IOQM Test 8

Questions
12. We have a 9 by 9 chessboard with 9 kings (which can move
Maths to any of 8 adjacent squares) in the bottom row. What is the
1. PQRS is a trapezium with PQ = 6cm, SR = 10 cm, PS = QR. minimum number of moves, if two pieces cannot occupy the
The distance between PQ and SR is 12 cm. If r is the radius same square at the same time, to move all the kings into an
of smallest circle covering PQRS. Find (9r2 )/13 X shape (a 5 × 5 region where there are 5 kings along each
diagonal of the X, as shown below)?
2. Let a, b, c and d be Integers with a < 2b, b < 3c, c < 4d, d <
100. If the largest possible value of a is m, find sum of digits
of m.

3. How many 5-digit natural numbers formed by 1, 2, 3, 4 such

13. Define the functions
that the number is divisible by 9 and contained exactly two 5 4 3 2
f (x) = x + 5x + 5x + 5x + 1
fours.
5 4 3 2
g (x) = x + 5x + 3x − 5x − 1

4. On holiday in the Pacific, Julie is about to send postcards to Find number of prime numbers p for which there exists a
friends in Australia, New Zealand and New Caledonia. In the whole number 0 ≤ x < p, such that both f(x) and g(x) are
local money, stamps for Australia cost 50c, for New Zealand divisible by p.
60c and for New Caledonia 80c. Half the postcards are going
to Australia and the total cost of stamps will be 14. AX, BY, CZ are the perpendiculars from the vertices of a
$14(1$ = 100c). How many postcards is she sending? ∆ABC upon the opposite sides. if R = 3r and the ratio of the
perimeters of the ∆ABC and ∆XYZ = k : 1. Find k. Here R and
5. What is the remainder, in base 10, when 247 + 3647 + 437 + r are the circumradius and inradius of ∆ABC.
127 + 17 is divided by 6?
15. let k(n) be the product of the digits of n excluding the digit 0.
(If n is single digit number then k(n) = n) Let S = k(1) + k(2) +
6. Let M, P1, P2, P3, P4, Q1, Q2, Q3, Q4, Q5 be ten points on a
k(3) + ……. + k(999). If P is the greatest prime factor of S, find
circle such that MP1P2P3P4 is a regular pentagon and
sum of digits of P.
MQ1Q2Q3Q4Q5 a regular hexagon, and Q1 on minor arc
MP1. Let Q5Q3 intersect Q1P2 at M1 and let Q5P3 intersect 16. Find the number paths that starts from white square in the
MQ3 at M2. Determine the degree measure of ∠ MM2M1. top row, uses only diagonal moves, and ends in white
square in the bottom row.
7. In a 5-digit number, the sum of the digits is 20. The sum of
the first and last digits is equal to the fourth digit form left.
Sum of first two digits is equal to that of the last two digits.
Finally, the sum of the second and fifth digits from left is
twice the sum of the first and fourth digit from left. What is
the middle digit of the number?

8. If N = 1015 + 1035 + ……. + 1995. Find the remainder when N

is divided by 4.

9. Yashica and Baani are running around a 300 m track.

17. In a unique hockey series between India and Pakistan, they
Yashica runs clockwise at 6 m/s and Baani runs
decide to play on till a team wins 6 matches. If “n” be the
counterclockwise at 9 m/s. They start at the same spot on
number of ways in which the series can be won by India, if
the track and run for 10 minutes. How many times do they
no match ends in a draw. Find n/6
meet each other after they start running?

18. Find number of integer solutions (x, y) of the equation 2x3 +

10. Triangle PQR satisfies ∠ Q > ∠ R. Let M be the midpoint of
QR, and let the perpendicular bisector of QR meet the xy – 7 = 0.
circumcircle of ∆ PQR at a point S such that points P, S, R,
and Q appear on the circle in that order. Given that ∠ PSM = 19. If N be the number of positive integers n < 500 such that n2
56° and ∠ SPR = 61°, find ∠ Q in degrees. + 5n + 6 is divisible by 6. Find Sum of digits of N.

11. Let ω be a circle, and let ABCD be a quadrilateral inscribed in 20. Compute the largest integer n such that 2n2 - 23n + 56 is a
ω . Suppose that BD and AC intersect at a point E. The positive prime number.
tangent to ω at B meets line AC at a point F, so that C lies
between E and F. Given that AE = 6, EC = 4, BE = 2, and BF =
12, find DA2/2.
21. Given the product of n – 4 consecutive positive integers is
equal to n!. For example, 5! = 4 . 5 . 6 written as product of
three consecutive integers. If the largest value of n is M, find
sum of digits of M.

