in collaboration with

### The CENTRE for EDUCATION

### in MATHEMATICS and COMPUTING

presents the

## Canadian Open

## Mathematics Challenge

### Wednesday, November 24, 2004

### Solutions

c

Part A

1. If x+ 2y= 84 = 2x+y, what is the value of x+y?

Solution 1

Since x+ 2y = 84 and 2x+y = 84, then adding these two equations together, we obtain 3x+ 3y= 168 or x+y= 56.

Solution 2

Since x+ 2y= 84, then x= 84−2y.

Substituting into the second equation, we get

2(84−2y) +y = 84
168_{−}3y = 84
84 = 3y

y = 28

Therefore,x= 84_{−}2(28) = 28 and so x+y= 56.

Solution 3

Since 2x+y= 84, then y= 84_{−}2x.
Substituting into the first equation, we get

x+ 2(84−2x) = 84
168_{−}3x = 84
84 = 3x

x = 28

Therefore,y= 84−2(28) = 28 and so x+y= 56.

Solution 4

SInce these two expressions are identical when x is replaced byy and y is replaced by x, then

x=y.

Therefore, 3x= 84 orx= 28 and so y= 28. Thus, x+y= 56.

2. Let S be the set of all three-digit positive integers whose digits are 3, 5 and 7, with no digit repeated in the same integer. Calculate the remainder when the sum of all of the integers inS

is divided by 9.

Solution 1

We can write down the elements ofS: 357, 375, 537, 573, 735, 753. The sum of these elements is 357 + 375 + 537 + 573 + 735 + 753 = 3330.

Since 3330 is divisible by 9 (because the sum of its digits is divisible by 9), the remainder when we divide by 9 is 0.

Solution 2

There are six numbers formed with the three given numbers.

Two of these numbers have a 3 in the 100s position, two have a 5 in the 100s position, and two have a 7 in the 100s position.

The same can be said about the distribution of numbers in the 10s and units positions. Therefore, the sum of the six numbers is

2(3 + 5 + 7)(100) + 2(3 + 5 + 7)(10) + 2(3 + 5 + 7)(1) = 3330

The remainder is 0 when 3330 is divided by 9.

Answer: _{0}

3. In the diagram, point E has coordinates (0,2), and B lies on
the positive x-axis so that BE = √7. Also, point C lies on the
positive x-axis so that BC = OB. If point D lies in the first
quadrant such that∠CBD= 30◦ _{and} _{∠}_{BCD} _{= 90}◦_{, what is the}
length ofED?

O x

y

B E

D

C

Solution

In order to find the length ofED, we will try to find the coordinates ofD. Let the coordinates of B be (b,0).

Since BE =√7 and the coordinates of E are (0,2), then

p

(b_{−}0)2_{+ (0}

−2)2 _{=} √_{7}

b2 + 4 = 7

b2 = 3

(Note that it would have also been possible to find the coordinates of B by using Pythagoras –OE2

+OB2

=EB2

so OB2

= 7−4 = 3 so OB =√3 and so B has coordinates (√3,0).) Since BC =OB, then C has coordinates (2√3,0).

Since ∠BCD = 90◦ _{and} _{D} _{lies in the first quadrant, then} _{D} _{has coordinates (2}√_{3}_{, d}_{), with}

d >0.

Since_{△}DBC has ∠BCD= 90◦ _{and} _{∠}_{CBD}_{= 30}◦_{, then it is a 30}◦_{−}_{60}◦_{−}_{90}◦ _{triangle. Since}

CB =√3 (and CB is opposite the 60◦ _{angle), then} _{DC} _{(which is opposite the 30}◦ _{angle) has}
length 1.

Therefore,D has coordinates (2√3,1).

O x

y

E

### (

0, 2### )

C

### (

2 3, 0### )

D

### (

2 3, 1### )

B

### (

3, 0### )

Thus, ED= q

(2√3−0)2_{+ (1}

−2)2 _{=}√_{12 + 1 =}√_{13.}

Answer: √_{13}

4. A function f(x) has the following properties:

i) f(1) = 1

ii) f(2x) = 4f(x) + 6

iii) f(x+ 2) =f(x) + 12x+ 12

Calculate f(6).

Solution 1

Using property ii) withx= 1,

f(2) = 4f(1) + 6 = 4(1) + 6 = 10

since f(1) = 1 by property i). Using property ii) withx= 2,

Using property iii) withx= 4,

f(6) =f(4) + 12(4) + 12 = 46 + 48 + 12 = 106

Therefore, the value of f(6) is 106.

