Olympiad 8 - Test 9

Questions
13. If n be number of ways to paint each of the integers 2, 3,
Maths ……., 9 either red, green, or blue so that each number has
1. In the triangle ABC, AB = 8, BC = 7 and CA = 6. Let E be the different color from each of its proper divisors. Find sum
point on BC such that ∠BAE = 3∠EAC. Find
4
AE
2
. of digits of n.
5

2. If n be the number of integers between 1 and 1000, 14. There are k people and n chairs in a row, where 2 < k < n.
inclusive, can be expressed as the difference of the There is a couple among the k people. The number of
squares of nonnegative integers. Find n/10. ways in which all k people can be seated such that the
couple is seated together is equal to the number of ways
3. Triangle ABC has BC = 1 and AC = 2. What is the maximum in which the (k - 2 ) people, without the couple present,
possible value of angle A in degree? can be seated. Find the smallest value of n

4. 15. Find the sum of positive solutions to

Find (x2 + y2)/2 if x and y are positive integers such that 1 1 2
+ − = 0
xy + x + y = 71 x2 −10x−29 x2 −10x−45 x2 −10x−69

x2y + xy2 = 880.

16. Consider trapezoid ABCD which has AB || CD with AB = 5
5. Two circles with centres A and B and radius 2 units touch and CD = 9. Moreover, ∠ C = 15° and ∠ D = 75°. Let M1 be
each other externally at ‘C’. A third circle with centre ‘C’ the midpoint of AB and M2 be the midpoint of CD. What is
and radius ‘2’ units meets other two at D.E. If area of the distance M1M2?
quadrilateral ABDE is √b here b is integer, find b.
17. There exists a point P inside an equilateral triangle ABC
6. A race is organized for two teams of 7 runners each. The such that PA = 3, PB = 4, and PC = 5. If the side length of
runner who arrives first scores 1 point for their team, the the equilateral triangle is √a + c√3, here a and c are
runner who arrives second scores 2 point for their team, integers, find a + c.
and so on, so that finally, the runner who arrives last
scores 14 points for their team. Two runners never arrive 18. ∆ ABC is right-angled at ‘C’. The internal bisectors of CA
at exactly the same time. The wining team is the team and CB meet BC and CA at P respectively. M, N are the feet
with the lower total score. How many different winning of perpendiculars from P and Q to AB. Find ∠ MCN in
scores are possible? degree.

7. In triangle ABC, AB = 4, BC = 6. and AC = 8. Squares ABQR 19. In the triangle ABC, a circle passes through the point A,
and BCST are drawn the midpoint E of AC, the midpoint F of AB and is tangent
external to and lie in the same plane as ∆ ABC. Compute to the side BC at D. Suppose
AB
+
AC
= 4

QT2.
AC AB

Determine the size of ∠ EAF in degrees.

8. Find all the positive integers n (not ending with zero) for 20. For a positive integer n, let p(n) denote the product of the
which the number obtained by erasing the last digit is a positive integer factors of n. Determine the number of
divisor for n. factors n of 2310 for which p(n) is a perfect square.

9. Let x and y be real numbers satisfying x4y5 + y4x5 = 810 21. Let the altitude of ∆ ABC from A intersect the circumcircle
and x3 y 6 + y3 x6 = 945. Evaluate 2x3 + (xy)3 + 2y3. of ∆ ABC at D. E is a point on line AD such that E ≠ A and
AD = DE. Let AB = 13, BC = 14, and AC = 15. If the area of
10. Two workers A and B are engaged to do a piece of work. quadrilateral BDCE = m/n, find m – 100n. Here m and n
Working alone, A takes 8 hours more to complete the are coprime.
work than if both worked together. On the other hand,
working alone, B would need 9/2 hours more to complete 22. In convex quadrilateral ABCD, ∠ ADC = 90° + ∠ BAC. Given
the work than if both worked together. How much time that AB = BC = 17, and CD = 16. If A is the maximum
would they take to complete the job working together? possible area of the quadrilateral, find [A/10]. Here [x]
represents greatest integer less than or equal to x.
11. The real numbers x, y, z satisfy
xyz = -4 23. Let γ 1, γ 2, γ 3 be three circles with radii 3, 4, 9,
(x + 1)(y + 1)(z + 1) = 7 respectively, such that γ 1 and γ 2 are externally tangent at
(x + 2)(y + 2)(z + 2) = -3. C, and γ 3 is internally tangent to γ 1 and γ 2 at A and B,
Find the absolute value of (x + 3)(y + 3)(z + 3). respectively. Suppose the tangents to γ 3 at A and B
intersect at X. The line through X and C intersect γ 3 at two
12. For how many primes p, there exist an integer n such that a √3
points, P and Q. If PQ = , find a + b. Here a and b are
p divides n3 + 3 and n5 + 5. b

