Download FIFTH GRADE First problem Let p be a four

FIFTH GRADE
First problem
Let p be a four-digit number and q be a three-digit number. We call a (p; q) pair CLOCK if
when p is divided by q we obtain a remainder which is equal with the fifth power of the
quotient. How many CLOCK pairs exists?
Solution:
Let p = abcd , q = xyz , D = Q ∙ I + R, R < I , where D is the dividend, I is the divisor,
C is the quotient and R is the remainder............................................................................. 1p
abcd = xyz ∙ Q + R; R = Q5; abcd = xyz ∙Q + Q5  Q ≠ 0 and Q5 < xyz ≤ 999< 1024 = 210 =
(22)5 = 45 ............................................................................................................................ 1p
Q5 < 45 and Q ≠ 0  Q  {1, 2, 3} ..................................................................................0,5p
1) If Q = 1
abcd = xyz ∙1 + 15 = xyz + 1  xyz + 1 ≥1000 | -1  xyz ≥ 999
Because x ≤ 999  xyz = 999  q = 999 and p = 1000  (p, q) = (999, 1000) =>
 1 CLOCK pair .......................................................................................................... 1p
2) If Q = 2
abcd = xyz ∙2 + 25 = xyz ∙2 + 32
1000 ≤ abcd ≤ 9999  1000 ≤ xyz ∙2 + 32 ≤ 9999 | -32  968 ≤ 2∙ xyz ≤ 9967 | :2 
xyz  {484, 485, ..., 999}  (999 – 483) = 516  516 numbers q = xyz  516 CLOCK
pairs ......................................................................................................................... 1,5p
3) If Q = 3
abcd = xyz ∙3 + 35 = xyz ∙3 + 243
1000 ≤ abcd ≤ 9999  1000 ≤ xyz ∙3 + 243 ≤ 9999 | -243  757 ≤ 3∙ xyz ≤ 9756 | :3 
xyz  {253, 254, ..., 999}  (999 – 252) = 747 numbers q = xyz  747 CLOCK pairs 1,5p
So the number of CLOCK pairs is: 1 +516 + 747 = 1264 .....................................0,5p
FIFTH GRADE
Second problem
Let
a) Show that a is a perfect square.
b) Find the tens digit of
.
SOLUTION
a)
………………………………………………………….. 2p
b)
…………………………………………………………………………………………………………… 1p
Let ZU(x) be the last two digits of x.
……………………..……………………………………………………………………………………… 2p
Because
So the tens digit of
we have
………………………………………. 1p
is 4................................................................................ 1p
FIFTH GRADE
Third problem
At „Clock-Tower School”, instead of the well-known austrian clock has been
installed, initially, an electronic,”singing”, clock: „each time the digit 2 appears on the
clock’s screen, it sings while the digit remains on the screen”. Knowing that the clock shows
the time from 00:00 to 23:59, how long does the daily concert take?
SOLUTION
The clock sings for 4 hours: while it shows the time from 20:00 to 24:00……………...2p
The clock sings an hour: each time it shows 02:xy; 12:xy …………………………….2p
The clock sings 15 minutes each time it shows 00:xy, 01:xy, 03:xy, …, 09:xy, 10:xy, 11:xy;
13:xy;14:xy; 15:xy, 16:xy ; 17:xy ; 18:xy and 19:xy . …………………………….…….2p
The concert takes 10 hours and 15 minutes. ………………………………………….1p
FIFTH GRADE
Fourth problem
In this classroom are 25 students at the contest. After the worksheets are checked, at
this problem, each student has a score that represents a digit not greater than 7.
a) Show that in the classroom are, at least, four students with the same score.
b) Show that exists, in the classroom, at least two students which have the same
number of friends in that group of 25 students (if A is friend with B, then B is
also friend with A).
SOLUTION
a) There are 8 possible scores: 0, 1, 2, …, 7 ……………………………………..….. 2p
Using pigeonhole principle, there are 4 students with the same score…………….. 2p
b) We imagine 25 boxes, representing the number of friends which any student can
have.
……………………
0
1
2
3
24
If there is a student which has no friends, then there aren’t any students with 24
friends. …………………………………………………………………………………1p
If there is a student with 24 friends, then there aren’t any students with zero friends……1p
Applying pigeonhole principle for 25 students and 24 possibilities (regarding the
number of friends) , we obtain that there are two students with the same number of friends. 1p
SIXTH GRADE
Second problem
The ABC triangle is right-angled isosceles, with m(Â) = 900. Let D (AC) and E in
the opened half-space determined by line AC and point B, such that EC  AC and
[EC][AD]. Prove that:
a) [AE][BD];
b) Angles DBE and AEB are complementary;
c) Angles DBE and AEB can’t be congruent.
SOLUTION
a) ABDCAE (C.C.)  AE = BD
……………………………………………………….2p
b)
B  EAC and because m(B) + m(D) = 900  m(D) + m(EAC) = 900 
m(I) = 900  m(DBE) + m(AEB) = 900 …………………………………….….3p
c) We prove by reductio ad impossibile.
We assume that m(DBE)  m(AEB) = 450  IB = IE, but DB = AE  ID = IA 
 m(D)  m(IAD)  AC = CE = AD  D  C (false because D (AC)) …..…..2p
SIXTH GRADE
Third problem
For p a prime number, p > 3 we consider the set
1.
Ap  
k  p2
| k  N , k  2009
24

