кцгда фс

417
SKOLIAD
No. 128
Lily Yen and Mogens Hansen
Please send your solutions to problems in this Skoliad by 1 May 2011. A
opy of CRUX with Mayhem will be sent to one pre-university reader who
sends in solutions before the deadline. The deision of the editors is nal.
Our ontest this month is the Mathematis Assoiation of Quebe Contest, Seondary level, 2010. Our thanks go to Mar Bergeron, Cegep
de
Sainte-Foy, Quebe, for providing us with this ontest and for permission
to publish it.
Mathematis Assoiation of Quebe Contest, 2010
Seondary level
3 hours allowed
. An alphameti is a small mathematial puzzle onsisting of an equation
in whih the digits have been replaed by letters. The task is to identify the
value of eah letter in suh a way that the equation omes out true. Dierent letters have dierent values, dierent digits are represented by dierent
letters, and no number begins with a zero. For example, the alphameti
PAPA + PAPA = MAMAN has the solution P = 7, A = 5, M = 1, and
N = 0, yielding 7575 + 7575 = 15150.
Find the solution to this \reversing" alphameti:
1
NOMBRE ×
3
5
=
. Find all polynomials of the form p(x)
integers.
2
ERBMON .
= x3 + mx + 6
whose roots are
units √from the centre of a circle of
. A line is loated at 22 units from the entre of a
irle of radius 1, separating it into two parts. What
is the area of the smaller part?
3
. The gure shows a map of a ity. In how many
ways an you travel along the roads of the ity from
point A to point B if you an only travel east and
south (right and down in the gure)?
4
A ..s..................................
... ... ... ...
...........................................
.. . .. .
.. .. .. ...
.............................................................................
... ...
... ... ... ...
...............
................................................
... ..
.
...........................................................................................
.
.... .... .... ....
........................................s
B
418
. (a) How many zeroes are at the right-hand end of the number
3 × · · · × 52?
5
1×2×
(b) What is the rightmost nonzero digit of 1 × 2 × · · · × 52? (For example,
the rightmost nonzero digit of 1 × 2 × · · · × 12 = 479 001 600 is 6.)
6. Juliette and Philippe play the following game: At
the beginning of the game, eah orner of a square is
overed with a number of hips. In turn, eah player
hooses one side of the square and removes as many
hips as (s)he wants from the endpoints of that side
provided (s)he takes at least one hip. It is not neessary to remove the same number of hips from eah
endpoint. The player who removes the last hip wins.
At the beginning of the game on the square ABCD
there are 10 hips on orner A, 11 hips on B , 12
hips on C , and 13 hips on D. If Juliette begins,
how should she play?
. Find all funtions F
numbers x.
7
:R→R
A....................................................................... B
...
...
...
...
...
..
...
..
...
...
...
...
...
..
...
...
...
...
...
..
...
...
...
..
...
.......................................................................
D
C
suh that F (x) + xF (−x) = 1 for all real
Conours de l'Assoiation mathematique
du Quebe,
2010
Ordre seondaire
3 heures a permis
. Un alphametique
est un petit asse-t^ete mathematique
qui onsiste en
une equation
ou les hires sont remplaes
par des lettres. Le resoudre
onsiste a trouver quelle lettre orrespond a quel hire pour que l'equation
soit vraie. Dans le probleme,
le m^eme hire ne peut e^ tre represent
e par
deux lettres dierentes
et une lettre represente
toujours le m^eme hire.
Bien entendu, un nombre ne doit jamais ommener par zero.
Par exemple,
l'alphametique
PAPA + PAPA = MAMAN a pour solution P = 7, A = 5,
M = 1 et N = 0. Ainsi, en remplaant les lettres par les hires, on a bien
7575 + 7575 = 15150.
Trouvez la solution de l'alphametique
\renversant" suivant:
1
NOMBRE × 35
=
ERBMON .
. Trouvez tous les polynomes
^
de la forme p(x) = x3 + mx + 6 dont tous
les zeros
sont des nombres entiers.
