LOGICAL THINKING

1. Find the average of the sequence

407, 428, 449, . . . , 2003, 2024, 2045.

Answer: 1226
Solution. Arranging the terms in the sequence will give an arithmetic sequence with
common difference 21. Hence, the average a of the finite sequence is given by

1 1
a = (a1 + an ) = (407 + 2045) = 1226.
2 2

2. Define the operation symbol ⊗ as follows:

5 ⊗ 11 = 5 × 7 × 9 × 11
4 ⊗ 14 = 4 × 6 × 8 × 10 × 12 × 14
8 ⊗ 20 = 8 × 10 × 12 × 14 × 16 × 18 × 20.

It is known that n! = n × (n − 1) × (n − 2) × · · · × 3 × 2 × 1. Find the value of

2024!
.
7! × (13 ⊗ 2023)(10 ⊗ 2024)

Answer: 792
Solution. Observe that

2024! 11 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= = 792.
7! × (13 ⊗ 2023)(10 ⊗ 2024) 7×6×5×4×3×2×1

3. Use undecimal number system (base-11) to represent quinary number 2024035 .

Answer: 4A63 or 4A6311

TIMO 2024 FRR Senior Secondary Set 1

 Solution. Quinary is a base-5 numeral system. It uses the digits 0, 1, 2, 3 and 4 to represent
any real number. While undecimal is a positional system that represents numbers using
a base of 11. It uses numbers 0 − 9 and the letter A. Convert first 2024035 into decimal:

2024035 = 2 · 55 + 0 · 54 + 2 · 53 + 4 · 52 + 0 · 51 + 3 · 50
= 6250 + 0 + 250 + 100 + 0 + 3
= 660310

Now, to convert decimal to undecimal,

6603 ÷ 11 = 600 with remainder 3 =⇒ 3

600 ÷ 11 = 54 with remainder 6 =⇒ 6
54 ÷ 11 = 4 with remainder 10 =⇒ A
4 ÷ 11 = 0 with remainder 4 =⇒ 4

Then we just write down the remainders in the reverse order to obtain the undecimal
number which is 4A63 or 4A6311 .

4. International Day of Mathematics is celebrated annually on the 14th of March. This year,
March 14 is Thursday. Which day of the week will the International Day of Mathematics
on 2059 be?
Answer: Friday
Solution. A normal year has 365 days while a leap year has 366 days. There are 27 normal
years and 8 leap years from 2024 to 2059 excluding 2024. Observe that 365 mod 7 ≡
1 mod 7 and 366 mod 7 ≡ 2 mod 7. Thus,

(27 · 1 + 8 · 2) mod 7 ≡ 43 mod 7 ≡ 1 mod 7.

Since the remainder is 1, the International Day of Mathematics on 2059 must be a day
after Thursday, which is Friday.
√
5
5. Find the nearest integer of 20242024.
Answer: 29
Solution. Assuming that k 5 ≈ 20242024, we know that k is a positive integer. So we are
looking for the 5th root of 20242024. Observe that

205 = 3200000 < 20242024 < 24300000 = 305 .

Note that 24300000 − 20242024 = 4057976 < 17042024 = 20242024 − 3200000. This

TIMO 2024 FRR Senior Secondary Set 1

 means that k is nearer to 30 than to 20. Hence, we test 26,27,28,and 29.

At k = 26 =⇒ k 5 = 11881376
At k = 27 =⇒ k 5 = 14348907
At k = 28 =⇒ k 5 = 17210368
At k = 29 =⇒ k 5 = 20511149
√
5
Clearly, the nearest integer of 20242024 is 29.

6. Find the largest 8-digit number that has exactly 3 positive factors.
Answer: 99460729
Solution. If a number N has exactly 3 factors, it means N is the square of a prime
number p. That is, the factors of N are 1, p and p2 . Observe that 100000000 = 100002
is the smallest 9-digit perfect square number but 10000 is not prime. Hence, 10000 has
more than 3 factors. Moreover, we need to find the largest 4-digit prime number less than
10000. By inspection, 9973 is such number. Therefore, the largest 8-digit number that
has exactly 3 factors is 99732 = 99460729.

TIMO 2024 FRR Senior Secondary Set 1

 ALGEBRA
p √
89 + 12 55.
7. Calculate
√ √ √ √
Answer: 2 11 + 3 5 or 3 5 + 2 11
Solution. Observe that
√ √
q q
89 + 12 55 = 89 + 2 · 6 55
√
q
= 89 + 2 1980
√
q
= 44 + 2 44 · 45 + 45
q√ √ √ √
= ( 44)2 + 2 · 44 · 45 + ( 45)2
q√ √
= ( 44 + 45)2
√ √
= 44 + 45
√ √
= 2 11 + 3 5.

