1
Problem: Define 𝑓 = 1−𝑥. Compute 𝑓 1999 (2000).
Solution:
Get my hands dirty:
𝑥 = 2000
𝑓 2 (𝑥)
1999
2000
𝑓(𝑥)
1
−
1999
1
𝑓 3 (𝑥)
2000
𝑓 4 (𝑥)
1
−
1999
𝑓 5 (𝑥)
1999
2000
1999
The pattern is 𝑓 4𝑘 (2000) = − 1999 , 𝑓 4𝑘+2 (2000) = 2000 , 𝑓 4𝑘+3 (2000) = 2000
→ 𝑓 1999 (2000) = 2000.
Problem(Putnam 1990): Let 𝑇0 = 2, 𝑇1 = 3 , 𝑇2 = 6, 𝑎𝑛𝑑 𝑓𝑜𝑟 𝑛 ≥ 3:
𝑇𝑛 = (𝑛 + 4)𝑇𝑛−1 − 4𝑛𝑇𝑛−2 + (4𝑛 − 8)𝑇𝑛−3
The first few terms are: 2, 3, 6, 14, 40, 152, 784, 5168, 40576, 363392, …
Find a formula for 𝑇𝑛 of the form 𝑇𝑛 = 𝐴𝑛 + 𝐵𝑛 , 𝑤ℎ𝑒𝑟𝑒 (𝐴𝑛 ) 𝑎𝑛𝑑 (𝐵𝑛 ) 𝑎𝑟𝑒 𝑤𝑒𝑙𝑙 − 𝑘𝑛𝑜𝑤𝑛 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒𝑠.
Solution:
152 and 784 and kinda like 125 and 720, which is 5! 𝑎𝑛𝑑 6!.
So we have 𝑇𝑛 = 𝑛! + 2𝑛 . We prove it by induction. The first few ones are obvious. Assume the
argument is true with 𝑛 = 𝑘 − 1(𝑘 > 0). We have:
𝑇𝑘 = (𝑘 + 4)𝑇𝑘−1 − 4𝑘𝑇𝑘−2 + (4𝑘 − 8)𝑇𝑘−3
= (𝑘 + 4) ((𝑘 − 1)! + 2𝑘−1 ) − 4𝑘 ((𝑘 − 2)! + 2𝑘−2 ) + (4𝑘 − 8) ((𝑘 − 3)! + 2𝑘−3 )
= 2𝑘 + 𝑘!.
Case closed.
Problem: Consider a function 𝑓 that satisfies 𝑓(1) = 1, 𝑓(2𝑛) = 𝑓(𝑛), 𝑓(2𝑛 + 1) = 𝑓(2𝑛) + 1 for all
𝑛 ∈ 𝑁. Find a nice algorithm for 𝑓(𝑛). Your algorithm should be a single sentence long, at most.
Solution:
𝑓(1) = 1, 𝑓(2) = 1, 𝑓(3) = 2, 𝑓(4) = 1, 𝑓(5) = 2,
𝑓(6) = 2, 𝑓(7) = 3, 𝑓(8) = 1, 𝑓(9) = 2, 𝑓(10) = 2,
𝑓(11) = 3, 𝑓(12) = 2
So, 𝑓(2𝑘 ) = 1, 𝑓(3𝑘) = 2, 𝑓(2𝑘) = 2(𝑤𝑖𝑡ℎ 𝑘 ≠ 2𝑘 ). It’s obvious that our function can not be a
polynomial. What can it be? What if it doesn’t exist such an algorithm? It’s clear that f is 𝑁 → 𝑁. I’ll
leave it for now.
Problem: For each integer n > 1, find distinct positive integers 𝑥 and 𝑦 such that
1 1 1
+ =
𝑥 𝑦 𝑛
Solution:
First we see that 𝑥 and 𝑦 can not be smaller than n.
1
1
Lets transform the expression: 𝑛(𝑥 + 𝑦) = 1.
1
𝑥
1
𝑦
But again by BCS: 𝑛 ( + ) ≥
4𝑛
𝑥+𝑦
→ 𝑥 + 𝑦 ≥ 4𝑛
We observe that 𝑥 + 𝑦 ≠ 4𝑛 for 𝑛 > 1. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑥 + 𝑦 > 4𝑛
Thus one number must be greater than 2𝑛. By that we can also obtain that one number must
smaller than 2n. (1)
But we’re stuck here. Let’s get our hands dirty.
