Logical Thinking
1.) Independence Day in the Philippines is celebrated every 12th of June. This year, June 12 is
Wednesday. Which day of the week will the Independence Day of 2050 be?
Answer: Sunday
Solution:
Use the chart beside:
Now we list what we need to find, which is.
12 𝐽𝑢𝑛𝑒, 2050
In using this chart, the year 2050 will be split into two: 20 and 50. This will be useful for later
calculation. Thus we have.
12 𝐽𝑢𝑛𝑒, 20 ↔ 50
The formula in getting the exact day is
𝑛𝑡ℎ 𝑑𝑎𝑦+𝑚𝑜𝑛𝑡ℎ+𝑐𝑒𝑛𝑡𝑢𝑟𝑦+𝑦𝑒𝑎𝑟+𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑜𝑓 𝑦𝑒𝑎𝑟𝑠 𝑖𝑛𝑡𝑜 𝑤𝑒𝑒𝑘𝑠
.
7 (𝑑𝑎𝑦𝑠)
Now, we have:
12(𝑡ℎ 𝑑𝑎𝑦) + 4(𝐽𝑢𝑛𝑒) + 6(21𝑠𝑡 𝑐𝑒𝑛𝑡𝑢𝑟𝑦) + 50(50𝑡ℎ 𝑦𝑒𝑎𝑟) +
50
(𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑜𝑛𝑙𝑦)
4
12(𝑡ℎ 𝑑𝑎𝑦) + 4(𝐽𝑢𝑛𝑒) + 6(21𝑠𝑡 𝑐𝑒𝑛𝑡𝑢𝑟𝑦) + 50(50𝑡ℎ 𝑦𝑒𝑎𝑟) + 12 = 84
Divide the sum by 7 to get the remainder, that will determine the exact date of the week.
84
7
= 12 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 0.
By looking at the days chart, we can conclude that 12 𝐽𝑢𝑛𝑒, 2050 will be on Sunday.
2.) Use decimal number system to represent septimal number 20247.
Answer: 704
Solution:
Heptad is a base-7 numeral system. It uses the digits 0, 1, 2, 3, 4, 5, 6 to represent any real
number. To convert 20247 into decimal, we have.
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
3
2
1
0
20247 = [2 · 7 ] + [0 · 7 ] + [2 · 7 ] + [4 · 7 ]
20247 = 686 + 14 + 4
20247 = 704
Hence, heptad number 20247 is 70410 or 704 in the decimal number system.
3.) There are n lines that are not parallel with each other on a plane. There are no 3 intersecting
lines at a point. If they intersect 1035 times, find n.
Answer: 46
Solution:
The number of times n lines intersect such that these lines are not parallel with each other and no
3 lines intersect at a point is given by the formula:
𝑛
𝑛!
𝐶2 = 1035 ⇒ (𝑛−2)!×2! = 1035 ⇒
𝑛(𝑛−1)
2
= 1035
2
𝑛(𝑛 − 1) = 2070 ⇒ 𝑛 − 𝑛 − 2070 = 0 ⇒ (𝑥 − 46)(𝑥 + 45) = 0
Since n must be nonnegative, hence 𝑛 = 46.
4.) If 𝑎1 = 15 and 𝑎𝑚+1 = 𝑎𝑚 + 3𝑎1 for all 𝑚 ≥ 𝑛 and m,n are positive integers, find the value
of 𝑎2024.
Answer: 91050
Solution:
Observe these pattern based on the given condition:
𝑎𝑚+1 = 𝑎𝑚 + 3𝑎1 ⇒ 𝑎
2024
= 𝑎2023 + 3𝑎1
𝑎2024 = 𝑎2022 + 3𝑎1 + 3𝑎1 ⇒ 𝑎2024 = 𝑎2022 + 2 · 3𝑎1
𝑎2024 = 𝑎2021 + 3 · 3𝑎1
This patterns continues until:
𝑎2024 = 𝑎1 + 2023 · 3𝑎1
By this, we can now determine the value of 𝑎2024. Hence, we have:
𝑎2024 = 𝑎1 + 2023 · 3𝑎1
𝑎2024 = 15 + 2023(3 · 15) = 91, 050
The value of 𝑎2024 = 91, 050.