8. Find the value of positive integer x such that x3 + 11x2 = 122x + 1320.
Answer: 11
Solution. Rearranging and factoring the equation, we have

x3 + 11x2 = 122x + 1320

x3 + 11x2 + 122x + 1320 = 0
(x + 10)(x2 + x − 132) = 0
(x + 10)(x − 11)(x + 12) = 0.

Note that the other two factors will give negative integer solutions. Hence, the only
positive integer solution is x = 11.
1 2
9. If α and β are roots of x − 2x + 3 = 0, evaluate α3 β + 2α2 β 2 + αβ 3 .
7
Answer: 4116
Solution. Simplifying the given equation, we obtain x2 −14x+21 = 0. By Vieta’s formula,
α + β = 14 and αβ = 21. Hence, we have

α3 β + 2α2 β 2 + αβ 3 = αβ(α2 + 2αβ + β 2 )

= αβ(α + β)2
= (21)(14)2
α3 β + 2α2 β 2 + αβ 3 = 4116.

TIMO 2024 FRR Senior Secondary Set 1

 1 1
10. It is known that x + = 5. Find the value of x5 + 5 .
x x
Answer: 2525
1
Solution. Let x + = 5. Then
x

2  3
1

1 2 x+ = 53
x+ =5 x
x  
1 3 1 1
x2 + 2 + = 25 x +3 x+ + 3 = 125
x2 x x
1 1
x2 + 2 = 23 x3 + 3 = 110
x x

Observe that
  
2 1 1 1 1
x + 2 x + 3 = x5 + + x + 5
3
x x x x
 
5 1 1
(23)(110) = x + 5 + x +
x x
1
2530 = x5 + 5 + 5
x
1
2525 = x5 + 5
x

3
11. Solve the logarithmic equation log9 (2x2 + 512) + = xlogx 2 + 2 log9 x. (Answer in simplest
2
surd form)
√
Answer: 16 2
Solution. Rearranging the terms of the equation,

log9 (2x2 + 512) + log9 27 = 2 + 2 log9 x

log9 (2x2 + 512) + log9 27 − 2 log9 x = 2
log9 (2x2 + 512) + log9 27 − log9 x2 = 2
27(2x2 + 512)
log9 =2
x2
27(2x2 + 512)
= 92
x2
27(2x2 + 512) = 81x2
2x2 + 512 = 3x2
x2 = 512
√
x = ±16 2
√ √
However, we cannot have x = −16 2. Therefore, the only solution is x = 16 2.

TIMO 2024 FRR Senior Secondary Set 1

12. Factor 36x2 − 25y 2 − 288x − 250y − 49 completely.
Answer: (6x + 5y + 1)(6x − 5y − 49) or (6x − 5y − 49)(6x + 5y + 1)
Solution. Observe that

36x2 − 25y 2 − 288x−250y − 49 = 36(x2 − 8x) − (y 2 + 10y) − 49

= 36(x2 − 8x + 16) − 25(y 2 + 10y + 25) − 49 − 576 + 625
= [6(x − 4)]2 − [5(y + 5)]2
= [6(x − 4) + 5(y + 5)][6(x − 4) − 5(y + 5)]
= (6x + 5y + 1)(6x − 5y − 49)

Therefore, 36x2 − 25y 2 − 288x − 250y − 49 = (6x + 5y + 1)(6x − 5y − 49).

TIMO 2024 FRR Senior Secondary Set 1

 NUMBER THEORY

13. It is known that x is rational and

2024
x = −2 + .
2024
−2 +
2024
−2 +
2024
−2 +
−2 + · · ·

Find the minimum value of x.

Answer: −46
Solution. Observe that

2024
x = −2 +
2024
−2 +
2024
−2 +
2024
−2 +
−2 + · · ·
2024
x = −2 +
x
2
0 = x + 2x − 2024
0 = (x − 44)(x + 46).

Hence, there are two possible values of x, i.e., x = 44 and x = −46. Therefore, the
minimum value of x is −46.

14. Yesterday was Saturday. Which day of the week will it be after 20212024 days?
Answer: Thursday
Solution. Note that

20212024 mod 7 ≡ 52024 mod 7

≡ (−2)2024 mod 7
≡ 22024 mod 7
≡ 23·674+2 mod 7
≡ (23 )674 · 22 mod 7
≡ 8674 · 4 mod 7
≡ 1674 · 4 mod 7
≡ 4 mod 7.

Since the remainder is 4, we count 4 days after Sunday and that is Thursday.

TIMO 2024 FRR Senior Secondary Set 1

15. It is known that x > 5000. Find the least positive integral solution of system of congruence
equations 
9x ≡ 34 mod 13


7x ≡ 26 mod 17 .