Solution 2

Using property iii) withx= 1,

f(3) =f(1) + 12(1) + 12 = 1 + 12 + 12 = 25

since f(1) = 1 by property i). Using property ii) withx= 3,

f(6) = 4f(3) + 6 = 4(25) + 6 = 106

Therefore, the value of f(6) is 106.

Solution 3

Working backwards,

f(6) = 4f(3) + 6 (by property ii) with x= 3)

= 4(f(1) + 12(1) + 12) + 6 (by property iii) with x= 1) = 4f(1) + 4(24) + 6

= 4(1) + 102 (by property i))

Therefore, the value of f(6) is 106.

Answer: _{f}_{(6) = 106}

5. The Rice Tent Company sells tents in two different sizes, large and small. Last year, the Com-pany sold 200 tents, of which one quarter were large. The sale of the large tents produced one third of the company’s income. What was the ratio of the price of a large tent to the price of a small tent?

Solution

Since the Rice Tent Company sold 200 tents, of which one quarter were large, then they sold 50 large tents and 150 small tents last year.

LetL be the price of a large tent andS the price of a small tent.

From the given information,

50L = 1

3(50L+ 150S) 150L = 50L+ 150S

100L = 150S L

S =

150 100 =

3 2

Therefore, the ratio of the price of a large tent to the price of a small tent was 3 : 2.

Answer: _{3 : 2}

6. In the diagram, a square of side length 2 has semicircles drawn on each side. An “elastic band” is stretched tightly around the figure. What is the length of the elastic band in this position?

Solution

Label the four vertices of the square as W,X, Y, Z, in clockwise order.

Label the four midpoints of the sides of the square (that is, the centres of the four semicircles) asM, N, O, P, in clockwise order, starting with M being the midpoint ofW X.

In each semicircle, join the centre to the two points on that semicircle where the band just starts (or stops) to contact the circle. Label these eight points asA, B,C, D,E, F, G, andH

in clockwise order, starting withA and B on the semicircle with centre M.

W X

Z Y

M

N

O P

A B

C

D

E F

G H

By symmetry, the four straight parts of the band will be equal in length (that is, BC =

DE = F G = HA) and the four arc segments of the band will be equal in length (that is,

AB=CD =EF =GH).

Therefore, the total length of the band is 4(Length of arcAB) + 4(Length of BC).

Now, BC will actually be tangent to the two semicircles (with centres M and N) where it initially just touches them.

SinceM B =N C = 1 (because they are radii of the semicircles and each semicircle has diame-ter 2), thenM BCN must actually be rectangle, so BC is equal and parallel to M N.

SinceM and N are the midpoints of the sides of the square of side length 2, thenM Y =Y N = 1, so M N =√2, so BC =√2.

Next, we determine the length of AB. Previously, we saw that M BCN is a rectangle, so

BC was parallel toM N. Similarly, HA is parallel toP M.

But P M is perpendicular to M N, so HAis perpendicular to BC.

Therefore, ∠AM B = 90◦_{, ie.} _{AB} _{is one-quarter of the circumference of a circle with radius 1}
or 1

4(2π(1)) = 1 2π.

Therefore, the total length of the band is 4(1

2π) + 4(

3b_{, then multiplying these two equations together, we get}

a2 =ab_{·}a3b =a4b.

Dividing both sides bya2

(since a_{6}= 0), we get a4b−2
= 1.
Since a >1, then 4b_{−}2 = 0 or b= 1

2.

Substituting back into the first equation, we get 1 2a=a

1/2

=√a or a= 2√a. Squaring both sides gives a2

= 4a ora2

Since a >1 and b >0, we can take logarithms of both sides of both equations. In the first equation, using log rules on log(ab) = log ab

gives log(a) + log(b) =blog(a).
In the first equation, using log rules on log a_{b}

= log a3b

gives log(a)−log(b) = 3blog(a).
Adding these two new equations gives 2 log(a) = 4blog(a) or (4b_{−}2) log(a) = 0.

Substituting this back into the first log equation gives log(a) + log 1 2

= 1

2log(a) or 1

2log(a) = −log 1 2

= log(2) or log(a) = 2 log(2) = log(4), soa= 4.