coprime.
24. Ninety-four bricks, each measuring 4" × 10" × 19", are to 27. Number of real values of k such that r4 + kr3 + r2 + 4kr +
be stacked one on top of another to form a tower 94 16 = 0 is true for exactly one real number r.
bricks tall. Each brick can be oriented so it contributes 4"
or 10" or 19" to the total height of the tower. If n be 28. Find the number of pairs of non-negative integers (a, b)
number of different tower heights can be achieved using
such that the polynomials P(x) = x4 + 2ax2 + 4bx + a2 and
all ninety-four of the bricks, find sum of digits of n.
Q(x) = x3 + ax + b have two distinct common real roots.
25. Six musicians gathered at a chamber music festival. At
each scheduled concert some of the musicians played 29. An integer N is worth 1 point for each pair of digits it
while the others listened as members of the audience. contains that forms a prime in original order. For example
What is the least number of such concerts which would 6733 is worth 3 points (for 67, 73, and 73 again), and 2030
need to be scheduled so that for every two musicians each worth 2 points (for 23 and 03). If N be the smallest positive
must play for the other in some concert? integer that is worth exact 11 points, find sum of digits of
N. [Note: Leading zeros are not allowed in the original
26. Given a rational number, write it as a fraction in lowest integer.)
terms and calculate the product of the resulting
numerator and denominator. For how many rational 30. Ten positive integers are arranged around a circle. Each
numbers between 0 and 1 will 18! be the resulting number is one more than the greatest common divisor of
product? its two neighbors. What is the sum of the ten numbers?
Answer Key

1 2 3 4 5

27.00 75 30 73.00 27

6 7 8 9 10

25 40.00 23.00 89.00 6

11 12 13 14 15

28.00 2.00 9.00 12.00 13.00

16 17 18 19 20

2.00 37 45 60 27.00

21 22 23 24 25

41.00 26.00 79.00 15 4.00

26 27 28 29 30

64 2.00 0 14.00 28.00

Solutions

1. In general, we let AB = c, AC = b and BC = a. Let AD be the bisector of ∠ A. Using the angle bisector theorem, BD/CD= c/b.
Thus BD= ac/(b+c) and CD = ab/(b+c)
By Stewart’s theorem.
ab 2 ac 2 2 abc 2
( )c + ( )b = aAD + a ,
2
b+c b+c (b+c)

Which after solving AD2 gives

2
2 a
AD = bc [1 − ( ) ]
b+c

AD2 = 36
AD = 6 = AC
2(AE2 + CD2/4) = AD2 + AC2
4AE2 + 9 = 2(36 + 36)
4AE2/5 = 135/5 = 27

2. Notice that all odd numbers can be obtained by using (a + 1)2 – a2 = 2a + 1, where a is a nonnegative integer. All multiples of 4
can be obtained by using (b + 1)2 – (b – 1)2 = 4b, where b is a positive integer. Numbers congruent to 2 (mod 4) cannot be
obtained because squares are 0, 1 (mod 4). Thus, the answer is 500 + 250 = 750.

3.
∧
∧ ∧

B is on the circle with midpoint C and radius 1. Then Areaches its maximum value if AB is tangent to the circ le (sin A =
sin B

2
∧ ∧ ∧ ∧

reaches a maximum if sin B reaches a maximum, i.e., if B = 90o. ) In this case, sin A= 1/2, giving A = 30o.

4. Define a = x + y and b = xy. Then a + b = 71 and ab = 880. Solving these two equations
yields a quadratic: a2 – 71a + 880 = 0, which factors to (a – 16)(a – 55) = 0. Either
a = 16 and b = 55 or a = 55 and b = 16.
For the first case, it is easy to see that (x, y) can be (5,11) (or vice versa).
In the second case, since all factors of 16 must be ≤ 16, no two factors of 16 can sum greater than 32, and so there are no
integral solutions for (x, y). The solution is 52 + 112 = 146.

5.

∆ AEC and ∆ CDB are both equilateral and congruent with sides 2 units
∆ DCE is also equilateral triangle with side 2 cm
∴ ED = 2
But AB = AC + CB = 4
∴ area of ABDE = ∆ ACE + ∆ CED + ∆ BCD
1 √3
=3× × ( × 2) × 2 = 3√3 sq. unit.
2 2
6. The total of all individual scores is:
1 + 2 + _ _ _ + 14 = (14 x 15)/2 = 105:
So if one team score is x, then the other team score is 105 – x. If x is the winning score, then
x < 105 – x
2x < 105
x < 52.5
The minimum winning score is 1 + 2 ……+7 = 28 and the maximum winning score is 52. So the number of possible winning
scores is 52 – 27 = 25

7.