3x ≡ −5 mod 19


Answer: 8409
Solution. Observe that we can simplify the system to
  

 9x ≡ 34 mod 13 
9x ≡ 8 mod 13 
x ≡ 11 mod 13
  
7x ≡ 26 mod 17 =⇒ 7x ≡ 9 mod 17 =⇒ x ≡ 11 mod 17.

 
 

3x ≡ −5 mod 19 3x ≡ 14 mod 19 x ≡ 11 mod 19
  

Since all of the moduli of the simplified system are relatively prime, we know that by the
Chinese Remainder Theorem that this has a solution modulo the product of the moduli.
Now, 13 × 17 × 19 = 4199. We need to find the least positive integral solution x > 5000
with remainder 11 at modulo 4199, i.e., 2(4199) + 11 = 8409.

16. It is known that n! = n × (n − 1) × (n − 2) × · · · × 2 × 1. Find the number of 0’s at the

end of 202403!.
Answer: 50596
Solution. To determine the number of 0’s at the end of 2075!, we determine how many
multiples of 5 are there in the numbers from 1 to 2075. To do that, consider the following
algorithm:
 
1 202403
5 : = ⌊40480.6⌋ There are 40480 factors of 5.
51
 
202403
52 : = ⌊8096.12⌋ There are additional 8096 factors of 5.
52
 
202403
53 : = ⌊1619.22⌋ There are additional 1619 factors of 5.
53
 
202403
54 : = ⌊323.84⌋ There is an additional 323 factors of 5.
54
 
202403
55 : = ⌊64.77⌋ There is an additional 64 factors of 5.
55
 
202403
56 : = ⌊12.95⌋ There is an additional 12 factors of 5.
56
 
202403
57 : = ⌊2.59⌋ There is an additional 2 factors of 5.
57

Therefore, 202403! has 40480 + 8096 + 1619 + 323 + 64 + 12 + 2 = 50596 trailing 0’s.

TIMO 2024 FRR Senior Secondary Set 1

17. Find the remainder when 2024172024 is divided by 17.
Answer: 1
Solution. Observe that

2024172024 mod 17 ≡ (17 · 11906 + 15)2024 mod 17

≡ 152024 mod 17
≡ (−2)2024 mod 17
≡ 22024 mod 17
≡ 24·506 mod 17
≡ (24 )506 mod 17
≡ (16)506 mod 17
≡ (−1)506 mod 17
≡ 1 mod 17.

Therefore, the remainder when 2024172024 is divided by 17 is 1.

18. There are 3 distinct prime numbers. Their product is 31 times of their sum. Find the
least prime number among them.
Answer: 3
Solution. Let p, q, and r be the 3 distinct prime numbers. Observe that based on the
given, pqr = 31(p + q + r), one among p, q, and r mus be equal to 31. Without loss of
generality, we assume that p = 31. Now,

pqr = 31(p + q + r) =⇒ 31qr = 31(31 + q + r)

=⇒ qr = 31 + q + r
=⇒ qr − q − r = 31
=⇒ (q − 1)(r − 1) = 32.

It can be directly observed that the only prime solutions are (31, 17, 3) up to permutation.
Therefore, the least among them is 3.

TIMO 2024 FRR Senior Secondary Set 1

 GEOMETRY
 
3 2
19. Find the shortest distance from the point − , to the straight line 5y = 12x + 63.
4 5
Answer: 4
Solution. The minimum distance d from a point (x0 , y0 ) to a line Ax + By + C = 0 is
given by the formula
|Ax0 + By0 + C|
d= √ .
A2 + B 2
Hence, we have
   
3 2
12 − −5 + 63
4 5 | − 9 − 2 + 63| 52
d= √ = = = 4.
122 + 52 13 13

20. Find the area enclosed by the x-axis, the y-axis, and the straight line 22x−23y−1012 = 0.
Answer: 1012
Solution. Note that the region obtained is a right triangle with two legs at the coordinate
axes. Also, the intercepts of the lines are at (46, 0) and (0, −44). Observe that the height
of the triangle corresponds to the absolute value of the y coordinate of the y-intercept,
i.e., h = | − 44| = 44.Moreover, the base of the triangle is the absolute value of the x
coordinate of the x-intercept, i.e., b = 46. Therefore, the area is given by

1
A= · 44 · 46 = 1012 sq. units.
2

21. Find the radius of the inscribed circle of a rhombus with diagonals of lengths 96 and 110.
2640 12
Answer: or 36
73 73
Solution. Let d1 = 96 and d2 = 110 be the lengths of the diagonals and a the side of the
rhombus and r be the radius of the inscribed circle. Using the Pythagorean Theorem, we
can compute for a as follows:
s 2 2
√

d1 d2
a= + = 482 + 552 = 73.
2 2

Note that we can express the area of the rhombus in two different ways:
 
d1 d2 1
Arhombus = and Arhombus = 4 ar .
2 2

Combining the two,

 
d1 d2 1 d1 d2 96 · 110 10560 2640
=4 ar =⇒ d1 d2 = 4ar =⇒ r= = = = .
2 2 4a 4 · 73 292 73

