Problem 1. Determine
a) the smallest number
b) the biggest number
n ≥ 3 of non-negative integers x1 , x2 , ... , xn , having the sum 2011 and satisfying:
x1 ≤ | x2 − x3 | , x2 ≤ | x3 − x4 | , ... , xn−2 ≤ | xn−1 − xn | , xn−1 ≤ | xn − x1 | and
xn ≤ | x1 − x2 | .
Dorel Mihe’t
Solution:
Let x1 , x2 , ... , xn non-negative integers satisfying the conditions from the statement. Let us imagine them as being arranged on a circle, in this order. We denote
with M the biggest of the numbers x1 , ... , xn . Without loss of generality, we may
suppose that x1 = M . Obviously M 6= 0. From x1 ≤ | x2 − x3 | and the choice of
x1 it follows that {x2 , x3 } = {M, 0}. Therefore any M is followed by an M and a
0 on the circle, either in the succession M − M − 0 or the succession M − 0 − M .
By induction after the last M known on the circle, we find that on the circle there
are only the numbers 0 and M . Moreover
• there are no the neighboring zeros (or else the number preceding them would
also be a 0 and, inductively, all the numbers on the circle would be zeros, which
contradicts the fact that their sum is 2011)
• there are no three consecutive M -s.
Since 2011 is prime and x1 + x2 + ... + xn = kM , where k is the number of the M
-s, we find that M = 1 and k = 2011.
a) The smallest value of n is obtained when the number of zeros is the smallest
possible. Out of three consecutive numbers on the circle, at least one is 0, therefore we have at least 1006 zeros, which means at least 3017 numbers. A possible
configuration fulfilling the conditions of the statement is x3k = 0, k = 1, 2, ..., 1005,
x3017 = 0, and the rest of the xj = 1.
b) The biggest n is obtained when the number of zeros is the largest possible. Since
there can be no neighboring zeros, we have at most 2011 zeros. This happens if we
place alternatively 0 and 1 on the circle, configuration that satisfies the conditions
from the statement. Thus, the biggest value of n is 4022.
Problem 2. We consider an n × n (n ∈ N, n ≥ 2) square divided into n2 unit
squares. Determine all the values of k ∈ N for which we can write a real number
in each of the unit squares such that the sum of the n2 numbers is a positive number, while the sum of the numbers from the unit squares of any k × k square is
a negative number.
Solution:
We will prove that the desired numbers k are those that are not factors of n.
If k | n that we can tile the n × n square with k × k squares and the total sum
1
should simultaneously be positive and negative.
If k 6 | n then n = kq + r, where 0 < r < k. We fill the unit squares with a (to be
chosen conveniently later on) in the positions (ik, jk) with i, j = 1, ..., q and with
1 in the other positions. Every k × k square contains exactly one unit square of
the form (ik, jk), therefore the sum in every k × k square is a + k 2 − 1. The total
2
2
2
sum
q . We will choose a arbitrarily from the non-empty interval
is 2q a + n −
n
1 − 2 , 1 − k2 .
q
Remark: For the case k 6 | n there are many other ways of choosing the numbers
from the unit squares. Another choice is to fill all the unit squares of the columns
jk, j = 1, ..., q, with a convenient a and the other ones with 1.
Problem 3. a) Prove that if the sum of the non-zero digits a1 , a2 , ... , an is a multiple of 27, then it is possible to permute these digits in order to obtain an n-digit
number that is a multiple of 27.
b) Prove that if the non-zero digits a1 , a2 , ... , an have the property that every
n-digit number obtained by permuting these digits is a multiple of 27, then the
sum of these digits is a multiple of 27.
Andrei Eckstein
Solution:
.
a) Obviously n ≥ 3. If n = 3 then a1 = a2 = a3 = 9 and 999 .. 27. Suppose n ≥ 4.
Having the sum of its digits a multiple of 9, each of the numbers formed with the
digits a1 , a2 , ... , an is a multiple of 9, therefore division with 27 will yield one of
the remainders 0, 9 or 18.
• If all the digits give the same remainder r when divided by 3:
N
a1 a2 an
− if r = 0 then we choose N = a1 a2 ...an . We have that
=
...
has the
3
3 3
3
N
sum of its digits a multiple of 9, therefore the number
is a multiple of 9, hence
3
N is a multiple of 27.
.
− if r ∈ {1, 2}, then from a1 + a2 + ... + an .. 27 and ai ≡ 0 (mod 3), ∀i it follows
that nr ≡ 0 (mod 3), hence n is a multiple of 3.
Then:
a1 a2 ...an − (a1 + a2 + ... + an ) = 9 a1 a1 ...a1 + a2 a2 ...a2 + ... + an−1 .
| {z } | {z }
n−1 cifre
n−2 cifre
{z
}
|
M
Then M ≡ (n − 1)a1 + (n − 2)a2 + ... + an−1 ≡ r (n − 1) + (n − 2) + ... + 2 + 1 ≡
n(n − 1)
r·
≡ 0 (mod 3). Therefore, in this case, a1 a2 ...an and a1 + a2 + ... + an
2
give the same remainder at the division by 27.
• If not all of the digits a1 , a2 , ... , an have the same reminder when divided by
2
3, then we choose three of them such that their sum is not a multiple of 3. Let
a, b, c be these three digits. Then the numbers x = ...abc, y = ...bca and z = ...cab
give different reminders at the division by 27. Indeed, x − y = abc − bca =
108a − 81b − 9(a + b + c) = M27 − 9(a + b + c) 6= M27 . Since the possible reminders
at the division by 27 are only 0, 9 and 18, it follows that (exactly) one of the
numbers x, y, z is a multiple of 27.
b) The differece between two numbers obtained one from the other by swapping
two neighboring digits a and b is 9(a − b) · 10k and needs to be a multiple of 27. We
obtain that each of the digits a1 , ..., an has to give the same remainder when divided by 3. From here one continues as by a) in order to show that, in this case, the
numbers a1 a2 ...an and a1 +a2 +...+an give the same remainder when divided by 27.
b of the acute triangle ABC is 60◦ , and
Problem 4. The measure of the angle A
HI = HB, where I and H are the incenter and the orthocenter of the triangle
b
ABC. Find the measure of the angle B.
Solution:
We have m(BIC) ≡ m(BHC) = 120◦ , hence B, H, I, C are situated on a
circle;
1
If m(B) > 60◦ then m(HBI) = m(ABI) − m(ABH) = · m(B) − 30◦ .
2
But m(HBI) = m(HIB) = m(HCB) = 90◦ − m(B), hence m(B) = 80◦ .
It is easy to see that the case m(B) ≤ 60◦ is not possible.
3