Find the cardinality of the set
SOLUTION
For p – prime, p>3  p2 – 1 = M24 (1) …………………………................................3p
Because p – prime  p  M24+1, M24+5, M24+7, M24+11, M24+13, M24+17, M24+19,
M24+23
If
p = M24+1  p2 – 1 = (M24 + 1)2 – 1 = M24
p = M24+5  p2 – 1 = (M24 + 5)2 – 1 = M24 + 25 – 1 = M24
p = M24+7  p2 – 1 = (M24 + 7)2 – 1 = M24 + 49 – 1 = M24
p = M24+11  p2 – 1 = (M24 + 11)2 – 1 = M24 + 120 = M24
…………………………………………………………………
p = M24+23  p2 – 1 = (M24 + 23)2 – 1 = M24 + 528 = M24
(1) 
………………… .................. 1p
But

if (k + 1) 24 (2) …………………………............... 1p
, 2010 =
and according to (2)  k + 1
.......................................................................................1p
Because 1
So the cardinality of
Ap
…………………………..……….........................1p
SIXTH GRADE
Fourth problem
On a board are written the numbers 1, 2, 3, ..., 2009. We erase any three number and,
instead of them, we write their sum. We repeat the procedure until there is jut one number
written on the board.
a) The last number is odd or even? Justify!
b) How many times must the procedure must be repeated?
Solution
b)
At a procedure we “lose” 2 numbers.
2009 = 2·1004 + 1 so the procedure must be repeated 1004 times…………………. 2p
a)
At any procedure we may erase:
Where i is an odd number and p is an even number.
(p,p,p)  p+p+p = p  number of even numbers remains the same
(p,p,i)  p + p + i = i  number of even numbers remains the same
(p, i, i)  p + i + i = p  we lose 2 even numbers
(i, i, i)  i + i + i = i  we lose 2 even numbers ……………….…………2p
So the number of even numbers either remains the same, or decreases with 2. That means that
its parity remains the same …………………………………………………………..…….1p
But in the string 1, 2, 3, ..., 2009 we have 1005 even numbers………………………….1p
So, after we repeat the procedure 1004 times, the last number is even (if it would be odd, then
we have zero even numbers (false))……………………………………………………...1p
I suggest you another solution for b). After each procedure the sum of the written
numbers remains the same. Therefore the last number on the board is the sum of all the
numbers written on the board, that means that the last number is 1+2+3+..+2009=
=2019045 so the last number is even.
SEVENTH GRADE
First problem
Among the non-empty sets M a  { x  Z x 
an  100
, n  N}, a  Z , find the ones with
n2
the minimum cardinality.
SOLUTION
Let Card M be the cardinality of M
Because
an1  100 an2  100