1
1
1 1
1
1
1
1
1
+ 2 = 3 + 6 , 3 = 4 + 12 , 4 = 5 + 20 , …
We see a pattern here: (𝑥, 𝑦) = (𝑛 + 1, 𝑛2 + 𝑛)
Is that true? We have
(𝑛 + 1)2
1
1
1
+ 2
=
= .
2
𝑛 + 1 𝑛 + 𝑛 𝑛(𝑛 + 2)
𝑛
Case closed !
ANOTHER QUESTION: Is there another pair?
Assume that there is another pair of (𝑥, 𝑦).
Let 𝑥 = 𝑛 + 𝑎, 𝑎 < 𝑛.
We have
1 1
1
𝑎
= −
=
𝑦 𝑛 𝑛 + 𝑎 𝑛(𝑛 + 𝑎)
→ 𝑎|𝑛(𝑛 + 𝑎)
So our problem becomes: Given a natural number n. With 1 < 𝑎 < 𝑛, can 𝑎|𝑛(𝑛 + 𝑎)?
We have
𝑛(𝑛 + 𝑎) = 𝑛2 (𝑚𝑜𝑑 𝑎)
However, with 1 < 𝑎 < 𝑛 then n is not divisible by a. DONE.
Problem: For each positive integer n, find positive integers solutions 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 to the equation:
1
1
1
1
+ + ⋯+ +
=1
𝑥1 𝑥2
𝑥𝑛 𝑥1 𝑥2 … 𝑥𝑛
Solution:
No ideas pop up yet.
Lets get our hands dirty.
𝑛 = 2: (𝑥1 , 𝑥2 ) = (2,3)
𝑛 = 3: (𝑥1 , 𝑥2 , 𝑥3 ) = (2,3,7)
𝑛 = 4: (𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 ) = (2,3,7,43)
Til here we can predict that 𝑥𝑘 is the smallest prime number that is not smaller than 𝑥1 𝑥2 … 𝑥𝑘−1 .
But then it doesn’t work with n=5.
Problem(BMO 1996): Define
𝑛
𝑞(𝑛) = [
].
[√𝑛]
Determine all positive integers for which 𝑞(𝑛) > 𝑞(𝑛 + 1).
Solution:
Well, lets get our hands dirty:
𝑞(1) = 1, 𝑞(2) = 2, 𝑞(3) = 3, 𝑞(4) = 2, 𝑞(5) = 2, 𝑞(6) = 3, 𝑞(7) = 3, 𝑞(8) = 4, 𝑞(9) = 3
So we predict that 𝑛 = 𝑘 2 − 1 then 𝑞(𝑛) > 𝑞(𝑛 + 1)
We have: 𝑞(𝑘 2 − 1) = [
𝑘 2 −1
[√𝑘 2 −1]
𝑘 2 −1
] = [ 𝑘−1 ] = 𝑘 + 1, 𝑎𝑛𝑑 𝑞(𝑘 2 ) = 𝑘.
One side finished. Now we prove that this is the only solution.
Conjecture: 𝑞(𝑘 2 ) ≤ 𝑞(𝑘 2 + 𝑎) 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑘 2 + 𝑎 < (𝑘 + 1)2 − 1. 𝑇ℎ𝑖𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑎𝑠𝑖𝑙𝑦 𝑝𝑟𝑜𝑣𝑒𝑑.
Case closed.
Problem: Find infinitely many positive integer solution to the equation
𝑥2 + 𝑦2 + 𝑧2 = 𝑤 2
Solution:
1
Problem(Germany 1995): Let 𝑥 be a real number such that 𝑥 + 𝑥 is an integer. Prove that 𝑥 𝑛 + 𝑥 −𝑛
is an integer, for all positive integers 𝑛.
Solution:
We have:
𝑛−1
𝑥 𝑛 + 𝑥 −𝑛
1 𝑛
𝑛
= (𝑥 + ) − ∑ ( ) 𝑥 2𝑘−𝑛
𝑥
𝑘
𝑘=1
𝑛−1
We prove ∑
𝑘=1
𝑛
𝑛
2
2
(𝑛𝑘)𝑥 2𝑘−𝑛 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟. 𝐵𝑒𝑠𝑖𝑑𝑒𝑠, ([𝑛−𝑖]) 𝑥 −𝑖 + ([𝑛+𝑖]) 𝑥 𝑖 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟.
By the hypothesis of induction we completed the proof.