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
5.) There are 23 pairs of white chopsticks, 20 pairs of yellow chopsticks, and 24 pairs of brown
chopsticks mixed together. Close your eyes. If you want to get 3 pairs of chopsticks with
different colors, at least how many pieces(s) of chopsticks is/are needed to be taken?
Answer: 96
Solution:
In order to obtain the least number of chopsticks needed to be taken, we need to get the two
groups of colored chopsticks with the most number of pairs and simply add a pair of chopsticks
from the least colored group. In this case, we need at least
2(24)+2(23)+2= 96
Hence, we need at least 96 pieces of chopsticks to get 3 pairs with different colors.
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
Algebra
6.) Find the value of x such that 𝑥 = 𝑥 − 552.
Answer: 576
Solution:
Observe that when we rearrange the equation, it can be seen as quadratic in 𝑥. That is,
𝑥 = 𝑥 − 552 ⇒ 0 = 𝑥 − 𝑥 − 552
Then Factorize (base from the constant − 552), we have,
( 𝑥 + 23)( 𝑥 − 24) = 0
Now, 𝑥 + 23 will give you a complex solution. Thus 𝑥 − 24 is the only real value for x.
Hence, the value is:
𝑥 − 24 = 0 → 𝑥 = 24
𝑥 = 576
2
7.) Given that x is a real number, find the maximum value of − 𝑥 − 48𝑥 + 1148.
Answer: 1724
Solution:
Get the vertex of the equation:
2
− 𝑥 − 48𝑥 + 1148
𝑏
ℎ =− 2𝑎 =
−(−48)
2(−1)
2
=− 24 ; 𝑘 =
4𝑎𝑐−𝑏
4𝑎
2
=
4(−1)(1148)−(−48)
4(−1)
= 1, 724
Based on the answer, the 𝑥 𝑎𝑥𝑖𝑠 =− 24 and 𝑦 𝑎𝑥𝑖𝑠 = 1, 724 (− 24, 1724). This means the
graph is downward. Hence, the maximum value of the given equation is 1,724.
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
8.) Let x,y,z and w be real numbers satisfying the equations
2
2
Find the value of 𝑧 + 5𝑤 .
Answer: 613
Solution:
2
2
2
2
Focus on the equation 𝑧 + 5𝑤 . This equation has a similarities with 𝑥 + 5𝑦 . With this logic,
we can multiply both of them, so that we can expand and derive equivalent. Thus we have:
Then, we group.
(𝑧2 + 5𝑤2)(𝑥2 + 5𝑦2) ⇒ 𝑥2𝑧2 + 5𝑦2𝑧2 + 5𝑤2𝑥2 + 25𝑤2𝑦2
(𝑧2 + 5𝑤2)(𝑥2 + 5𝑦2) ⇒ (𝑥2𝑧2 + 25𝑤2𝑦2) + (5𝑦2𝑧2 + 5𝑤2𝑥2)
Then, we complete the square of both expressions, such that it is factorable.
(𝑧2 + 5𝑤2)(𝑥2 + 5𝑦2) ⇒ (𝑥2𝑧2 + 10𝑤𝑥𝑦𝑧 + 25𝑤2𝑦2) + (5𝑦2𝑧2 − 10𝑤𝑥𝑦𝑧 + 5𝑤2𝑥2)
(𝑧2 + 5𝑤2)(𝑥2 + 5𝑦2) ⇒ (𝑥𝑧 + 5𝑤𝑦)2 + (5𝑦𝑧 − 5𝑤𝑥)2
(𝑧2 + 5𝑤2)(𝑥2 + 5𝑦2) ⇒ (𝑥𝑧 + 5𝑤𝑦)2 + 5(𝑥𝑤 − 𝑦𝑧)2
2
2
2
2
Since (𝑥 + 5𝑦 ) = 8, we simply substitute the other values to find (𝑧 + 5𝑤 ). Hence, we
have:
2
(𝑧2 + 5𝑤2) · 8 = ( 2024) + 5(− 24)2
= 613
(𝑧2 + 5𝑤2) = 2024+2880
8
2
2
Therefore, 613 is the value of (𝑧 + 5𝑤 ).
