Problem: [BRITAIN 1984/4]
x^2 - |_x^2_| = ( x - |_x_| )^2.
Count the number of solutions 1<=x<=n.
Solution:
Let x = m + d, where m = |_x_| and d = x - |_x_| which is 0<=d<1.
Plugging x = m + d into x^2 - |_x^2_| = ( x - |_x_| )^2 we get
(m + d)^2 - |_(m + d)^2_| = d^2 and after simplification
2md = |_2md + d^2_|. Therefore 2md is an integer k.
Case m = 0:
Then every 0<=d<1 is possible. Therefore all 0 <= x < 1 are valid.
Case m > 0:
Then 2md = k. That is d = k/(2m). As 0<=d<1 we have 0 <= k/(2m) < 1.
So possible values are k in {0, 1, 2, ..., 2m-1 }. The corresponding
x values are m, m + 1/(2m), m + 2/(2m), ..., m + (2m-1)/(2m).
Let L(a,b) = { a<=x** | n/a - n/b | < 1 => | 1/a - 1/b | < 1/n
Since this is true for all n, 1/a = 1/b.
Problem: [Jan87, EdM P964]
Find all real pairs (a,b) such that for all positive integers n
a |_b n_| = b |_a n_|.
Solution: [BrF88, EdM P964]
It is clear that a |_b n_| = b |_a n_| for all natural numbers n
if either ab = 0, or if a=b, or if a and b are both integers.
We show that this condition is also necessary. Thus we suppose
a |_b n_| = b |_a n_| for all n, ab != 0, and a != b. Then,
taking n = 1, we have bm = ak, where m = |_a_| and k = |_b_|.
Thus 2m <= 2a < 2m + 2, so that either 2m <= 2a < 2m + 1 or
2m + 1 <= 2a < 2m + 2. Similarly, either 2k <= 2b < 2k + 1 or
2k + 1 <= 2b < 2k + 2. Taking n = 2 we conclude that in fact
|_2a_| = 2m and |_2b_| = 2k. (Each of the other possibilities
contradicts one of our hypotheses. E. g. assume |_2a_| = 2m + 1
and |_2b_| = 2k + 1 then b(2m+1) = a(2k+1) and as bm = ak we
have the contradiction b = a.) Repeating this argument we
inductively establish that |_2^r a_| = 2^r m and |_2^r b_| = 2^r k,
so that m <= a < m + 1/2^r and k <= b < k + 1/2^r for all
natural numbers r. Thus a = m and b = k, and our asertion is proven.
Problem: (Sillke)
Find all real pairs (x,y) such that
x |_y_| = y |_x_|.
Problem: Komal F3232
Let b(n) denote the minimum value of expression k + n/k,
where k is a positive integer.
Prove that for any natural number n,
|_b(n)_| = |_sqrt(4n+1)_|
Problem: Komal Gy2047
Solve the equation |_sqrt(|_x_|)_| = |_sqrt(sqrt(x))_|
on the set of real numbers.
Solution:
First observe that |_sqrt(|_x_|)_| = |_sqrt(x)_|.
Second set x = z^4 and get the nicer looking equation |_z^2_| = |_z_|.
Case z < 0: |_z_| <= z < 0 <= |_z^2_|. No solution.
Case z >= tau = (sqrt(5) + 1)/2: As z^2 >= tau*z >= z + 1 we get
|_z^2_| >= |_z + 1_| = |_z_| + 1 > |_z_|. No solution.
Case 0 <= z < tau: Analyze the range of |_z_| which is {0, 1}.
So there are only two cases left.
Case |_z^2_| = 0 = |_z_|: solutions for 0 <= z < 1.
Case |_z^2_| = 1 = |_z_|: solutions for 1 <= z < sqrt(2).
Collecting the results we get the solution 0 <= x < 4.
Problem: Ouardini Problem 2-10
Solve the equation |_x^(1/2)_| = |_x^(1/3)_|
on the set of real numbers.
Solution:
First set x = z^6 and get the nicer looking equation |_z^3_| = |_z^2_|.
Case z < 0: |_z^3_| <= z^3 < 0 <= |_z^2_|. No solution.
Case z >= 3/2: As z^3 >= 3/2 z^2 >= z^2 + 1 we get
|_z^3_| >= |_z^2 + 1_| = |_z^2_| + 1 > |_z^2_|. No solution.
Case 0 <= z < 3/2: Analyze the range of |_z^2_| which is {0, 1, 2}.
So there are only three cases left.
Case |_z^3_| = 0 = |_z^2_|: solutions for 0 <= z < 1.
Case |_z^3_| = 1 = |_z^2_|: solutions for 1 <= z < 2^(1/3).
Case |_z^3_| = 2 = |_z^2_|: solutions for 2^(1/2) <= z < 3^(1/3).
Collecting the results we get the solution 0 <= x < 2 and 8 <= x < 9.
Problem: SSM 3696
Solve the equation |_sqrt(x)_| = |_x/k_|
on the set of real numbers, where k is an integer.
Problem: 20th MMO 9.1.2
Solve the equation x^3 - |_x_| = 3
on the set of real numbers.
Solution:
Rearranging the equation gives x^3 = 3 + |_x_|.
Therefore the right hand side is an integer.
Case x>=2: x^2 >= 4 => x^3 >= 4x >= 3 + x >= 3 + |_x_|. No solution.
Case 2>x>=1: x^3 = 3 + |_x_| = 4 => x = 4^(1/3) = 1.587401
Case 1>x>=0: x^3 = 3 + |_x_| = 3 => x > 1. No solution.
Case 0>x>=-1: x^3 = 3 + |_x_| = 2 => x > 1. No solution.
Case x < -1: x^2 >= 1 => x^3 < x < 2 + x < 3 + |_x_|. No solution.
Problem: Ouardini Problem 2-12
Solve the equation 1 + sin^2(x) + sin^2(x - |_sqrt(x)_|) = cos(x)
on the set of real numbers.
Solution:
This equation looks difficult by the obeservation
1 + sin^2(x) + sin^2(x - |_sqrt(x)_|) >= 1 >= cos(x)
makes it rather simple. So we are looking for the common solution of
the three equations: cos(x) = 1, sin(x) = 0, and sin(x - |_sqrt(x)_|) = 0.
The first has the solutions 2Pi*Z, the second Pi*Z. So we have only to
check x = x_n = 2Pi*n with n in Z. But x_m = x_n - k for an integer k
has only one solution k = 0 as Pi is irrational. As k=0 means x=0
we have only one solution for the original equation.
Problem: Komal C596
Solve the equation |_1/(1-x)_| = |_1/(1.5-x)_| on the set of
real numbers.
Solution:
case x < 1:
0 < 1-x < 1.5-x
k <= 1/(1.5-x) < 1/(1-x) < k+1
k=0: x<0
case x > 1.5:
1-x < 1.5-x < 0
k=-1: x>=2.5
Problem: Komal C605 (Dec 2000)
Solve the equation
1 1
----- + --------- = x
|_x_| x - |_x_|
Solution:
Multiply the equation by |_x_| and x - |_x_|. This gives
x = x |_x_| (x - |_x_|)
Case x=0: This don't solve the original equation.
Case x!=0: Cancel x
1 = |_x_| (x - |_x_|) => x = |_x_| + 1/|_x_|.
So for each integer n>=2 we get a valid
solution x = n + 1/n.
Problem:
Determine the number of real solutions a of the equation
|_a/2_| + |_a/3_| + |_a/5_| = a.
Solution:
There are 30 solutions.
Since |_a/2_|, |_a/3_|, and |_a/5_| are integers, so is a.
Now write a = 30p + q for integers p and q, 0 <= q < 30.
Then
|_a/2_| + |_a/3_| + |_a/5_| = a
<=> 31p + |_q/2_| + |_q/3_| + |_q/5_| = 30p + q
<=> p = q - |_q/2_| - |_q/3_| - |_q/5_|.
Thus, for each value of q, there is exactly one value of p
(and one value of a) satisfying the equation. Since q can equal
any of thirty values, there are exactly 30 solutions as claimed.
reference:
- 37th International Mathematical Olympiad (July 1996) Mumbai India
Problems proposed by not used.
reprint: Crux Mathematicorum 23:8 (Dec 1997) 452
problem 18:
Find all positive integers a and b for which
|_a^2/b_| + |_b^2/a_| = |_(a^2+b^2)/(ab)_| + ab.
- 20th Moskau Mathematical Olympiad (1957)
Class 9 Round 1 Problem 2.
Solve the equation x^3 - |_x_| = 3
on the set of real numbers.
- Britain 1984/4:
British Mathematical Olympiad
- Komal Gy2006
- Komal Gy2047
- Komal F3232
May 1998
- Komal C596
October 2000
- SSM 3696
Douglas E. Scott
- W. Janous;
Problem 964
Elemente der Mathematik 42:3 (1987) 80 proposal by W. Janous
Elemente der Mathematik 43:? (1988) 93 solved by J. L. Brenner, L. L. Foster
- Abderrahim Ouardini;
Mathématiques de Compétition, 112 problèmes corrigés,
Ed. Ellipse, 2000, Paris, ISDN 2-7298-0125-1
- Arthur Engel;
Mathematische Olympiadeaufgaben aus der UDSSR,
Klett Verlag, Stuttgart, 1965, ISBN 3-12-710300-X
- Arthur Engel;
Problem-solving strategies,
Problem Books in Mathematics. New York, NY: Springer. x, 403 p. (1998)
ISBN 0-387-98219-1/hbk
Zbl 887.00002
Chap 14.6 Integer Function
--
http://www.mathematik.uni-bielefeld.de/~sillke/
mailto:Torsten.Sillke@uni-bielefeld.de
**