2
419
units from the
centre of a circle of
√
. Une droite est situee
a 22 unites
du entre
en deux parties.
d'un erle de rayon 1, le separant
Quelle est l'aire de la plus petite partie?
3
. La grille suivante represente
le plan d'une ville.
En partant du point A, ombien y a-t-il de hemins
(ourts) distints se rendant a B ? (Un hemin ourt
ne va jamais vers le haut de la grille ni vers la
gauhe.)
4
A ..s..................................
... ... ... ...
............................................
.. .. ... ..
.. .. .. ...
...............................................................................
... ...
... .... .... ....
..............
........................................
.. ..
... .. .. ..
... ...
...................................................................................
.... .... .... ....
....................................s
a la n de 1 × 2 × 3 × · · · × 52?
. (a) Combien y a-t-il de zeros
B
5
(b) Quel est le dernier hire (i.e. le plus a droite) non nul de l'expansion
deimale
de 1 × 2 × 3 × · · · × 52? (Par exemple, le dernier hire non
nul de 1 × 2 × 3 × · · · × 12 = 479 001 600 est 6.)
6. Juliette et Philippe jouent au jeu suivant. Au
debut
de la partie, haque oin d'un arre est reouvert d'un ertain nombre de jetons. A tour de role,
^
haque joueur hoisit un ot
^ e du arre et retire autant
de jetons qu'il veut des deux oins qui limitent e ot
^ e,
pourvu qu'il en enleve
en tout au moins un. Il n'est
pas neessaire
de retirer le m^eme nombre de jetons a
haun des oins. Le premier joueur qui se retrouve
devant un arre dont tous les oins sont vides a perdu.
Au debut
de la partie, sur le arre ABCD, il y a 10
jetons sur le oin A, 11 sur le oin B , 12 sur le oin C
et 13 sur le oin D. Si Juliette ommene, omment
devrait-elle jouer?
7. Quelles sont les fontions F
pour tout x reel?
: R → R
veriant
A....................................................................... B
...
....
...
..
...
...
...
...
...
...
...
...
...
...
..
..
...
...
...
...
...
..
...
..
...
.......................................................................
D
C
F (x) + xF (−x) = 1,
Next are solutions to the 27th New Brunswik Mathematis Competition, 2009, Grade 9, Part C, given in Skoliad 122 at [2010 : 1{3℄.
. If you write all integers from 1 to 100, how many even digits will be
written? (When you write the number 42, two even digits are written.)
(A) 50
(B) 71
(C) 80
(D) 89
(E) 91
1
Solution by Lena Choi, student, Eole
Dr. Charles Best Seondary Shool,
.
You an simply ount the even digits, whih is most onveniently done
Coquitlam, BC
420
by setting up a table like this:
1{10
11{20
21{30
31{40
41{50
5 even
6 even
14 even
6 even
14 even
digits,
digits,
digits,
digits,
digits,
51{60
61{70
71{80
81{90
91{100
6 even
14 even
6 even
14 even
6 even
digits,
digits,
digits,
digits,
digits,
for a total of 91 even digits.
Also solved by BILLY SUANDITO, Palembang, Indonesia.
Alternatively, you an notie that every seond ones digit is even; that ontributes
digits.
The
Finally,
100
20's, 40's, 60's,
and
80's
eah ontribute ten even tens digits; that is
ontributes a single even tens digit. Again, the grand total is
91
40
50
digits.
even digits.
2. In a farm there are hens (no hump, two legs), amels (two humps, four
legs) and dromedaries (one hump, four legs). If the number of legs is four
times the number of humps, then the number of hens divided by the number
of amels will be?
(A) 12
(B) 1
(C) 32
(D) 2
(E) Not enough
information
Solution by Billy Suandito, Palembang, Indonesia.