8. Let b be the last digit of the number n and let a be the number obtained from n by erasing the last digit b. Then n = 10a + b.
Since a is a divisor of n, we infer that a divides b. a is a digit and n is one of the numbers 11, 12, .... 19, 22, 24, 26, 28, 33, 36,
39, 44, 48, 55, 66, 77, 88 or 99.

9. let a = xy and b = x + y. Then a4b = 810….(i)

2

Notice that x3 + y3 = (x + y)(x2 – xy + y2) = b(b2 – 3a). Now, if we divide the second equation by the first one, we get 7/6= ;
b −3a

a
2 3
5
then . Therefore, a . Substituting a equation (i) gives us b ;
b 25 6 2 3
= = b =
a 6 25 2

Then a3 = 33 x 2
2b(b2 – 3a) + a3 = 35 + 54 89.

10. If A & B together take time t hr. If A work alone take t + 8 hr. to complete the work
Work/hr by A is . Similarly for B work/hr is
1 1

t+8 t+9/2

When A & B work together then work/hr

1 1
+
t+8 t+9/2

Work will be done in t hr.

1 1
t( + ) = 1
t+8 t+9/2

By solving
t=6

11.

12. It is easy to see that p divides n15 + 125 and n15 + 243. So p| 118 which means p = 2, 59

13. The 5 and 7 can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of 5
or 7. There are 3 ways to paint each, giving us 9 ways to paint both. The 2 is the most restrictive number. There are 3 ways to
paint 2, but without loss of generality, let it be painted red 4 cannot be the same color as 2 or 8, so there are 2 ways to paint
4, which automatically determines the color for 8, 6 cannot be painted red, so there are 2 ways to paint 6, but WLOG, let it be
painted blue. There are 2 choices for the color for 3, which is either red or green In this case Lastly, there are 2 ways to
choose the color for 9.
9 . 3 . 2 . 2 . 2. 2 = 432.

14. \[\left( {n - 1} \right) \times 2 \times \left( {\begin{array}{*{20}{c}}

{n - 2} \\
{k - 2}
\end{array}} \right) \times \left( {k - 2} \right)! = \left( {\begin{array}{*{20}{c}}
n \\
{k - 2}
\end{array}} \right) \times \left( {k - 2} \right)!\] Then 2 =
n

(n−k+1)(n−k+2)

Thus is 2n2 − (4k − 6) n + (2k

2
− 6k + 4 − n) = 0 . We can solve
√8k−7−5
n = k + (n > k)
4

Note that the square of any odd number has the form 8k–7. Choose k so that √8k − 7 − 5 = 4

i.e., k = 11. Then n = 12

15. We could clear out the denominators by multiplying, though that would be unnecessarily tedious.
To simplify the equation, substitute a = x2 – 10x – 29 (the denominator of the first fraction). We can rewrite the equation as
. Multiplying out the denominators now, we get:
1 1 2
+ − = 0
a a−16 a−40

(a – 16)(a – 40)+ a(a – 40)– 2(a)(a – 16) = 0

Simplifying, –64a + 40 × 16 = 0, so a = 10. Re-substituting,
10 = x2 – 10x – 29 ⇔ 0 = (x – 13) (x + 13). The positive root is 13.

16.

17.

We will use the 60o clockwise rotation around A. Let P’ be the image of P through this rotation. Since BP’ = CP = 5 and PP’ = AP
= 3, PP’B is a right triangle with ∠ BPP’ = 90o. Also, in the equilateral triangle AP’P, the measure of the angle ∠ P’PA is 60o;
hence ∠ APB = 150o. Applying the law of cosines in the triangle APB, we obtain AB2 = 32 + 42 + 2. 3. 4 . √ 3/2; hence AB =
√25 + 12√3 .

18. Since ∠ BNQ = ∠ BCQ ⇒ ∠ BNQ = 90o

∴ BCQN are concylic
So ∠ CNQ = ∠ NCQ
B
=
2

Hence ∠ MCP =
A B o
+ ∠M CN + = ∠ACB = 90
2 2

We have ∠MCN = 45o

19.
20.

21.

22.