Answer: _{4}

8. A rectangular sheet of paper,ABCD, hasAD = 1 andAB =r, where 1< r <2. The paper is folded along a line throughAso that the edge ADfalls onto the edge AB. Without unfolding, the paper is folded again along a line through B so that the edge CB also lies on AB. The result is a triangular piece of paper. A region of this triangle is four sheets thick. In terms of

r, what is the area of this region?

Solution

Start with the rectangular sheet of paper,ABCD, withA in the top left andB in the bottom left.

Fold AD across so thatAD lies along AB. Let D′ _{be the point were} _{D} _{touches} _{AB} _{and let} _{E}
be the point onDC where the fold hits DC.

Since AD′ _{is the old} _{AD}_{, then} _{AD}′ _{= 1.}

Since D′_{E} _{is perpendicular to} _{D}′_{A} _{(since} _{DC} _{was perpendicular to} _{AD}_{) then} _{D}′_{E} _{is parallel}
toBC, so D′_{E} _{= 1 as well.}

A D

B C

E 1

r_{−}_{1}
′

D

We can also conclude that D′_{B} _{=}_{EC} _{=}_{r}_{−}_{1, since}_{AB} _{has length} _{r}_{.}

Next, we fold the paper so that BC lies along AB. Unfold this paper and lay it flat so that we can see the crease.

SinceBC is folded onto AB, then the crease bisects ∠ABC, that is the crease makes an angle
of 45◦ _{with both} _{AB} _{and} _{BC}_{.}

A

B C

E ′

D

X

Y

Now when the paper had been folded the second time (before we unfolded it!), the only way to obtain a region four sheets thick was to fold a region two sheets thick on top of a region which is also two sheets thick.

Since_{△}XY E is the only part of the paper “below” the second crease which is two sheets thick,
and, when the second fold is made, it lies entirely over another region which is two sheets thick,
then the desired area is the area of _{△}XY E.

Since ∠ABX = 45◦_{, then} _{△}_{BD}′_{X} _{is isosceles and right-angled, so} _{D}′_{X} _{=}_{D}′_{B} _{=}_{r}_{−}_{1.}
Thus, EX = 1−D′_{X} _{= 1}_{−}_{(}_{r}_{−}_{1) = 2}_{−}_{r}_{.}

Since∠D′_{XB} _{= 45}◦_{, then}_{∠}_{Y XE} _{= 45}◦_{. Also, since} _{△}_{AD}′_{E} _{is right isosceles, then}_{∠}_{Y EX} _{=}
45◦_{, so} _{△}_{XY E} _{is isosceles and right-angled.}

Y

X E

s s

2−r

Suppose XY =s. Then√2s=XE = 2_{−}r or 2s2

= (2_{−}r)2
.
The area of _{△}XY E is 1

2s 2

or 1

4(2−r) 2

. Therefore, the area of the desired region is 1

4(2−r) 2

.

Answer: 1

Part B

1. The points A(−8,6) and B(−6,_{−}8) lie on the circle x2
+y2

= 100.

(a) Determine the equation of the line through A and B.

Solution

First, we determine the slope of the line segment AB. The slope is 6−(−8)

−8−(−6) =−7.
We could now proceed to find the equation of the line in several different ways.
Using the point-slope form, we obtain y_{−}6 = −7(x_{−}(−8)) or y=−7x_{−}50.

(b) Determine the equation of the perpendicular bisector of AB.

Solution

Since the slope of AB is −7, then the slope of the perpendicular bisector of AB is 1 7. Also, the perpendicular bisector passes through the midpoint of AB, which is

1

2((−8) + (−6)), 1

2(6 + (−8))

= (_{−}7,_{−}1).

Therefore, the equation of the perpendicular bisector isy_{−}(−1) = 1

7(x−(−7)) ory= 1 7x.

(c) The perpendicular bisector of AB cuts the circle at two points, P in the first quadrant and Qin the third quadrant. Determine the coordinates of P and Q.

Solution 1

y

x P

Q A

B

We want to find the points of intersection of the circle x2 +y2

obtain

We want to find the points of intersection of the circle x2 +y2

= 100 and the liney= 1 7x. Substituting y= 1

7x into the equation of the circle we obtain

x2

(d) What is the length of P Q? Justify your answer.

Solution 1

The points P and Q are joined by the line y= 1

7x, which passes through the origin. Since the origin is the centre of the circle, then P Q must be a diameter of the circle. Since the circle has equation x2

+y2

= 100 = 102

, then its radius is 10, so its diameter is 20.

Therefore, P Q= 20.

Solution 2

P Q by direct calculation:

Therefore, the length of P Qis 20.

