*****************
Solutions
*****************
Problem 121. Prove that any integer
greater than or equal to 7 can be written as a sum of two relatively prime integers, both greater than 1.
(Two integers are relative prime if they share no common positive divisor other than 1. For example, 22 and 15 are relatively prime, and thus 37 = 22 + 15 represents the number 37 in the desired way.) (Source: Second Bay Area Mathematical Olympaid)
Solution. CHAO Khek Lun Harold (St. Paul’s College, Form 6), CHIU Yik Yin (St. Joseph’s Anglo-Chinese School, Form 5), CHONG Fan Fei (Queen’s College, Form 4), CHUNG Tat Chi (Queen Elizabeth School, Form 4), LAW
Siu Lun (Ming Kei College, Form 6), NG Cheuk Chi (Tsuen Wan Public Ho Chuen
Yiu Memorial College), WONG Wing
Hong (La Salle College, Form 3) & YEUNG Kai Sing (La Salle College,
Form 4).
For an integer n≥7, n is either of the form 2j + 1 (j > 2) or 4k(k > 1) or 4k + 2(k > 1). If n = 2j + 1, then j and j + 1 are relatively prime and n = j + (j + 1). If n = 4k, then 2k - 1 (>1) and 2k + 1 are relatively prime and n = (2k - 1) + (2k + 1). If n = 4k + 2, then 2k - 1 and 2k + 3 are relatively prime and n = (2k - 1) + (2k + 1).
Other commended solvers: HON Chin
Wing (Pui Ching Middle School, Form 6), LEUNG Wai Ying (Queen Elizabeth
School, Form 6), NG Ka Chun
Bartholomew (Queen Elizabeth School,
Form 6) & WONG Tak Wai Alan (University of Toronto).
Problem 122. Prove that the product of
the lengths of the three angle bisectors of a triangle is less than the product of the lengths of the three sides. (Source: 1957 Shanghai Junior High School Math Competition).
Solution. YEUNG Kai Sing (La Salle College, Form 4).
Let AD, BE and CF be the angle bisectors of ABC∆ , where D is on BC, E is on CA and F is on AB. Since ∠ADC = ∠ABD + ∠BAD > ∠ABD, there is a point K on CA such that ∠ADK=∠ABD. Then
ABD
∆ is similar to ∆ADK. So AB/AD = AD/AK. Then AD = 2 AB⋅AK < CA AB⋅ . Similarly, BE2 <BC⋅ABand . 2 CA BC CF < ⋅ Multiplying these in-equalities and taking square roots, we getAD⋅BE⋅CF< AB⋅BC⋅CA. Other commended solvers: CHAO
Khek Lun Harold (St. Paul’s College,
Form 6), CHIU Yik Yin (St. Joseph’s Anglo-Chinese School, Form 5), HON
Chin Wing (Pui Ching Middle School,
Form 6) & LEUNG Wai Ying (Queen Elizabeth School, Form 6).
Problem 123. Show that every convex
quadrilateral with area 1 can be covered by some triangle of area at most 2. (Source: 1989 Wuhu City Math Competition)
Solution. CHAO Khek Lun Harold (St. Paul’s College, Form 6), CHUNG Tat
Chi (Queen Elizabeth School, Form 4) & LEUNG Wai Ying (Queen Elizabeth
School, Form 6).
Let ABCD be a convex quadrilateral with area 1. Let AC meet BD at E. Without loss of generality, suppose AE≥EC. Construct ,∆AFG where lines AB and AD meet the line parallel to BD through C at F and G respectively. Then ∆ABE is similar to ∆AFC. Now AE≥EC implies .AB≥BF Let [XYZ] denote the area of polygon XYZ, then [ABC]
≥ [FBC]. Similarly, [ADC] ≥ [GDC]. Since [ABC] + [ADC] = [ABCD] = 1, so [AFG] = [ABCD] + [FBC] + [GDC] ≤ 2[ABCD] = 2 and AFG∆ covers ABCD.
Problem 124. Find the least integer n
Mathematical Excalibur, Vol. 6, No. 2, Apr 01 – May 01 Page 3
A E K C F B D D A E B C G F
such that among every n distinct numbers , , , , 2 1 a an a chosen from [1,1000], there always exist a ,i aj such that
. 3 1
0<ai−aj< + 3aiaj
(Source: 1990 Chinese Team Training Test)
Solution. CHAO Khek Lun Harold (St. Paul’s College, Form 6), CHUNG Tat
Chi (Queen Elizabeth School, Form 4) & LEUNG Wai Ying (Queen Elizabeth
School, Form 6).