Say the farm has A hens, B amels, and C dromedaries. Then the hens
ontribute 0 humps and 2A legs, the amels ontribute 2B humps and 4B
legs, while the dromedaries ontribute C humps and 4C legs for a total of
2B + C humps and 2A + 4B + 4C legs. Sine the number of legs is given to
be four times the number of humps, 2A+4B +4C = 4(2B +C) = 8B +4C .
2B
A
=
= 2.
Thus 2A + 4B = 8B , so 2A = 4B , so A = 2B . Therefore, B
B
Also solved by LENA CHOI, student, Eole
Dr. Charles Best Seondary Shool, Coquitlam, BC; MONICA HSIEH, student, Burnaby North Seondary Shool, Burnaby, BC; and ELLEN
CHEN, WEN-TING FAN, VICKY LIAO, JUSTIN MIAO (all students at Burnaby North Seondary
Shool, Burnaby, BC), and LISA WANG, student, Port Moody Seondary Shool, Port Moody,
BC (in ollaboration).
. A ubi box of side 1 m is plaed on the oor. A seond ubi box of side
m is plaed on top of the rst box so that the entre of the seond box is
diretly above the entre of the rst box. A painter paints all of the surfae
area of the two boxes that an be reahed without moving the boxes. What
is the total area of surfae that is painted?
3
2
3
(A)
49
9
m2
(B)
57
9
m2
(C)
61
9
m2
(D)
72
9
m2
(E) None of these
.
The total area of ve faes of the large ube is 5 × 1 m × 1 m = 5 m2 .
The total area of ve faes of the small ube is 5 × 23 m × 23 m = 20
m2 . The
9
small ube hides a 23 m × 23 m square of one fae of the large ube.
Solution by Billy Suandito, Palembang, Indonesia
421
m2 − 49 m2 =
Therefore, the painted area is 5 m2 + 20
9
61
9
m2 .
Also solved by LENA CHOI, student, Eole
Dr. Charles Best Seondary Shool, Coquitlam,
BC; MONICA HSIEH, student, Burnaby North Seondary Shool, Burnaby, BC; ELLEN CHEN,
WEN-TING FAN, VICKY LIAO, JUSTIN MIAO (all students at Burnaby North Seondary Shool,
Burnaby, BC), and LISA WANG, student, Port Moody Seondary Shool, Port Moody, BC (in
ollaboration); and one anonymous solver.
4
. What is the ones digit of 22009 ?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
Solution by Lena Choi, student, Eole
Dr. Charles Best Seondary Shool,
.
You an easily nd the ones digit of the rst few powers of 2: 21 ≡ 2,
22 ≡ 4, 23 ≡ 8, 24 ≡ 6, 25 ≡ 2, 26 ≡ 4, 27 ≡ 8, 28 ≡ 6, et. mod 10. So
the pattern is 2, 4, 8, 6, 2, 4, 8, 6, . . . , whih repeats in groups of four. Now,
2009 is not divisible by 4, but 2008 ÷ 4 = 502. Therefore, 22008 ends in a 6,
so 22009 ends in a 2, beause the next number in the pattern is 2.
Coquitlam, BC
Also solved by MONICA HSIEH, student, Burnaby North Seondary Shool, Burnaby,
BC; and ELLEN CHEN, WEN-TING FAN, VICKY LIAO, JUSTIN MIAO (all students at Burnaby North Seondary Shool, Burnaby, BC), and LISA WANG, student, Port Moody Seondary
Shool, Port Moody, BC (in ollaboration).
28 ≡ 6 mod 10"
The notation \
28
and
6
remainder
(read
28
is equivalent to
leave the same remainder when divided by
10;
6
modulo
10)
means that
and, indeed, they both leave the
6.
One ought to
prove the pattern, but it follows easily from the fat that the ones digit of
a produt depends only on the ones digits of the fators.
. The numbers 1, 2, 3, 4, 5, and 6 are to be arranged in a row. In how
many ways an this be done if 2 is always to the left of 4, and 4 is always to
the left of 6? (For example 2, 5, 3, 4, 6, 1 is an arrangement with 2 to the
left of 4 and 4 to the left of 6.)