2. (a) Determine the two values of x such thatx2

−4x_{−}12 = 0.

Solution

Factoring the given equation x2

−4x_{−}12 = 0, we obtain (x_{−}6)(x+ 2) = 0.
Therefore, the two solutions are x= 6 andx=−2.

(b) Determine the one value of x such thatx_{−}√4x+ 12 = 0. Justify your answer.

Solution

We first eliminate the square root by isolating it on one side and squaring:

x_{−}√4x+ 12 = 0

x = √4x+ 12

x2 = 4x+ 12

x2_{−}4x_{−}12 = 0
(x_{−}6)(x+ 2) = 0

Therefore, the two possible solutions are x= 6 and x=_{−}2. (Since we have squared both
sides, it is possible that we have introduced an extraneous root, so we should verify both
of these.)

If x= 6, then 6_{−}p4(6) + 12 = 6_{−}√36 = 0.

(c) Determine all real values of csuch that

x2

−4x_{−}c_{−}√8x2

−32x_{−}8c= 0

has precisely two distinct real solutions for x.

Solution

We start by attempting to solve this equation and then seeing what conditions on carise. Since 8x2

Let’s look at these last two equations. First, we look at x2 quadratic equation is (−4)2

−4(−(c+ 8)) = 48 + 4c. Therefore, this equation has

• zero solutions if 48 + 4c <0, so c <_{−}12,

• exactly one solution if 48 + 4c= 0, so c=−12, and

• two distinct solutions if 48 + 4c > 0, so c >_{−}12.

We see also that any value of xthat satisfies one of these two equations cannot satisfy the other, since we cannot have both x2

−4x_{−}c = 0 andx2

−4x_{−}c= 8. (In other words,
no roots overlap between these two equations.)

We make a table to combine our observations:

c <_{−}12 c=_{−}12 _{−}12< c <_{−}4 c=_{−}4 c >_{−}4

x2

−4x_{−}c= 0 0 solutions 0 solutions 0 solutions 1 solution 2 solutions

x2

−4x_{−}c= 8 0 solutions 1 solution 2 solutions 2 solutions 2 solutions
Total solutions 0 solutions 1 solution 2 solutions 3 solutions 4 solutions

3. A map shows all Beryl’s Llamaburgers restaurant locations in North America. On this map, a line segment is drawn from each restaurant to the restaurant that is closest to it. Every restaurant has a unique closest neighbour. (Note that if A and B are two of the restaurants, then A may be the closest toB without B being closest toA.)

(a) Prove that no three line segments on the map can form a triangle.

Solution 1

We start by assuming that three line segments on the map do form a triangle, and show that this is in fact impossible.

Notice that if restaurants X and Y are joined by a line segment, then either X is the closest restaurant to Y or Y is the closest restaurant to X (or both).

Assume that A,B and C are the three points on the map connect by segments.

B

A C

To begin, we focus on the segment joining A to B. Let’s assume that A is the closest restaurant to B. (It doesn’t matter which direction we assume here.) This means thatC

is not the closest restaurant to B, so BA < BC.

But B and C are connected and C is not the closest restaurant to B. Therefore, B is the closest restaurant to C, which means CB < CA.

But C and A are also connected andAis not the closest restaurant to C. Therefore, C is the closest restaurant to A, which meansAC < AB.

But this means that BA < BC, BC < AC and AC < BA. This cannot be the case. Therefore, it is impossible for three line segments to form a triangle.

Solution 2

We prove this by showing that constructing a triangle is impossible. We start by considering two locations A and B and the line segmentAB.

Since A and B are connected, we can assume without loss of generality that A is closest to B. (The caseB closest to A involves interchanging A and B, and the case of Aand B

closest to each other is included in the case of A closest to B.)

If A is closest to B and we add a new location C which is connected to B, then B

If we join C to A, then eitherC is closest to A orA is closest to C. But A can’t be closest to C since B is closest to C.

Therefore, we must have C closest to A.

But thenAC is shorter thanAB, along withABbeing shorter thanBC(sinceAis closest toB), which means thatAC is shorter than BC orAis closer to C thanB is, which isn’t true. This is a contradiction.

Therefore, we can’t construct a triangle.

(b) Prove that no restaurant can be connected to more than five other restaurants.

Solution

We start by assuming that one restaurant can be connected to six others and show that this is impossible. From this we can conclude that no restaurant can be connected to more than five other restaurants (for if it could be joined to 8 others, say, then we could consider six of them only and reach a contradiction).