For ,n≤10 let ai =i3 (i=1,2 , ,n). Then the inequality cannot hold since
3 3 0<i − j implies i− j≥1 and so 3 3 j i − = (i− j)3 + 3ij(i− j) ≥1+3ij. For n = 11, divide [1,1000] into intervals
] ) 1 ( , 1 [k3+ k+ 3 for k = 0, 1, …, 9. By pigeonhole principle, among any 11 distinct numbers a1,a2,,a11 in [1, 1000], there always exist a ,i aj , say
j i a
a > , in the same interval. Let 3 i a x= and y=3aj , then 0 < x - y < 1 and 0 < ai−aj = x3−y3 = (x−y)3 + ) ( 3xy x−y < 1 + 3xy = 1 + 33aiaj. Other commended solvers: NG Cheuk
Chi (Tsuen Wan Public Ho Chuen Yiu
Memorial College), NG Ka Chun
Bartholomew (Queen Elizabeth School,
Form 6), WONG Wing Hong (La Salle College, Form 3) & YEUNG Kai Sing (La Salle College, Form 4).
Problem 125. Prove that
$ $
$
$ tan 3 tan 5 tan 89
1
tan2 + 2 + 2 + + 2 is an integer.
Solution. CHAO Khek Lun (St. Paul’s College, Form 6).
For θ=1$,3$ ,5$ ,,89$, we have cosθ 0
≠ and cos90θ =0. By de Moivre’s theorem, cos90θ+isin90θ = (cosθ+
90
) sinθ
i . Taking the real part of both sides, we get
∑
= − − = 45 0 2 2 90 90 2 cos sin ) 1 ( 0 k k k k kC θ θ .Dividing by cos90θ on both sides and letting x=tan2θ , we get
∑
= − = 45 0 90 2 ) 1 ( 0 k k k kC x .So tan21$ ,tan23,tan25$, ,tan289$
are the 45 roots of this equation. Therefore, their sum is C8890=4005.
Olympiad Corner
(continued from page 1) How many of these permutations will have no terms of the sequence s , …, 1
100
s divisible by three?
Problem 3. Let A = (a1 ,a2 , ,a2000) be a sequence of integers each lying in the interval [-1000, 1000]. Suppose that the entries in A sum to 1. Show that some nonempty subsequence of A sums to zero.
Problem 4. Let ABCD be a convex
quadrilateral with ADB CBD= ∠ ∠ 2 , ABD ∠ = 2∠CDB and AB = CB. Prove that AD = CD.
Problem 5. Suppose that the real
numbers a1,a2,,a100 satisfy ≥ ≥ ≥ ≥ 2 100 1 a a a 0, a1+a2≤100 and a3+a4++a100≤100.
Determine the maximum possible value of a12+a22++a1002 , and find all possible sequences a1,a2,,a100 which achieve this maximum.
Base n Representations
(continued from page 2) Now try an example, say n = 12345. We can write n in more than one ways in the form .a+10b Remember we want a, b to be unique in the set X. Now for b in X,
b
10 will shift the digits of b to the left one space and fill the last digit with a 0. Now we can try writing n = 12345 = 10305 + 10(204). So if we take X to be the positive integers whose even position digits from the end are 0, then the problem will be solved for n = a + 10b. How about n=a+2b? If the reader examines the reasoning in the case
, 10b
a+ it is easy to see the success
comes from separating the digits and observing that multiplying by 10 is a shifting operation in base 10. So for
, 2b
a+ we take X to be the set of positive integers whose base 2 even position digits from the end are 0, then the problem is solved for positive integers.
How about the original problem with integers? It is tempting to let X be the set of positive or negative integers whose base 2 even position digits from the end are 0. It does not work as the example 1 + 2⋅1 = 3 = 5 + 2 ( -1) shows uniqueness fails. Now what other ways can we describe the set X we used in the last paragraph? Note it is also the set of positive integers whose base 4 representations have only digits 0 or 1. How can we take care of uniqueness and negative integers at the same time? One idea that comes close is the Bachet weights.
The brilliant idea in the official solution of the 1996 USAMO is do things in base
). 4
(− That is, show every integer has a unique representation as ( 4), 0 i k i i c −
∑
=where each c = 0, 1, 2 or 3 and i ck ≠0. Then let X be the set of integers whose base (−4) representations have only
0 = i
c or 1 will solve the problem. To show that an integer n has a base (−4) representation, find an integer m such that
n m≥ + + + 2 2 0 4 4 4 and express n + 3 (41+43++42m−1) in base 4 as
∑
= m i i i b 2 0 4 . Now set c2i =b2i and .c2i−1=3−b2i−1 Then . ) 4 ( 2 0 i m i i c n=∑
− =To show the uniqueness of base (−4) representation of n, suppose n has two distinct representations with digits c 's i and d 's. Let j be the smallest integer i such that cj ≠dj. Then
∑
= − − = − = k j i i i i d c n n ( )( 4) 0would have a nonzero remainder when divided by 4j+1, a contradiction.