 n1  n2 , card M a is equal with the number of natural numbers n
n1  2
n2  2
for which n  2 an  100 …………………………………………………………………..… 2p
We have n  2 an  100  n  2 an  2a   an  100    n  2 2a  100  n  2  D2a 100  2,  … 1p
Then Card M a is minimum if 2a 100 is prime ………………………………………. 1p
Because 2a  100  2a  50  , Card M a is minimum for a   49, 51 ……………………….. 1p
Then 2a  100  1 and n  2  2  n  0 …………………………………….....…….... 1p
So M a  50 , set with only one element ………………………………………….… 1p
SEVENTH GRADE
Second problem
How many right-angled triangles with the sides’ lengths natural numbers exists, such that their
area is 2008?
SOLUTION
Let x, y and z be the sides’ lengths of the triangle.
We have x 2  y 2  z 2 (*)……………………………………………… 1p
Because x 2 , y 2 , z 2  M 4 , M 4  1, according to (*) we have x  2 or y 2 ……. 1,5p
Because x 2 , y 2 , z 2  M 3 , M 3  1, according to (*) we have x 3 or y 3 ………. 1,5p
Conclusion: xy6 ……………………….………………………………… 1p
The area of the triangle is A 
xy
 2008  xy  4016 …………………. 1p
2
Because 6 is not a divisor of 4016 we obtain that there aren’t any right-angled
triangles which holds the hypothesis …………………………………….. 1p
SEVENTH GRADE
Third problem
Let ABC be a triangle and P a point inside the triangle. We construct the parallelograms
APCB’, BPCA’ and BPAC’.
a) Prove that the ABC triangle is congruent with A’B’C’ triangle;
b) Show that the linest AA’, BB’ and CC’ are intersected (have a joint point)
SOLUTION
If the points O1, O2, O3 are the middles of the segments [BC], [AC], [AB]  points A’, B’, C’
are the reflections of P over O1, O2 and O3………………………………………………. 1p
Because O2O3 is central median in PB’C’  B’C’ = 2 O1O2
Because O2O3 is central median in ABC BC = 2 O1O2 …………………………... 1p
So B’C’ = BC; analogous AB = A’B’ and AC = A’C’ …………………………...……. 1p
According to S.S.S. criteri,. ABC  A’B’C’ ……………………………………..……. 1p
Because AB’ PC and AB’ = PC and (PC BA’ si PC = BA’  AB’ BA’ and AB’ = BA’
 AB’A’B parallelogram  BB’ passes through the middle of [AA’] …………………. 1p
Analogous CC’ passes through the middle of [AA’]……………………………..……. 1p
So AA’, BB’, CC’ are intersected (have a common point) ………………………………. 1p
SEVENTH GRADE
Fourth problem
Let  ABC an acute-angled triangle and points M  ( AB), N  ( AC ) such that
MB NA 1

 .
AB AC 3
If ( MB  MS )( NC  NS )  0 for any point S which lies on the segment (BC), prove that the
ABC triangle is isosceles with AB=AC .
BAREM
If AB>AC and AD  BC, D  BC  BD  DC  3x  3 y  x  y …………1p
Let P be the symmetrical of N over A and Q the middle of [MA]
Let PP1  BC , QQ1  BC  P1D  y  x  Q1D  P1  (Q1D)
We choose S  (Q1P1 ) …………………………………………………………………….. 1p
In BQ1Q , right-angled, MQ1  MB  MS  MQ1  MB ……………………………... 1p
In CP1P right-angled, NP1  NC  NS  NP1  NC ………………………………….. 1p
So MB  MS   0 and NC  NS   0  MB  MS NC  NS   0 false ………….1p
Analogous for AB<AC ………………………………………………………... 1p
Conclusion: AB  AC  ABC isosceles …………………………………… 1p
EIGHTH
First problem
Let x, y, z be real numbers such that x 2  y 2  z 2  2( xy  yz  zx) .
a) Prove that xy  0, yz  0, zx  0.
b) Can x, y, z be the lengths of the sides of a triangle? Justify!
SOLUTION
a) x 2  y 2  z 2  2( xy  yz  zx)  ( x 2  y 2  2 xy)  z 2  2 z ( x  y)  0 ……………….…..1p
So ( x  y) 2  [ z 2  2 z ( x  y)  ( x  y) 2 ]  ( x  y) 2  0 …………………………………… 1p
So
( x  y)2  ( x  y) 2  ( x  y  z ) 2  4 xy  ( x  y  z ) 2  0 ……………..…….. 1p
Analogous for yz  0, zx  0 ……………………………………………………..………… 1p
b) From the symmetry of the relation, between x, y and z we can assume x  y  z and
according to 1a) we have:
( x  y)2  ( x  y  z )2  ( x  y)2  ( x  y  z  x  y)2 2( x  y  z )( x  y)  (2 x  z )2 …... 2p
So x  y  2 x  z  y  z  x and then x, y, z can’t be the lengths of the sides of a triangle 1p
EIGHTH GRADE
Second problem
Find the minimum of the expression
E
y2
x2
z2
, if x, y, z > 0 and


x y yz zx
xy  yz  zx  1
SOLUTION
E  x
xy
yz
zx
xy
yz
zx
 y
z
 x yz


x y
yz
zx
x y yz zx
…………………….1p
We use x  y  2 xy and the analogues ……………………………………..1p
Then
xy
xy
and the analogues………………………………………..1p

x  y 2 xy
 xy
yz
zx 
1


 x  y  z  ……………………..1p

2
2 
2
 2
We use the inequality a 2  b 2  c 2  ab  ac  bc for a  x , b  y , c  z ,
We obtain E  x  y  z  
We obtain x  y  z  xy  yz  xz  1 …………………………………….2p
1
2
Then E  , minimum E 
1
1
being obtained for x  y  z  …………….1p
2
3
EIGHTH GRADE
Third problem
Let ABCDA’B’C’D’ be a rectangular parallelepiped. Let E and F be the
middles of [BC] and [CD], and P and Q the projections of point A on A’B and
A’D. If AB =3√2 cm, AA’= 8 cm, and A’F  DE prove that:
a) DE  AF and ABCD is a square;
b) A’C  AP;
c) Calculate the distance from the point A’ to the plane (APQ).
SOLUTION
a) DE  A’F, DE  AA’=> DE  (A’AF) ………………………………………. 0,5 p
DE  (A’AF) => DE  AF …………………………………………………… 0,5 p
DE  AF =>  DAF   CDE
 DAF   CDE => tg DAF = tg CDE ………………………………………. 0,5 p
tg DAF = tg CDE => [AD]  [DC]
ABCD rectangular and [AD]  [DC] => ABCD square……………………….. 0,5 p
b) BC  AB, BC  AA’ => BC  (A’AB)
BC  (A’AB) => BC  AP ……………………………………………………. 0,5 p
AP  A’B, AP  BC => AP  (A’BC)
AP  (A’BC) => AP  A’C ……………………………………………………. 0,5 p
c) A’C  AQ ………………………………………………………………………. 1 p
A’C  AP, A’C  AQ => A’C  (APQ) ………………………………………..0,5 p
Let {R} = A’C  (APQ) => A’R = d [A’, (PAQ)] ……………………………… 1 p
R  (PAQ), A  (PAQ) => AR  (PAQ)
A’C  (PAQ) => A’C  AR …………………………………………………….0,5 p
32
AC = 6 cm, A’C = 10cm, AR = d =
cm = 6,4 cm ……………………………1 p
5
EIGHTH GRADE
Fourth problem
In a Cartesian coordinate system xOy there are considered the points
with
Determine the number of triangles which have as vertexes three of the given
points.
SOLUTION
Representation of the points on the Cartesian coordinate system ....... 1p
The 16 points determines
A14
A13
A12
A11
A24
A23
A22
A21
A34
A33
A32
A31
A44
A43
A42
A41
triplets formed from such as points …………. 1p
To find the number of triangles we eliminate the triplets formed by collinear points. ……. 1p
On each line there are 4 such as collinear triplets……………….…………………………. 1p
Analogous for each column and for the two of the diagonals of the formed square..……... 1p
There are still 4 triplets of collinear points on the “small diagonals”-example
.. 1p
Conclusion: There can be formed 560 - (16 + 16 + 8 + 4) = 516 triangles…………………. 1p