Assume that restaurant A can be connected to restaurants B, C, D, E,F, andH, where these restaurants are listed in clockwise order of their line segments joining to A.

B

C

D E

F H

A

Consider restaurants A, B and C.

We know that B andC are both connected toAand both cannot be the closest neighbour to A. Thus, A must be the closest neighbour to one of these, say B. (It doesn’t matter which we choose).

Since A is the closest restaurant to B, then BA < BC. Now consider the line joining C toA.

If C is the closest neighbour to A, then AC < AB, so AC < AB < BC.

If A is the closest neighbour to C, then CA < CB so CA < CB and BA < CB.

In either case, BC is the (strictly) longest side in _{△}ABC, and so must be opposite the
(strictly) largest angle.

But we can reapply this reasoning to conclude that ∠CAD, ∠DAE, ∠EAF, ∠F AH,
and ∠HAB are each greater than 60◦_{. But the sum of these six angles is 360}◦_{, since they}
will form a full circle around A, and six angles, each greater than 60◦_{, cannot add to 360}◦_{.}
So we have a contradiction.

Therefore, it is impossible for a restaurant to be connected to more than five other restau-rants.

4. In a sumac sequence, t1, t2, t3, . . ., t_{m}, each term is an integer greater than or equal to 0.
Also, each term, starting with the third, is the difference of the preceding two terms (that is,

t_{n}+2 =tn−tn+1 forn ≥1). The sequence terminates at tm iftm−1−tm <0. For example, 120,
71, 49, 22, 27 is a sumac sequence of length 5.

(a) Find the positive integer B so that the sumac sequence 150, B, . . . has the maximum possible number of terms.

Solution

Suppose that we have a sumac sequence with t1 = 150 and t2 = B. Let’s write out the next several terms (assuming that they exist) in terms of B:

t3 = 150−B t4 = 2B−150 t5 = 300−3B t6 = 5B−450

t7 = 750−8B t8 = 13B−1200 t9 = 1950−21B t10= 34B−3150 In order to maximize the length of this sumac sequence, we would like to chooseB so that as many terms as possible starting from t1 are non-negative. (When we reach the first negative “term”, we know that the sequence terminated at the previous term.)

For t2 ≥0, B ≥0.

For t3 ≥0, 150−B ≥0 or B ≤150. For t4 ≥0, 2B−150≥0 or B ≥75. For t5 ≥0, 300−3B ≥0 or B ≤100. For t6 ≥0, 5B−450≥0 or B ≥90.

For t7 ≥0, 750−8B ≥0 or B ≤ 750_{8} = 936_{8}.
For t8 ≥0, 13B −1200 ≥0 or B ≥ 1200_{13} = 92_{13}4.
For t9 ≥0, 1950−21B ≥0 or B ≤

1950 21 = 92

18 21.

Therefore, since B must be a positive integer, if we choose B = 93, then we can ensure that each of t1 through t8 are non-negative. This will maximize the number of terms starting from the beginning, since B must satisfy 924

13 ≤B ≤93 6

8 in order for at least the first eight terms to be non-negative. (Note that t9 will in fact be negative when B = 93.)

(b) Let m be a positive integer with m _{≥} 5. Determine the number of sumac sequences of
length m with t_{m} _{≤}2000 and with no term divisible by 5.

Solution

We begin our solution by making some observations about sumac sequences.

• A sumac sequence is completely determined by its first two terms. This is true since the first two terms give us the third, the second and third give us the fourth, and so on. The sequence will terminate when the “next term” would be negative.

• In a sumac sequence, since for every (valid) n we have tn+2 = tn−tn+1, then tn =

tn+1+tn+2. This means that we can “reverse engineer” a sumac sequence – if we know terms (n+ 1) and (n+ 2), then we can determine term n. Thus, if we know the final two terms in a sumac sequence, then we can determine all of the previous terms.

• From the first observation, the first two terms of a sumac sequence completely de-termines the sequence. Is the same true of the last two terms? No. When we start looking at a sumac sequence from the back, every new term as we proceed towards the front will always be non-negative (since we are adding non-negative terms). Thus, there is no “stopping condition” as there is when we work forwards. (For example, 3, 1, 2 is a sumac sequence ending with 1, 2, as is 4, 3, 1, 2.)

• However, if we know the final two terms and the length of the sequence, this completely determines the sumac sequence (and we will always be able to find such a sequence).