Problem(Putnam 1995): Let S be a set of real numbers that is closed under multiplication ( that is, if
𝑎 and 𝑏 are in S, then so is 𝑎𝑏). Let T and U be disjoint subsets of S whose union is S. Given that the
product of any 𝑡ℎ𝑟𝑒𝑒 (not necessarily distinct) elements of T is in T and that the product of any
𝑡ℎ𝑟𝑒𝑒 (not necessarily distinct) elements of U is in U. Show that at least one of the two subsets T, U
is closed under multiplication.
Solution:
Assume that U has element 𝑎 and 𝑏 such that 𝑎𝑏 is not in U. And T has element 𝑐 and 𝑑 such that 𝑐𝑑
is not in T.
Because T and U are disjoint subsets of S and S is closed under multiplication, 𝑎𝑏 and 𝑐𝑑 are in T
and U, respectively. We multiply three elements of T: 𝑐. 𝑑. 𝑎𝑏 = 𝑎𝑏𝑐𝑑 and three elements of U:
𝑎. 𝑏. 𝑐𝑑 = 𝑎𝑏𝑐𝑑. Therefore, T and U has a common element ( contradiction ).
Problem(Russia 1995): Is it possible to place 1995 different natural numbers along a circle so that
for any two of these numbers, the ratio of the greatest to the least is a prime?
Solution:
99% no.
Let the circle be placed with 𝑎0 , 𝑎1 , 𝑎2 , … , 𝑎1995 (we added 𝑎0 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑎0 = 𝑎1995 )
𝑎𝑘−1
𝑎𝑘
is either a prime or the reciprocal of a prime, suppose the former occurs 𝑚 times and the latter
occurs 1995 − 𝑚 times. The product of all these ratios is
𝑎0
𝑎1995
= 1. This can only occur when 𝑚 =
1995 − 𝑚(absurd).
Therefore we’re done.
Problem: Consider a 21999 × 21999 𝑠𝑞𝑢𝑎𝑟𝑒, 𝑤𝑖𝑡ℎ 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 1 × 1 square removed. Show that no
matter where the small square is removed, it is possible to tile this “giant square minus tiny square”
with ells.
Solution:
Problem(IMO 1997): An 𝑛 × 𝑛 matrix (square array) whose entries come from the set 𝑆 =
{1,2, … ,2𝑛 − 1} is called a silver matrix if, for each 𝑖 = 1,2, … , 𝑛; the 𝒊𝑡ℎ row and the 𝒊𝑡ℎ column
together contain all the members of S. Show that silver matrices exist for infinitely many values of 𝑛.
Solution:
Problem: Let 𝑎0 be any real number that’s smaller than 1 and greater than 0. Define (𝑎𝑛 ) by 𝑎𝑛 =
√1 − 𝑎𝑛−1 , 𝑛 = 1,2,3, … Prove that lim 𝑎𝑛 =
√5−1
.
2
After a few dirty-hand things, we separate 𝑎𝑛 into 𝑎2𝑘 and 𝑎2𝑘+1 . We have
𝑎2𝑘 − 𝑎2𝑘−2 = √1 − 𝑎2𝑘−1 − √1 − 𝑎2𝑘−3 = (𝑎2𝑘−3 − 𝑎2𝑘−1 )/ …
𝑎2𝑘+1 − 𝑎2𝑘−1 = √1 − 𝑎2𝑘 − √1 − 𝑎2𝑘−2 = (𝑎2𝑘−2 − 𝑎2𝑘 )/ …
So with induction we can prove that (𝑎2𝑘 ) , (𝑎2𝑘+1 ) converges. Thus 𝑎𝑛 converges.
We can easily obtain lim 𝑎𝑛 =
√5−1
.
2
Problem: An evil wizard has imprisoned 64 math geeks. The wizard announces, "Tomorrow I will
have you stand in a line, and put a hat on each of your heads. The hat will be colored either white or
black. You will be able to see the hats of everyone in front of you, but you will not be able to see your
hat or the hats of the people behind you. (You are not allowed to tum around.) I will begin by asking
the person at the back of the line to guess his or her hat color. If the guess is correct, that person will
get a cookie. If the guess is wrong, that person will be killed. Then I will ask the next person in line,
and so on. You are only allowed to say the single word 'black' or 'white ' when it is your tum to
speak, and otherwise you are not allowed to communicate with each other while you are standing in
line. Although you will not be able to see the people behind you, you will know (by hearing) if they
have answered correctly or not."
The geeks are allowed to develop a strategy before their ordeal begins. What is the largest number
of geeks that can be guaranteed to survive?
Solution:
Think outside the box.
The geeks can all agree that the first person to be asked will says “white” if the number of black hats
is odd and “black” if it is even. That first person has 50% percent to survive.