9.) It is known that x is real, 𝑥 > 0, and 𝑥 = 20 +
24
24
20+ 20+...
. Find the value of x.
Answer: 10+2 sqrt(31)→ 10 + 2 31
Solution:
Since this is a nested fraction, substitute x to the repeating elements of the fraction. Thus, we
have:
𝑥 = 20 +
24
24
⇒ 𝑥 · 𝑥 = 20 + 𝑥
𝑥
2
2
· 𝑥 ⇒ 𝑥 = 20𝑥 + 24
𝑥 − 20𝑥 − 24 = 0
There is no factor of 24 such that when add up it equals to 20, so we are completing the
square/quadratic formula. We have:
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
2
𝑥 − 20𝑥 + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 24 ; 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 =
2
𝑏 2
2
( )
2
𝑥 − 20𝑥 + 100 = 24 + 100 → (𝑥 − 10) = 124
(𝑥 − 10)
2
= 124 ⇒ 𝑥 − 10 = 2 31
Therefore, 10 + 2 31 is the value of 𝑥, in which the condition is a positive solution.
2
2
10.) Factor 𝑥 − 𝑦 + 26𝑥 + 34𝑦 − 120 completely.
Answer: (𝑥 + 𝑦 − 4)(𝑥 − 𝑦 + 30) 𝑜𝑟 (𝑥 − 𝑦 + 30)(𝑥 + 𝑦 − 4)
Solution:
This is a circle equation, and we can do the following to factorize it completely.
2
2
𝑥 − 𝑦 + 26𝑥 + 34𝑦 − 120
2
2
(𝑥 + 26𝑥 + 1𝑠𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) − (𝑦 − 34𝑦 + 2𝑛𝑑 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) − 120 − 1𝑠𝑡 + 2𝑛𝑑
2
2
(𝑥 + 26𝑥 + 169) − (𝑦 − 34𝑦 + 289) − 120 − 169 + 289
2
2
(𝑥 + 13) − (𝑦 − 17) + 0 ⇒ (𝑥 + 13 − 𝑦 + 17)(𝑥 + 13 + 𝑦 − 17)
2
2
(𝑥 + 13) − (𝑦 − 17) + 0 ⇒ (𝑥 − 𝑦 + 30)(𝑥 + 𝑦 − 4)
Thus, the equation is completely factored.
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
Number Theory
2024
11.) Now it is April. Which month will it be after 2025
Answer: January
Solution:
months?
We use modulo arithmetic, we have:
2024
2025
2024
𝑚𝑜𝑑 12 ≡ (2025 − [12 × 168])
2024
𝑚𝑜𝑑 12
2024
2025
𝑚𝑜𝑑 12 ≡ 9
𝑚𝑜𝑑 12
However, regardless of what the exponent of 9 is, the remainder of the number modulo 12 will
always be equal to 9. For example:
2
9 𝑚𝑜𝑑 12 ≡ 81 𝑚𝑜𝑑12 ≡ (81 − (12 × 6))𝑚𝑜𝑑12 ≡ 9 𝑚𝑜𝑑 12
3
9 𝑚𝑜𝑑 12 ≡ 729 𝑚𝑜𝑑12 ≡ (729 − (12 × 60))𝑚𝑜𝑑12 ≡ 9 𝑚𝑜𝑑 12
4
9 𝑚𝑜𝑑 12 ≡ 6561 𝑚𝑜𝑑12 ≡ (6561 − (12 × 546))𝑚𝑜𝑑12 ≡ 9 𝑚𝑜𝑑 12
2024
…… 𝑡ℎ𝑒 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑒𝑠 𝑢𝑛𝑡𝑖𝑙 9
Thus, we count 9 months after April, and that is January.
2024
12.) Find the remainder when 2039
Answer: 1
Solution:
𝑚𝑜𝑑 12
is divided by 24.
By using the modular arithmetic:
2024
2039
24
1
≡ 𝑎 𝑚𝑜𝑑 24 ⇒
2039
24
2024
≡ (2039 − 2016) × 1
𝑚𝑜𝑑 24
2024
23 × 1
Hence, the remainder is 1.
𝑚𝑜𝑑 24 ≡ (529 − 528)𝑚𝑜𝑑 24 ≡ 1 𝑚𝑜𝑑 24
6
13.) If 𝑘 = 1, 291, 467, 969, find the positive value of k.
Answer: 33
Solution:
6
Assuming that 𝑘 = 1, 291, 467, 969, we know that k is a positive integer. So we are looking
for the 6th root of 1, 291, 467, 969. We take the first four digits and observe that 1,291 is the
number. Find the number less than and greater than, we have:
6
6
3 < 1291 < 4 ↔ 729 < 1291 < 4096
Notice that these numbers are not in billions, so we put the tens digit, which will also represent
the same, but in larger quantities. Also we put 6 zeros to the middle number. We have:
6
6
30 < 1, 291, 467, 969 < 40 ↔ 729, 000, 000 < 1, 291, 467, 969 < 4, 096, 000, 000
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
Hence, k is between 30 and 40 but k is closer to 30 based on the inequality above. Now, we take
a look at the last digit of the powers of 31,32,....39 below (Note, if they have unit digit of 9, we
stop listing them):
𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 31 𝑢𝑛𝑖𝑡 𝑑𝑖𝑔𝑖𝑡: 1
1
31 = 31
2
31 = 961
3
31 = 29, 791; 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠
𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 32 𝑢𝑛𝑖𝑡 𝑑𝑖𝑔𝑖𝑡: 2, 4, 8, 6
1
32 = 32
2
32 = 1, 024
3
32 = 32, 768
4
32 = 1, 048, 576
5
32 = 33, 554, 432; 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠
𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 33 𝑢𝑛𝑖𝑡 𝑑𝑖𝑔𝑖𝑡: 3, 9, 7, 1
1
33 = 33
2
33 = 1, 089
3
33 = 35, 937
4
33 = 1, 185, 921
5
33 = 39, 135, 393; 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠
Since the last digit of 1, 291, 467, 969 is 9, we conclude that its 6 th root is 33, i.e. 𝑘 = 33.
360
14.) What is the largest integral value n that satisfies the inequality 𝑛
Answer: 117
Solution:
540
≤ 24
Observe that the exponent is divisible by 180, we have:
360
540
𝑛
≤ 24
2
𝑛 ≤
→𝑛
360
180
≤ 24
540
180
2
⇒ 𝑛 ≤ 24
3
3
24 ⇒ 𝑛 ≤ 24 24 ⇒ 48 6
Now we approximate 6, which is:
6 > 4 ⇐ 𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑠𝑞𝑢𝑎𝑟𝑒
Then we use this formula:
𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑠𝑞𝑢𝑎𝑟𝑒 +
4+
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑖𝑛𝑠𝑖𝑑𝑒 𝑎 𝑟𝑎𝑑𝑖𝑐𝑎𝑙 (𝑎>𝑏)
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟× 𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑠𝑞𝑢𝑎𝑟𝑒
6−4
2
1
→ 2 + 2×2 → 2 + 2 = 2. 5
(6−4)× 4
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
Lastly, we multiply 2.5 to 48 to approximate the value. Thus, we have.
48 × 2. 5 = 120 ⇒ 120 − 3 = 117
117 is the largest value of n that will satisfy the inequality.
15.) Find the largest 4-digit positive integral solution of the congruence equations
20𝑥≡1 𝑚𝑜𝑑 11
{24𝑥≡6 𝑚𝑜𝑑 19
Answer: 9,828
Solution:
20𝑥≡1 𝑚𝑜𝑑 11
𝑥≡𝑎 𝑚𝑜𝑑 11
We must express {24𝑥≡6 𝑚𝑜𝑑 19 into {𝑥≡𝑏 𝑚𝑜𝑑 19 . In this case, we have the following:
a. To make 20𝑥 ≡ 1 𝑚𝑜𝑑 11 into 𝑥 ≡ 𝑎 𝑚𝑜𝑑 11, we must multiply something to x such
that its remainder will be 1 when divided by 11. The number that we can multiply to 20 is
actually 5→(20 · 5 = 100). When divided by 11, we have the remainder 1. Therefore
the value of 𝑎 = 5.
b. To make 24 ≡ 6 𝑚𝑜𝑑 19 into 𝑥 ≡ 𝑏 𝑚𝑜𝑑 19, we must multiply something to x such that
its remainder will be 6 when divided by 19. The number that we can multiply to 24 is
actually 5→(24 · 5 = 120). When divided by 19, we have the remainder 6. Therefore
the value of 𝑏 = 5.
Thus, we have *by using Chinese Remainder Theorem (CRT):
𝑥≡5 𝑚𝑜𝑑 11
{𝑥≡5 𝑚𝑜𝑑 19
Now, we know that every time, a number divided by 11 or 19 will have a remainder 5. Thus, we
must find the LCM of both prime numbers whose role is as a divisor.
𝑥 ≡ 5 𝑚𝑜𝑑 (𝐿𝐶𝑀 [11, 19]) ⇒ 𝑥 ≡ 5 𝑚𝑜𝑑 209
To find the largest 4-digit positive integral solution of the congruence equations, we must
multiply something to 209 and add 5 (because of their remainder congruence).
a. The smallest 4-digit positive integral solution of the congruence equations is 1,050.
[209 · 5 + 5 = 1, 050]
b. The largest 4-digit positive integral solution of the congruence equations is 9,828.
[209 · 47 + 5 = 9, 828]
Since we were looking for the largest 4-digit positive integral solution of the congruence
equations, therefore 9,828 is the largest 4-digit positive integral solution.
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
Geometry
16.) If the area of a triangle with side lengths 37, 38, and 39 is 57 𝑛, find the value of n.
Answer: 120
Solution:
Using Heron’s Formula, we have:
𝐴𝑟𝑒𝑎∆(𝑔𝑖𝑣𝑒𝑛 𝑎𝑙𝑙 𝑠𝑖𝑑𝑒𝑠) = 𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐); 𝑠𝑒𝑚𝑖𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟(𝑠) =
𝑎+𝑏+𝑐
2
Find the semiperimeter first, before substituting the area equation.
𝑠𝑒𝑚𝑖𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟(𝑠) =
37+38+39
2
= 57
Then, simplify we have.
𝐴𝑟𝑒𝑎∆(𝑔𝑖𝑣𝑒𝑛 𝑎𝑙𝑙 𝑠𝑖𝑑𝑒𝑠) = 𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) → 57(57 − 37)(57 − 38)(57 − 39)
𝐴𝑟𝑒𝑎∆(𝑔𝑖𝑣𝑒𝑛 𝑎𝑙𝑙 𝑠𝑖𝑑𝑒𝑠) = 57(20)(19)(18) → 19 · 3 · 5 · 2 · 2 · 19 · 2 · 3 · 3
𝐴𝑟𝑒𝑎∆(𝑔𝑖𝑣𝑒𝑛 𝑎𝑙𝑙 𝑠𝑖𝑑𝑒𝑠) = 57 120
Therefore, the value of n is 120.
17.) How many time(s) is the exterior angle of a regular 24-sided polygon as much as the interior
angle?
1
Answer: 11
Solution:
The measure of each exterior angle of the 24-sided polygon is
°
360
24
°
= 15 . On The other hand,
°
180(𝑛−2)
180(24−2)
=
= 165
𝑛
24
15
1
= 11 times as the interior angle.
165
the measure of each interior angle of the 24-sided polygon is
Therefore, the exterior angle of the 24-sided polygon is
18.) Find the minimum value of 2 5𝑠𝑖𝑛𝑥 + 2 6𝑐𝑜𝑠𝑥 + 2024 11.
Answer: 2022sqrt(11)→ 2022 11
Solution:
2
(2
2
Using Cauchy-Schwarz inequality [𝐴𝑀 + 𝐵𝑁] ≤ 𝐴 + 𝐵
)(𝑀2 + 𝑁2), observe that:
2
(2 5𝑠𝑖𝑛𝑥 + 2 6𝑐𝑜𝑠𝑥) ≤ (20 + 24)(𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥)
2
2
(2 5𝑠𝑖𝑛𝑥 + 2 6𝑐𝑜𝑠𝑥) ≤ 44 ⇒ (2 5𝑠𝑖𝑛𝑥 + 2 6𝑐𝑜𝑠𝑥) ≤ 44
2
(2 5𝑠𝑖𝑛𝑥 + 2 6𝑐𝑜𝑠𝑥) ≤ 44
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
− 2 11 ≤ 2 5𝑠𝑖𝑛𝑥 + 2 6𝑐𝑜𝑠𝑥 ≤ 2 11 → 𝑎𝑑𝑑 2024 11 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠.
− 2 11 + 2024 11 ≤ 2 5𝑠𝑖𝑛𝑥 + 2 6𝑐𝑜𝑠𝑥 + 2024 11 ≤ 2 11 + 2024 11
2022 11 ≤ 2 5𝑠𝑖𝑛𝑥 + 2 6𝑐𝑜𝑠𝑥 + 2024 11 ≤ 2026 11
Hence, the minimum value is 2022 11.
19.) Find the area enclosed by the x-axis, y-axis, and the straight line
11𝑥 − 13𝑦 = 286.
Answer: 286
Solution:
Solve for x and y-intercept, we have:
𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠:
Equation A: (0, − 22)
11𝑥 − 13𝑦 = 286 ⇒ (0) − 13𝑦 = 286
− 13𝑦 = 286
−13𝑦
−13
286
= −13 ⇒ 𝑦 =− 22
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠:
Equation B: (0, − 22)
11𝑥 − 13𝑦 = 286 ⇒ (0) + 11𝑥 = 286
11𝑥 = 286
11𝑥
11
=
286
⇒ 𝑥 = 26
11
Now, it form a right triangle, since we are looking for the area, thus we have:
2
1
1
1
𝐴∆ = || 2 𝑏 · ℎ|| = || 2 (− 22 · 26)|| = || 2 (− 572)|| = 286𝑢
Therefore, the area enclosed by the x-axis, y-axis, and the straight line 11𝑥 − 13𝑦 = 286 is 286
unit squared.
20.) Find the area of the quadrilateral in the Cartesian plane formed by the points
𝐴(8, 6), 𝐵(− 4, − 3), 𝐶(− 5, 4), 𝐷(11, − 2).
Answer: 108
Solution:
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
Combinatorics
21.) If x and y are positive integers such that 11𝑥 + 23𝑦 = 2024, find the maximum value of
𝑥 + 𝑦.
Answer: 172
Solution:
Convert the 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 (standard form of line equation) into 𝑦 = 𝑚𝑥 + 𝑏
(slope-intercept form), we have:
11𝑥 + 23𝑦 = 2024 ⇒ 23𝑦 = 2024 − 11𝑥
23𝑦
23
=
2024−11𝑥
11
⇒ 𝑦 = 88 − 23 𝑥
23
Since 11 and 23 are coprime, x must be divisible by 23 for y to be an integer. Also, since
𝑥, 𝑦 > 0,
11
11
𝑦 = 88 − 23 𝑥 > 0 → 23 𝑥 < 88 ⇒ 11𝑥 < 2024
𝑥 < 184
Since we are looking for the maximum value of 𝑥 + 𝑦, we need to find the maximum integer
𝑥 < 184 that is divisible by 23 (because we are looking for the maximum value). The value of x,
in which will satisfy the condition 𝑥 < 184 is 161. We need to obtain y, hence:
11
𝑦 = 88 − 23 (161) = 11
Therefore, the maximum value of 𝑥 + 𝑦 = 161 + 11 = 172.
2
22.) Find the number of integers x such that 44𝑥 > 𝑥 + 480.
Answer: 3
Solution:
By rearranging and factoring the inequality, we have:
2
2
44𝑥 > 𝑥 + 480 ⇒ 𝑥 − 44𝑥 + 480 < 0
(𝑥 − 20)(𝑥 − 24) < 0 ⇒ 𝑥 = 20, 24
Here, we consider two cases:
Case 1:
𝑥 − 20 > 0 𝑎𝑛𝑑 𝑥 − 24 < 0 ⇒ (20, 24)
Case 2:
𝑥 − 20 < 0 𝑎𝑛𝑑 𝑥 − 24 > 0 ⇒ (⊘)
Hence, the solutions for the inequality lie in the interval (20, 24)→ 21, 22, 23. Therefore, the
number of the integral solution for the inequality is 3.
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)
23.) A fair 6-sided die is thrown 3 times. Find the probability that the sum of numbers obtained is
8.
7
Answer: 72
Solution:
3
Note that the total number of possible outcomes is 6 = 216. Now, the following are the
outcomes with sum of 8:
(4, 3, 2)6 𝑤𝑎𝑦𝑠 , (5, 2, 1)6 𝑤𝑎𝑦𝑠, (6, 1, 1)3 𝑤𝑎𝑦𝑠, (4, 2, 2)3 𝑤𝑎𝑦𝑠, (3, 3, 2)3 𝑤𝑎𝑦𝑠
Hence, the probability is:
𝑃(𝑠𝑢𝑚 𝑜𝑓 8) =
6+6+3+3+3
216
21
7
= 216 = 𝑃(𝑠𝑢𝑚 𝑜𝑓 8) = 72
24.) Find the number of sets of negative integral solutions of 𝑎 + 𝑏 >− 2024.
Answer: 2,045,253
Solution:
Counting the number of sets of negative integral solutions of 𝑎 + 𝑏 >− 2024 is the same as
counting the number of sets of positive integral solutions of − 𝑎 − 𝑏 < 2024. Now, using the
stars and bars formula where 𝑛 = 2024 and 𝑘 = 2 the number of variables, the number of sets
of positive integral solutions of − 𝑎 − 𝑏 < 2024 is given by:
𝑛−1
𝐶𝑘
2024−1
= 𝐶2
2023
= 𝐶2
2023!
⇒ 2021!×2! = 2, 045, 253
25.) Suppose that 4 cards are drawn from an ordinary poker deck of 52 playing cards one by
one without replacement. Find the probability of getting four of the same letter cards.
4
Answer: 270725
Solution:
The letter cards in a standard deck of cards are A, J, Q, and K with 4 different suits each. Thus,
the probability of drawing four the same letter cards without replacement is:
16
⇒ 𝑡ℎ𝑒 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑙𝑒𝑡𝑡𝑒𝑟 𝑐𝑎𝑟𝑑
52
3
⇒ 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 3 𝑐ℎ𝑜𝑖𝑐𝑒𝑠: 𝐽, 𝑄, 𝐾
51
2
⇒ 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 2 𝑐ℎ𝑜𝑖𝑐𝑒𝑠: 𝐽, 𝑄, 𝐾
50
1
⇒ 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 1 𝑐ℎ𝑜𝑖𝑐𝑒𝑠: 𝐽, 𝑄, 𝐾
49
Hence,
16
3
2
1
96
𝑃(𝑓𝑜𝑢𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑙𝑒𝑡𝑡𝑒𝑟 𝑐𝑎𝑟𝑑𝑠) = 52 × 51 × 50 × 49 = 6497400
4
𝑃(𝑓𝑜𝑢𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑙𝑒𝑡𝑡𝑒𝑟 𝑐𝑎𝑟𝑑𝑠) = 270725
Hong Kong International Mathematical Olympiad Heat Round 2024 (Senior Secondary Set 2)