22. Number of integers value of n such that n4 – n2 + 37 is the

square of an integer

23. In a regular hexagon ABCDEF, P, Q, R, S, T and U be the

midpoints of sides AB, BC, CD, DE, EF, and AF, respectively.
The segments AQ, BR, CS , DT , EU and F P bound a
¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯ ¯
¯¯¯
¯¯¯¯ ¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯ ¯¯
¯¯¯
¯¯¯

smaller regular hexagon. Let the ratio of the area of the

smaller hexagon to the area of ABCDEF be expressed as a
fraction m/n where m and n are relatively prime positive
integers. Find m + n.

24. Find the number of positive integers x = 2n3m where n and

m are non-negative integers, for which x6 is not a divisor of
6 x.

25. find x in degree.

26. We define n! as the product of the integers from 1 to n, and

0! Is defined as 1. Note that 40585 = 4! + 0! + 5! + 8! + 5!
There is one three-digit number that can be written in this
way. The sum of its three digits is

27. In an isosceles triangle PQR, PQ = QR, ∠Q = 20° M, N are on

PQ and QR respectively such that ∠MRP = 60°, ∠NPR = 50°.
Find ∠NMR in degrees.

28. Let x and y be real numbers satisfying x4y5 + y4x5 =

810 and x3y6 + y3x6 = 945. Evaluate 2x3 + (xy)3 + 2y3.

29. Number of six-digit numbers (a1a2a3a4a5a6)10 formed by

using the digits 1, 2, 3, 4, 5, 6 once each such that the
number (a1a2…ak)10 is divisible by k for 1 ≤ k ≤ 6.

30. Let P be an interior point of triangle ABC and extend lines

from the vertices through P to the opposite sides. Let a, b, c,
and d denote the lengths of the segments indicated in the
figure. Find the value of abc/21 if a + b + c = 43 and d = 3.
Answer Key

1 2 3 4 5

37.00 18.00 0.00 24.00 0

6 7 8 9 10

12.00 2.00 0.00 30.00 95.00

11 12 13 14 15

84.00 18.00 2.00 3.00 4.00

16 17 18 19 20

24.00 77.00 4.00 9.00 9.00

21 22 23 24 25

11.00 2.00 11.00 6.00 20

26 27 28 29 30

10 30.00 89.00 2.00 21.00

Solutions

1. The smallest circle covering PQRS is the circumcircle of PQRS. Locate the midpoint of SR at the origin. By, symmetry the
circumcenter of PQRS is located at a point (0, y) that is equidistant from Q and R. Hence the circumradius r of PQRS satisfies
2 2 2
r = (12 − y) + 3

= y
2
− 24y + 12
2
+ 3
2
(1)
r
2
= y
2
+ 5
2
(2)
, by (1) – (2)
2 2 2
∴ 0 = −24y + 12 + 3 − 5

24y = 122 –16

2 16
y = 6 − =
3 3

2
16
∴ r2 = ( ) + 5
2

2
1 2 2
= ( ) (16 + 15 )
3

2
1
= ( ) (256 + 225)
3

2
1
= ( ) .481
3

1
∴r = √481
3

2
(9r )/13 = 37

2. If d < 100. then d ≤ 99. Then c < 4d ≤ 4 • 99 = 396. so c ≤ 395. Then b < 3c ≤ 3 • 395 = 1185, so b ≤ 1184. Finally, a < 2b ≤ 2 • 1184
= 2368, so a ≤ 2367. Furthermore, the values a = 2367, b = 1184, c = 395, and d = 99 satisfy the conditions in the problem, so the
maximum value of a is 2367.

3. Let S(n) be the sum of the digits of n. For any admissible n we observe that 11 ≤ S(n) ≤ 17 and hence there is no value of S(n) that
is a multiple of 9. Thus no such n exists.

4. Say that Julie is sending c postcards to New Caledonia and z to New Zealand. Then she is sending c + z to Australia. The total cost
will be 50(c + z) + 60z + 80c = 130c +110z cents. This must equal $14 = 1400 cents, so we have 130c + 110z = 1400, or equivalently
13c + 11z = 140.
Observe that 13 – 11 = 2 and 10 .13 = 130, so that we have
140 = 5(13 – 11) +10 . 13
= 15 . 13 – 5 . 11
= (15 – 11t)13 + (–5 + 13t)11.
Hence, the general solution over the integers is
C = 15 – 11t, z = –5 + 13t,
Where t ∈ Z, for which only t= 1 gives c and z positive. Thus, we find the only solution is c = 4 and z = 8 so the total number of
stamps is 4 + 8 + 12 = 24

5. 247 + 3647 + 437 + 127 + 17 = 2.7 + 4 + 3.72 + 6.7 + 4 + 4.7 + 3 + 1.7 + 2 + 1

Remainder of (2 + 4 + 3 + 6 + 4 + 4 + 3 + 1 + 2 + 1)/6 = 0
6. Note that MQ3 is a diameter of the circle. As a result, P2, P3 are symmetric with respect to MQ3 as are Q1, Q5. Therefore, Q1P2 and
Q5P3 intersect along line MQ3, so in fact, Q1,P2,M1,M2 are collinear.
We now have
ˆ ˆ ∘ ∘
M Q1 −Q3 P2 60 −36 ∘
∠M M2 M1 = ∠M M2 Q1 = = = 12
2 2

7. Let the four digit number be abcde.

a +b + c + d + e = 20…(i)
a + e = d…(ii)
a + b = d + e…(iii)
b + e = 2(a + d)….(iv)
From (ii) and (iii), b = 2e.
From (ii) and (iv), 3b = 4d + e ⇒ 5e = 4d
Now e can be 4 or 8. But if e = 8, then d = 10 is not possible. So e = 4 which gives d = 5.
b = 8, a = 1, c = 2

8. (2n – 1)5 ≡ (2n – 1) mod 4

100

∑ (2r − 1) ( mod 4)
r=51

≡ 1002 – 502 mod 4 ≡ 0 mod 4

∴ Remainder is 0.

9. Yashica runs 6 × 60 × 10 = 3600 m, while Baani run 5400 m. Because they run in opposite directions, they pass each other every
time they run 300 m combined. Thus, they meet each other (3600 + 5400)/300 = 30 times.

10.

Extend SM to hit the circumcircle at E. Then, note that since PSEQ is a cyclic quadrilateral, ∠ PQE = 180° – ∠ PSE = 180° – ∠ PSM =
180° – 56° = 124°.
We also have ∠ MER = ∠ SER = ∠ SPR = 61°. But now, since M is the midpoint of QR and since EM ⊥ QR, triangle QER is isosceles.
This implies that ∠ QEM = ∠ MER = 61°, and ∠ MQE = 90° – ∠ MQE = 90° – ∠ MEQ = 29°. It follows that ∠ Q = ∠ PQE – ∠ MQE =
124°–29° = 95°.

11.

12. The optimal X is centered and touching the bottom. Every king can only move up at most 1 tile per move. If we count the number
of "up" moves necessary, we find that we need at least 18 up moves (with equality if the X is touching the bottom). It is easily
achievable by just moving the
pieces greedily, hence the answer is 18.

13. The only such primes are p = 5,17 Note that

f (x) + g (x) = 2x (x + 1) (x + 4).
3

Thus if P divides f(x) and g(x), it divides either 2, x, x + 1 or x + 4 as well. Since f(0) = 1 and f(1) = 17, we can’t have p = 2. If p divides
x then f(x) ≡ 1 (mod p), also impossible. If p divides x + 1 then f(x) ≡ 5 (mod p), so p divides 5, and x = 4 works. If p divides x + 4
then f(x) ≡ 17 (mod p), so divides 17, and x = 13 works.
14.

Clearly AO⊥ZY, Similarly OB⊥ZX and OC⊥XY

1 1 1
[ABC] = Δ = AO. ZY + BO. ZX + CO. XY
2 2 2

1
Δ = (ZX + XY + ZY ). R
2

2Δ
ZX + XY + ZY =
R

(ZX+XY +ZY ) 2Δ r
= =
AB+BC+AC 2sR R

15. Suppose we write each number in the form of a three-digit number (so 5 ≡ 005), and since our k(n) ignores all of the zero-digits,
replace all of the 0’s with ’s. Now note that in the expansion of (1 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)(1 + 1 + 2 + 3 + …… + 9)(1 + 1 + 2
+ 3 + ……..+ 9)
we cover every permutation of every product of 3 digits, including the case where that first 1 represents the replaced 0s.
However, since our list does not include 000, we have to subtract 1. Thus, our answer is the largest prime factor of
(l + l + 2 + 3 + - + 9)3 – l = 463 – l = (46 – 1)(462 + 46 + 1) = 15 • 3 • 103 • 21
Largest prime = 103
Sum of digits = 4