The others can count the hats in front of them and will have enough information due to the first
person’s answer to make the right choice.
So the answer is 63.
Problem: Assume an 8 x 8 chess board with the usual coloring. You may repaint all squares (a) of a
row of a column (b) of a 2 x 2 square. The goal is to obtain just one black square. Can you reach the
goal?
Solution:
A white square represents zero and a black square represents number one.
Let S be the sum of all square. At first, 𝑆 = 32.
(a) If we change a column or a row that means 𝑆 will gain or lose 4 unit. That means 𝑆 will be
invariant in module 4. Which means 32 = 1(𝑚𝑜𝑑 4)(absurd).
(b) Same to (a)
Problem: We start with the state (𝑎, 𝑏) where 𝑎 and 𝑏 are positive integers. To this initial state we
apply the following algorithm:
𝒘𝒉𝒊𝒍𝒆 𝑎 > 𝟎, 𝒅𝒐 𝒊𝒇 𝑎 < 𝑏 𝒕𝒉𝒆𝒏 (𝑎, 𝑏) → (2𝑎, 𝑏 − 𝑎) 𝒆𝒍𝒔𝒆 (𝑎 − 𝑏, 2𝑏).
For which starting positions does the algorithm stop? In how many steps does it stop, if it stops?
If it stops? What can you tell about periods and tails
The same questions, when a, b are positive reals.
Soltion:
We have no clue ( besides that the algorithm stops when 𝑎 = 𝑏). So let’s just get our hands dirty.
(5,9) → (10,4) → (6,8) → (12,2) → (10,4) → ⋯ (𝑐𝑦𝑐𝑙𝑖𝑐)
(5,6) → (10,1) → (9,2) → (7,4) → (3,8) → (6,5) → (1,10) → (2,9) → ⋯ (𝑐𝑦𝑐𝑙𝑖𝑐)
(3,9) = (6,6).
Stops.
So if (𝑎, 𝑏) = (𝑡, 3𝑡)(𝑣𝑖𝑐𝑒 𝑣𝑒𝑟𝑠𝑎) then the algorithm stop after one step.
This can easily be proved be the comment that the algorithm stops when 𝑎 = 𝑏. Which means that
2𝑎 = 𝑏 − 𝑎 → 3𝑎 = 𝑏
We obtain the result.
This also works with positive real numbers.
Problem: There are 𝑎 white, 𝑏 black, 𝑐 red chips on a table. In one step, you may choose two chips of
different colors and replace them by a chip of third color. If just one chip will remain at the end, its color
will not depend on the evolution of the game. When can this final state be reached?
Solution:
Yes.
All three numbers a, b, c change their parity in one step. If one of the numbers has different parity from
the other two, it will retain this property to the end. This will be the one which remains.
Problem: There are 𝑎 white, 𝑏 black, 𝑐 red chips on a table. In one step, you may choose two chips of
different colors and and replace each one by a chip of the third color. Find conditions for all chips to
become of the same color. Suppose you have initially 13 white 15 black and 17 red chips. Can all chips
become of the same color? What states can be reached from these numbers?
Solution:
(a, b, c) will be transformed into one of the three triples (a + 2, b − 1, c − 1), (a − 1, b + 2, c − 1), (a − 1, b −
1, c + 2). In each case, I
a − b mod 3 is an invariant. But b − c
0 mod 3 and a − c
0 mod 3 are also invariant. So I
0 mod 3 combined with a + b + c
0 mod 3 is the condition for reaching a monochromatic state.
Problem:There is a positive integer in each square of a rectangular table. In each move, you may double
each number in a row (call this move 1) or subtract 1 from each number of a column(call this move 2).
Prove that you can reach a table of zeros by a sequence of these permitted moves.
Solution:
If there are numbers equal to 1 in the first column, then we double the corresponding rows and subtract
1 from all elements of the first column. This operation discussed the sum of the numbers in the first
column until we get a column of ones, which is changed to a column of zeros by subtracting 1. Then we
go to the next column, etc.
Problem: There is an integer in each square of an 8 × 8 chessboard. In one move, you may choose any 4
× 4 or 3 × 3 square and add 1 to each integer of the chosen square. Can you always get a table with each
entry divisible by (a) 2, (b) 3?
a) Let S be the sum of all the numbers on board except for those that are on 3rd and 6th column. S is
invariant mod 2.
b) Let S be the sum of all the numbers on board expect for those that are on 4th and 8th column. S is
invariant mod 3.