(A) 20
(B) 36
(C) 60
(D) 120
(E) 240
5
.
These are the possible arrangements of the numbers 2, 4, and 6:
Solution by Billy Suandito, Palembang, Indonesia
2
2
2
2
2
2
2
2
2
2
4
4
4
4
6
6
6
6
4
4
4
6
6
6
4
4
6
4
6
6
2
2
2
2
2
2
4
4
4
6
6
6
4
4
2
2
2
4
4
2
6
4
6
4
4
6
6
6
6
6
422
In eah of these twenty ases, it remains to plae the numbers 1, 3, and 5 in
three slots. You have three hoies for the plaement of 1; for eah of these,
you have two hoies left for the plaement of 3; and that leaves just one
slot for 5. Hene, in eah of the twenty ases, you an plae the numbers
1, 3, and 5 in 3 × 2 × 1 = 3! = 6 ways. Therefore, the total number of
arrangements is 20 × 6 = 120.
Also solved by LENA CHOI, student, Eole
Dr. Charles Best Seondary Shool, Coquitlam,
BC; and MONICA HSIEH, student, Burnaby North Seondary Shool, Burnaby, BC.
Our solver found that there are twenty ases by listing them.
6
If you are familiar with
ounting permutations and ombinations, you an also alulate that number: Of the six slots
you an hoose three for the numbers
2, 4,
and
6
in
3
=
6!
3!(6−3)!
= 20
ways.
have hosen three slots, the three numbers must go into those slots in the order
. The square ABCD is insribed in
a irle with diameter BD of length 2.
If AB is the diameter of the semiirle
on top of the square, what is the area
of the shaded region?
6
(A) 4 −4 π
(D) 1
2
(B) π −
(C)
4
(E) Not enough
information
1
2
One you
2, 4, 6.
..................................
......
.....
....
...
.
.
.
... .................................................. .....
.
A ....................................................................... B
...
.......
... ....
.... ... ....
..... ... ....
..... ....
.
.
.
.
... ...
.
....
... ..
... ....
.....
.
.
..... ....
.
.. ..
....
... ...
... ...
.
.
.
.
. 2
... ...
.
.
... ...
.
... ... ......
.. ..
... ... .....
... ...
.
...........
......................................................................
...
...
D ................................................. C
Solution by Lena Choi, student, Eole
Dr. Charles Best Seondary Shool,
.
2
2
By the Pythagorean Theorem, |AB|2 + |AD|
√ = |BD| , but |AD| =
2
|AB| and |BD| = 2, so 2|AB| = 4, so |AB| = 2. Therefore the area of
the square is 2. Clearly the radius of the irle is 1, so the area of the irle
Coquitlam, BC
square is 2. Clearly the radius of the circle is 1,
is π, whene the area of the shaded region
is π − 2. Thus the area of
whence the area of
quarter-segment,
2. Thus the areaa single
of a single
segment,
Sine |AB| =
2
.
,, is
is π −
4
√
2, the radius of the semiirle is
 √ ‹2
π
semiirle is 12 π . 22It follows
= . It
4
π
π−2
1
−
= .
4
4
2
√
2
,
2
so the area of the
follows
thatarea
the area
lune,
that the
of of
thethelune,
,, isis
Also solved by MONICA HSIEH, student, Burnaby North Seondary Shool, Burnaby,
BC; BILLY SUANDITO, Palembang, Indonesia; ELLEN CHEN, WEN-TING FAN, VICKY LIAO,
JUSTIN MIAO (all students at Burnaby North Seondary Shool, Burnaby, BC), and LISA WANG,
student, Port Moody Seondary Shool, Port Moody, BC (in ollaboration); and one anonymous
solver.
This issue's prize of one opy of CRUX with MAYHEM for the best so
lutions goes to Lena Choi, student, Eole
Dr. Charles Best Seondary Shool,
Coquitlam, BC. We look forward to reeiving more reader solutions.