16. We determine the number of ways to get to each white square in the grid obeying the given rules.
In the first row, there is 1 way to get to each white square: by starting at that square.
In each row below the first, the number of ways to get to an white square equals the sum of the number of ways to get to each of
the white squares diagonally up and to the left and up and to the right from the given square. This is because any path passing
through ay white square needs to come from exactly one of these white squares in the row above. In the second row, there are 2
ways to get to each white square: 1 way from each of two squares in the row above:
In the third row, there are 2 ways to get to each of the outside white squares and 4 ways to get to the middle white square.
Continuing in this way, we obtain the following number of ways to get to each white square:

Since there are 6,12 and 6 ways to get to the white squares in the bottom row, then there are 6 + 12 + 3 = 24 paths through the
grid that obey the given rules.
17. The last match has to be a win(w)
They can loose(L) at most 5 matches

Case Arrangements Number of ways

6w 6w 1

6w + 1L 5w + 1L 6
C5 = 6

(The las will be a win)

6w + 2L 5w + 2L 7C = 21
5

6w + 3L 5w + 3L 8C = 56
5

6w + 4L 5w + 4L 9C = 126
5

6w + 5L 5w + 5L 10C = 252
5

Total = 462

18. Since x(2x2 + y) = 7, the number x must be an integer divisor of number 7, that is, must be equal to one of the numbers 1, 7, –1, –
7. Substituting there values to the equation, we obtain for y the values 5, –97, –9, –99. Thus, our equation has four solutions in
integers, namely (1, 5), (7, –97), (–1, –9), (–7, –99).

19. n2 + 5n + 6 ≡ 0 (mod 6)
⇔ (n +2)(n +3) ≡ 0 mod 2 and (n + 2)(n + 3) ≡ 0 mod 3
⇔ (n + 2)(n + 3) ≡ 0 mod 3
Which is true for all n except n = 3k + 2, here k is a whole number.
∴ Required number N = 500 - 167 = 333
Sum of Digits = 9

20. 2n2 - 293n + 56 = (2n - 7)(n - 8)

Because 2n – 7 and n – 8 are both integers, in order for their product to be prime, one factor must equal 1 or -1, so n = 3, 4, 7 or 9.
Checking these possibilities form the greatest downward, n = 9 produces 11.1 = 11, which is prime. So, the answer is 9.

21. Largest positive integer n for which n! can be expressed as the product of n – a consecutive positive integers is (a + 1)! – 1
For ex. largest n such that product of n – 6 consecutive positive integers is equal to n! is 7! - 1 = 5039.
Proof: let the largest of the n – a consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of
n – a consecutive positive integers will be less than n!.
Now, observe that for n to be maximum the smallest number (or starting number) of the n – a
consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is n + 1.
So the n – a consecutive positive integers are a + 2, a + 3, ... , n + 1
(n+1)!
So we have = n! ⇒ n + 1 = (a + 1)! ⇒ n = (a + 1)! – 1
(a+1)!

In our case a = 4, so n = 5! - 1 = 119

22. (n2 – 1)2 = n4 – n2 + 37 < n4 for n < -6 or n > 6

⇒ Possible values of n are -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6
⇒ n = -2, 2,
23.

Let M be the intersection of AH and BR

and N be the intersection of BR and CS.
Let O be the center.
Let BC = 2 (without loss of generality).
Note that ∠ BMQ is the vertical angle to an angle of regular hexagon, and so has degree 120° .
Because ∆ ABQ and ∆ BCR are rotational images of one another, we get that ∠ MBQ = ∠ QAB and hence ∆ ABQ ~ ∆ BMQ ~ ∆ BCR.
Using a similar argument, NR = MQ, and
MN = BR – NR – BM = BR - (BM + MQ)

Applying the law of cosines on ∆ BCR, BR = √22 + 1

2 ∘
− 2 (2) (1) (cos(120 )) = √7

BC+CR 3 BM +M Q
= =
BR √7 BQ

3BQ
BM + MQ =
3
=
√7 √7

3
MN = BR – (BM + MQ) = √7 −
4
=
√7 √7

2 2
Area of smaller hexagon MN 2 4
= ( ) = ( ) =
Area of bigger hexagon BC √7 7

Thus, the answer is 4 + 7 = 11.