Problem: There is a checker at point (1, 1) of the lattice (x, y) with x, y positive integers. It moves as
follows. At any move it may double one coordinate, or it may subtract the smaller coordinate from the
larger . Which points of the lattice can the checker reach?
Solution:
We know that gcd(𝑥, 𝑦) = gcd(𝑥, 𝑥 − 𝑦) so our invarience is the greatest common divisor of x and y.
And at each move, gcd(𝑥, 𝑦) either invariant or get doubled. So if gcd(𝑥, 𝑦) = 2𝑛 , 𝑛 = 1,2,3 … then (x,y)
can be reached.
Problem: Each term in a sequence 1, 0, 1, 0, 1, 0,...starting with the seventh is the sum of the last 6
terms mod 10. Prove that the sequence ..., 0, 1, 0, 1, 0, 1,... never occurs.
Solution:
Let 𝑆 = 2𝑥𝑖 + 4𝑥𝑖+1 + 6𝑥𝑖+2 + 8𝑥𝑖+3 + 10𝑥𝑖+4 + 12𝑥𝑖+5 , 𝑖 = 1,2,3, …
We obtain that 𝑥𝑖+6 = 𝑥𝑖 + 𝑥𝑖+1 + 𝑥𝑖+2 + 𝑥𝑖+3 + 𝑥𝑖+4 + 𝑥𝑖+5 (𝑚𝑜𝑑 10), 𝑖 = 1,2,3, … (1)
Consider the 𝒊𝑡ℎ number of the sequence, we have:
𝑆𝑖 − 𝑆𝑖−1 =
2𝑥𝑖 + 4𝑥𝑖+1 + 6𝑥𝑖+2 + 8𝑥𝑖+3 + 10𝑥𝑖+4 + 12𝑥𝑖+5 − 2𝑥𝑖−1 − 4𝑥𝑖 − 6𝑥𝑖+1 − 8𝑥𝑖+2 − 10𝑥𝑖+3 − 12𝑥𝑖+4
= 12𝑥𝑖+5 − 2𝑥𝑖+4 − 2𝑥𝑖+3 − 2𝑥𝑖+2 − 2𝑥𝑖+1 − 2𝑥𝑖 − 2𝑥𝑖−1
From (1) we have 𝑆𝑖 = 𝑆𝑖−1 (𝑚𝑜𝑑 10)
That’s our invariance.
Problem: A game for computing gcd(𝑎, 𝑏) and 𝑙𝑐𝑚(𝑎, 𝑏)
We start with 𝑥 = 𝑎, 𝑦 = 𝑏, 𝑢 = 𝑎, 𝑣 = 𝑏 and move as follows:
If 𝑥 < 𝑦 then, set 𝑦 → 𝑦 − 𝑥 and 𝑣 → 𝑣 + 𝑢
If 𝑥 > 𝑦, then set 𝑥 → 𝑥 − 𝑦 and 𝑢 → 𝑢 + 𝑣
The game ends with 𝑥 = 𝑦 = gcd(𝑎, 𝑏) and
𝑢+𝑣
2
= 𝑙𝑐𝑚(𝑎, 𝑏). Show this.
Solution:
The invariences are
i.
ii.
iii.
gcd(𝑥, 𝑦) = gcd(𝑥 − 𝑦, 𝑦) = gcd(𝑥, 𝑦 − 𝑥)
𝑥, 𝑦 > 0
𝑥𝑣 + 𝑢𝑦 = 2𝑎𝑏
Problem: Three integers a, b, c are written on a blackboard. Then one of the integers is erased and
replaced by the sum of the other two diminished by 1. This operation is repeated many times with the
final result 17, 1967, 1983. Could the initial numbers be (a) 2, 2, 2 (b) 3, 3, 3?
Solution:
Initially, if all components are greater than 1, then they will remain greater than 1. Starting with the
second triple the largest component is always the sum of the other two components diminished by 1. If,
after some step, we get (a, b, c) with a ≤ b ≤ c, then c
a + b − 1, and a backward step yields the triple (a, b, b − a + 1). Thus, we can retrace the last state (17,
1967, 1983) uniquely until the next to last step: (17, 1967, 1983) ← (17, 1967, 1951) ← (17, 1935, 1951)
← ··· ← (17, 15, 31) ← (17, 15, 3) ← (13, 15, 3) ← ··· ← (5, 7, 3) ← (5, 3, 3). The preceding triple should
be (1, 3, 3) containing 1, which is impossible. Thus the triple (5, 3, 3) is generated at the first step. We
can get from (3, 3, 3) to (5, 3, 3) in one step, but not from (2, 2, 2).
Problem: