Vista previa del material en texto
Mathematical Olympiads Group
P.O. Box 3, Ariel 44837, ISRAEL
1. A computer puts two queens on the chessboard randomly (not in
the same cell). What is the probability that the queens threaten
each other?
Answer. 13/36
Solution. Let’s assume we choose the place for the white queen
first and then for the black queen. Then there are 64 possible
places for the white queen and 63 possible places for the black
queen.
The probability p can be decomposed into 4 summands:
p = p– + p| + p/ + p\
which are probabilities of queens being in the same line, column,
diagonal orthogonal or diagonal parallel to the main diagonal
respectively.
Those events are mutually exclusive.
p– = p| = 7/63 = 1/9.
So it remains to compute p/ which is the same as p\.
There are 17 diagonals orthogonal to the main diagonal:
of 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1 cells respectively.
The probability of the white queen to get into specific diagonal of
length k is k / 64 and the probability of black queen to get into the
same diagonal is
(k – 1)/63.
To get p/ we should multiply these numbers and sum over all
diagonals.
1 0 1 1 2 2 3 ... 6 7 7 8 6 7 5 6 ... 1 2 0 1
64 63/
p
To compute this, we shall use a nice identity
1 2
1 2 2 3 ... 1
3
k k k
k k
which can be easily obtained by induction. Therefore
1 7 8 9 6 7 8 7 8 5 53 2
64 63 3 3 64 63 8 9 72/
p
1 1 5 5 2 5 8 5 13
9 9 72 72 9 36 36 36| \ + + +
/p p p p p
2. Solve the equation n5 + n4 = 7m – 1 in integer numbers.
Mathematical Olympiads Group
P.O. Box 3, Ariel 44837, ISRAEL
Answer. m = 0, n = 0 or -1
Solution. Surely m ≥ 0, otherwise non-integer number would be
integer.
7m = n5 + n4 + 1 = n5 + n4 + n3 + n2 + n + 1 – (n3 + n2 + n) =
= (n2 + n + 1)(n3 – n + 1)
Each common divisor of A = n3 – n + 1 and B = n2 + n + 1
should also be a divisor of C = (n – 1)B – A = n3 – 1 – A = n + 2
and of
D = B – (n – 1)C = B – (n2 + n – 2) = 3.
But 7m = AB, so A and B are both powers of 7. On the other
hand each their common divisor is divisor of 3, so A or B
should be 1.
So either A = 1, then n3 – n + 1 = 1, n(n – 1)(n + 1) = n3 – n = 0
that is
n = 0, 1 or -1 , of course 1 doesn’t fit,
or B = 1, then n2 + n + 1 = 1, and again n = 0 or -1
3. Prove that if X is a matrix such that trace(X) = 0, then there exist
two matrices A, B such that X = AB – BA.
(All matrices in this question may have complex coefficients).
Solution. We can transform the matrix X into its Jordan’s form, and
do the search for A and B in the Jordan basis of X.
So, we may assume that X will have nonzero only on two diagonals:
the main diagonal and the diagonal above the main.
The elements of X on the main diagonal will be called x1, x2, … , xn
and the elements on the above-the-main diagonal will be called y1,
y2, …, yn-1.
x1 + x2 + … + xn = trace(X) = 0
Let A be the matrix which has 1’s on above-the-main diagonal.
Mathematical Olympiads Group
P.O. Box 3, Ariel 44837, ISRAEL
Then it is easy to see that for any matrix B:
AB is the same as B shifted up by one row (first row delete, last row
filled by zeroes),
BA is the same as B shifted to the right by one column (first column
deleted, last column filled by zeroes).
Assume B has elements b1, b2, …, bn on the main diagonal (in this
order),elements c1, c2, …, cn-1 on below-the-main diagonal, and
zeroes elsewhere. Then AB – BA will have non-zero elements only
on the main and on above-the-main diagonal, namely b2 – b1, b3 –
b2, …, bn – bn-1,on above-the-main diagonal, and c1, c2 – c1, c3 – c2,
… , cn – cn-1, – cn on the main diagonal. It is easy to see that any
sequence of n numbers having 0 sum can be presented in the form
c1, c2 – c1, c3 – c2, … , cn – cn-1, – cn, and that any sequence of
numbers can be presented as b2 – b1, …, bn – bn-1.
4. Compute the area of { (x,y) | x ≤ 0 , y ≤ 0 , ex + ey ≥ 1 }.
Solution. Denote s = ex, t = ey.
If we change (x,y) by (z,t) then the domain of integration becomes
much simpler: a triangle { s ≤ 1 , t ≤ 1 , s + t ≥ 1 }
1 1 1 1 1
0 1 0 1 0
1 1 12 3 4 2 3 2 3
0 0 0
2
2
ln ln ln 1
... 1 ... 1 ...
2 3 4 2 3 4 2 3 4
1 1 11 ... ...
4 9 6
t t
dtds ds ds dt dtdxdy d t d s dt t
ts s s t t
t t t dt t t t t t tt dt dt
t
n