23.
24. We have the smallest stack, which has a height of 94 × 4 inches. Now when we change the height of one of the bricks, we
either add 0 inches, 6 inches, or 15 inches to the height. Now all we need to do is to find the different change values we can
get from 94 0s, 6's, and 15's. Because 0, 6, and 15 are all multiples of 3, the change will always be a multiple of 3, so we just
need to find the number of changes we can get from 0's, 2s, and 5's.
From here, we count what we can get:
0, 2 = 2, 4 = 2 + 2, 5 = 5, 6 = 2 + 2 + 2, 7 = 5 + 2, 8 = 2 + 2 + 2 + 2, 9 = 5 + 2 + 2,...
it seems we can get every integer greater or equal to four; we can easily deduce this by considering parity or using the
Chicken McNugget Theorem, which says that the greatest number that cannot be expressed in the form of 2m + 5n form, n
being positive integers is 5 × 2 – 5 – 2 = 3, But we also have a maximum change (94 × 5), so that will have to stop somewhere.
To find the gaps, we can work backwards as well. From the maximum change, we can subtract either 0s, 3's, or 5's. The
maximum we can’t get is 5 × 3 – 5 – 3 = 7, so the numbers 94 × 5 – 8 and below, except 3 and 1, work. Now there might be
ones that we haven’t counted yet, so we check all numbers between 94 × 5 – 8 and 94 × 5. 94 × 5 – 7 obviously doesn't work,
94 × 5 – 6 does since 6 is a multiple of 3, 94 × 5 – 5 does because it is a multiple of 5 (and 3), 94 × 5 – 4 doesn’t since 4 is not
divisible by 5 or 3,94 × 5 – 3 does since 3 = 3, and 94 × 5 – 2 and 94 × 5 – 1 don't, and 94 × 5 does.
Thus the numbers 0, 2, 4 all the way to 94 × 5 – 8, 94 × 5 – 6. 94 × 5 – 5, 94 × 5 – 3, and 94 × 5 work. That's 2 + (94 × 5 – 8 – 4 +
1) + 4 = 465 numbers. That's the number of changes you can make to a stack of bricks with dimensions 4 × 10 × 19, including
not changing it at all.

25.

26. For a fraction to be in lowest terms, its numerator and denominator must be relatively prime. Thus any prime factor that
occurs in the numerator cannot occur in the denominator, and vice-versa. There are seven prime factors of 18!, namely 2 ,3, 5,
7, 11, 13, and 17. For each of these prime factors, one must decide only whether it occurs in the numerator or in the
denominator. These seven decisions can be made in a total of 27 = 128 ways. However, not all of the 128 resulting fractions
will be less than L Indeed, they can be grouped into 64 pairs of reciprocals, each containing exactly one fraction less than 1.
Thus the number of rational numbers with the desired property is 64.

27.

28. Suppose x1 and x2 are real roots of P(x) and Q(x) (x1 ≠x2). They are also the roots of
T(x) = P(x) − xQ(x) = ax2 + 3bx+ a2.
Thus a ≠0 and the discriminant D of the trinomial T(x) must be positive: D = 9b2 −4a3 > 0.
Also x1 + x2 = −3b/a and x1x2 = a.
Since, by assumption, x1 ≠x2 and Q(x1) = Q(x2) = 0, we obtain
\[\begin{gathered}
{\text{0}} = \frac{{Q\left( {{x_1}} \right) - Q\left( {{x_2}} \right)}}{{{x_1} - {x_2}}} \\
{\text{ }} = \frac{{x_1^3 - x_2^3 + a{x_1} - a{x_2}}}{{{x_1} - {x_2}}} \\
{\text{ }} = x_1^2 + {x_1}{x_2} + x_2^2 + a \\
{\text{ = }}{\left( {{{\text{x}}_{\text{1}}}{\text{ + }}{{\text{x}}_{\text{2}}}} \right)^2} - {x_1}{x_2} + a \\
{\text{ = }}{\left( { - {\text{3b/a}}} \right)^2} \\
\end{gathered} \]
So b = 0 and 0 < D = −4a3, i.e., a < 0.
And conversely, if b = 0 > a, then the polynomials
P(x) = x4 + 2ax2 + a2 = (x2 + a)2 and Q(x) = x3 + ax = x(x2 + a)
Have the common roots √−a and −√−a.Thus the pairs sought are those of the form (a,0) with a < 0.
As a and b are non-negative integers, therefore no solution exist.
29.

30. First note that all the integers must be at least 2, because the greatest common divisor of any two positive integers is at least
1. Let n be the largest integer in the circle. The greatest common divisor of its two neighbors is n – 1. Therefore, each of the
two neighbors is at least n – 1 but at most n, so since (n – 1) ł n for n – 1 ≥ 2, they must both be equal to n – 1. Let m be one of
the numbers on the other side of n – 1 from n. Then gcd(n, m) = n – 2. Since n – 2 ≥ 0, (n – 2)|n only for n = 3 or 4. If n = 3,
each number must be 2 or 3, and it is easy to check that there is no solution. If n = 4, then it is again not hard to find that
there is a unique solution up to rotation, namely 4, 3, 2, 2, 3, 4, 3, 2, 2, 3. The only possible sum is therefore 28.