24. Substitute x = 2n3m into x6 and 6x, and find all pairs of non-negative integers (n, m) for which (2n3m)6 is not a divisor of 62
n m
3

Simplifying both expressions:

m m
n 3 n 3
2
6n
is not a divisor of 22
.3
6m
.3
2

Comparing both exponents (noting that there must be either extra powers of 2 or extra powers of 3 in the left expression):
n m n m
6n > 2 3 OR 6m > 2 3

Using the first inequality 6n and going case by case starting with n ∈
n m
> 2 3 {0, 1, 2, 3.....} :

n=0: 0 > 1 × 3m which has no solution for non-negative integers m

m
n=1: 6 > 2 × 3 which is true for m=0 but fails for higher integers ⇒ (1, 0)
n=2: 12 > 4 × 3m which is true for m=0 but fails for higher integers ⇒ (2, 0)
n=3: 18 > 8 × 3m which is true for m=0 but fails for higher integers ⇒ (3, 0)
n=4: 24 > 16 × 3m which is true for m=0 but fails for higher integers ⇒ (4, 0)
n=5: 30 > 32 × 3m which has no solution for non-negative integers m
There are no more solutions for higher n, as polynomials like 6n grow slower than exponentials like 2n.
Using the second inequality 6m > 2n 3m and going case by case starting with m ∈ {0, 1, 2, 3...}:
m=0: 0 > 2n × 1 which has no solution for non-negative integers
m=1: 6 > 2n × 3 which is true for n=0 but fails for higher integers ⇒ (0,1)
m=2: 12 > 2n × 9 which is true for n=0 but fails for higher integers ⇒ (0,2)
m=3: 18 > 2n × 27 which has no solution for non-negative integers n
There are no more solutions for higher m, as polynomials like 6m grow slower than exponentials like 3m.
Thus, there are six numbers corresponding to (1, 0), (2, 0), (3, 0), (4, 0), (0, 1), and (0, 2). Plugging them back into the original
expression, these numbers are 2, 4, 8, 16, 3, and 9, respectively. Their sum is 42.
25. Mark a point F on AB, such that AF = a and BF = b. Draw FG such that FG = FC.

Clearly BCF is equilateral triangle.

If we consider ∆ AGF and ∆ ADE
∆ AFG ≅ ∆ DAE by SAS
⇒ ∠ EDA = ∠ GAF = 20°

26.

27.

Introduce MK || PR and connect PK. Suppose that RM, PK intersect at O, then both triangles OMK and OPR are equilateral. We
have
∠ PNR = 180° - 80° - 50° = 50°,
OR = PR = RN.
∴ ∠ NRO = 80° - 60° = 20
∵∠ NOR = ½ (180° - 20°) = 80°,

∴ ∠ KON = 120° - 80° = 40°.

∵∠ OKN = 180° - 60° - 80° = 40°,

We obtain ON = KN, hence ∆ ONM ≅ ∆ KNM (S.S.S.), i.e.

1 ∘
∠N M C = ∠N M O = ∠KM O = 30
2

28. 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
b −3a
;
7
=
6 a
2
b 25 6
then . Therefore, a .
2
= = b
a 6 25
3
5
Substituting a in equation (i) gives us b3 =
2
;

Then a3
3
= 3 × 2

2b(b2 – 3a) + a3 = 35 + 54 = 89
29.

30. Call the cevians AD. BE, and CF. Using area ratios (PBC and ABC have the same base), we have:
d [P BC]
=
a+d [ABC]

[P CA] [P AB]
Similarily, and
d d
= = .
a+d [ABC] c+d [ABC]

[P BC] [P CA] [P AB] [ABC]

Then,
d d d
+ + = + + = = 1
a+d b+d c+d [ABC] [ABC] [ABC] [ABC]

Then identity is a form of Ceva’s Theorem.

d d d
+ + = 1
a+d b+d c+d

Plugging in d = 3, we get a+3

3 3 3
+ + = 1
b+3 c+3

3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a + 3)(b + 3)(c + 3)

3(ab + bc + ca) + l8(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9 (a + b + c) + 27
9(a + b + c) + 54 = abc = 441
abc/21 = 21