Now we proceed. Let m be a fixed positive integer with m_{≥}5.
Suppose that t1, t2, . . . , tm−1, tm is a sumac sequence of lengthm.

Because we are given a condition on the final term of the sequence, we will examine the sequence from the back.

Let x=tm and y=tm−1. Note that x, y and m determine the sequence.

Since x and y are the last two terms in the sumac sequence, then tm−1 −tm =y−x <0 or x > y.

Since we have m fixed, we would like to determine how many sumac sequences we can
form with t_{m} =x_{≤}2000, t_{m}_{−}1 =y < xand no term divisible by 5.

Let’s write out the last five terms of the sequence (in reverse order): x, y, x+y, x+ 2y,
2x+ 3y. (Sincem_{≥}5, we know that there are at least five terms in the sequence.)
Since we want no term divisible by 5, let us consider x and y modulo 5 to see what
hap-pens. (There are 25 possible pairs for (x, y) modulo 5.)

Since no term is divisible by 5, then we don’t wantx_{≡}0 (mod 5) ory_{≡}0 (mod 5). This
cuts us down to 16 possibilities for (x, y).

x y x+y x+ 2y 2x+ 3y

1 1 2 3 0

1 2 3 0

1 3 4 2 1

1 4 0

2 1 3 4 2

2 2 4 1 0

2 3 0

2 4 1 0

3 1 4 0

3 2 0

3 3 1 4 0

3 4 2 1 3

4 1 0

4 2 1 3 4

4 3 2 0

4 4 3 2 0

So the only possible pairs for (x, y) modulo 5 are (1,3), (2,1), (3,4) and (4,2).

If we start with (x, y) = (1,3) modulo 5, then the terms in the sequence modulo 5 are 1, 3, 4, 2, 1, 3, 4, 2, 1, . . ., ie. the terms cycle modulo 5 with no terms divisible by 5.

This similar cycling will happen with each of the other 3 pairs, so each of these 4 pairs give no terms divisible by 5.

So for each of these pairs, we need to determine the number of pairs of non-negative
integers (x, y) with x _{≤} 2000, y < x and congruent to the appropriate things modulo
5. Each such pair will give a sumac sequence of length m _{≥} 5 with no term divisible by
5. (Since the divisibility of the terms is independent of length, this also means that the
number of such sequences will be independent of m!)

Case 1: (x, y) congruent to (1,3) modulo 5 Here x can take the values 1996, 1991,. . ., 6, 1.

If x= 1996,y can be 1993, 1988, . . ., 8, 3. (399 possibilities) If x= 1991,y can be 1988, 1983, . . ., 8, 3. (398 possibilities)

This pattern continues, with one fewer possibility each time x decreases by 5, until we reach x= 6, where y= 3 is the only possibility.

Case 2: (x, y) congruent to (2,1) modulo 5 Here x can take the values 1997, 1992,. . ., 7, 2.

If x= 1997,y can be 1996, 1991, . . ., 6, 1. (400 possibilities) If x= 1992,y can be 1991, 1986, . . ., 6, 1. (399 possibilities)

This pattern continues, with one fewer possibility each time x decreases by 5, until we reach x= 2, where y= 1 is the only possibility.

Thus, there are 400 + 399 +_{· · ·}+ 2 + 1 possibilities for this case.

Case 3: (x, y) congruent to (3,4) modulo 5 Here x can take the values 1998, 1993,. . ., 8, 3.

If x= 1998,y can be 1994, 1989, . . ., 9, 4. (399 possibilities) If x= 1993,y can be 1989, 1984, . . ., 9, 4. (398 possibilities)

This pattern continues, with one fewer possibility each time x decreases by 5, until we reach x= 8, where y= 4 is the only possibility.

Thus, there are 399 + 398 +_{· · ·}+ 2 + 1 possibilities for this case.

Case 4: (x, y) congruent to (4,2) modulo 5 Here x can take the values 1999, 1994,. . ., 9, 4.

If x= 1999,y can be 1997, 1992, . . ., 7, 2. (400 possibilities) If x= 1994,y can be 1992, 1987, . . ., 7, 2. (399 possibilities)

This pattern continues, with one fewer possibility each time x decreases by 5, until we reach x= 4, where y= 2 is the only possibility.

Thus, there are 400 + 399 +_{· · ·}+ 2 + 1 possibilities for this case.

Therefore, overall there are

2(399+398+· · ·+2+1)+2(400+399+· · ·+2+1) = 399(400)+400(401) = 400(800) = 320 000