TIMO 2024 FRR Senior Secondary Set 1

22. An exterior angle of an n-sided regular polygon is (n − 9)◦ . Find the value of n.
Answer: 24
Solution. Recall that the sum of the exterior angles on the polygon is 360◦ . Since the
given polygon is regular and has n sides with each exterior angle measures (n − 9)◦ , thus

(n − 9)n = 360
n2 − 9n − 360 = 0
(n − 24)(n + 15) = 0

Thus, n = 24 or n = −15. But since n must be positive, then n = 24.

√
23. If x is a real number and −1 ≤ x ≤ 1, find the maximum value of x + 1 − x2 .
√
Answer: 2
Solution. Let x = cos θ where 0 ≤ θ ≤ π. Note that
√ √ h π   π i √  π
x+ 1− x2 = cos θ + sin θ = 2 cos θ cos + sin θ sin = 2 cos θ − .
4 4 4
π
Since the maximum value of the cosine function is 1, attained when θ = 4
, then the
√ √
maximum value of x + 1 − x2 is 2.
√
24. Find the maximum value of 2024 sin x + cos x − 2024.
Answer: −1979
Solution: Using Cauchy-Schwarz inequality, observe that
√ 2
≤ (2024 + 1) sin2 x + cos2 x


2024 sin x + cos x
√ 2
2024 sin x + cos x ≤ 2025
√
2024 sin x + cos x ≤ 45
√
2024 sin x + cos x − 2024 ≤ 45 − 2024
√
2024 sin x + cos x − 2024 ≤ −1979

TIMO 2024 FRR Senior Secondary Set 1

 COMBINATORICS

25. It is known that a, b, c, d, e are positive integers. Find the number of solution sets of
a + b + c + d + e = 24.
Answer: 8855
Solution. Using the stars and bars formula where n = 24 and k = 5 the number of
variables, the number of sets of integral solution for a + b + c + d + e = 24 is given by

n−1 Ck−1 = 24−1 C5−1 = 23 C4 = 8855.

26. Find the number of ways to assign 15 people into five groups evenly.
Answer: 1401400
Solution. We first calculate the number of permutations of 15 people, i.e., 15!. Observe
that the number of different ways to order the people within the 5 groups evenly is (3!)5 .
Moreover, the number of ways we can order 5 groups is 5!. Therefore, the number of ways
to assign 15 people into four groups evenly is

15!
= 1401400.
(3!)5 × 5!

27. Let xn be a positive integer for all integers n. How many number(s) of sets of solutions
do x1 + x2 + x3 + · · · + x28 ≤ 32 have?
Answer: 35960
Solution. Let xn be a positive integer for all integers n. We consider two cases:
Case 1: x1 + x2 + x3 + · · · + x28 < 32
Using the stars and bars formula where n = 32 and k = 28 be the number of variables,
the number of sets of integral solution for x1 + x2 + x3 + · · · + x28 < 32 is given by
     
n−1 32 − 1 31
= = = 4495.
k 28 28

Case 2: x1 + x2 + x3 + · · · + x28 = 32
Using the same formula where n = 32 and k = 28 be the number of variables, the number
of sets of integral solution for x1 + x2 + x3 + · · · + x28 = 32 is given by
     
n−1 32 − 1 31
= = = 31465.
k−1 28 − 1 27

Summing up the two cases, the number of sets of solution for x1 + x2 + x3 + · · · + x28 ≤ 32
is 4495 + 31465 = 35960.

TIMO 2024 FRR Senior Secondary Set 1

28. Find the number of permutations(s) arranging 4 identical white roses, 7 identical red
roses, and 4 identical yellow roses in a row.
Answer: 450450
Solution. There are a total of n = 15 roses to be arranged. Now, given that these roses
are identical by color with n1 = 4, n2 = 7, and n3 = 4, the number of ways to do this is

n! 15!
= = 450450.
n1 ! · n2 ! · n3 ! 4! · 7! · 4!

29. Find the number of integer(s) x such that x2 < 8x + 768.

Answer: 55
Solution. Rearranging the given inequality, we have

x2 < 8x + 768 =⇒ x2 − 8x − 768 < 0 =⇒ (x + 24)(x − 32) < 0.

Here, we consider two cases:

Case 1: x + 24 > 0 and x − 32 < 0 =⇒ x > −24 and x < 32 =⇒ Solution: (−24, 32)
Case 2: x + 24 < 0 and x − 32 > 0 =⇒ x < −24 and x > 32 =⇒ Solution: ∅
Hence, the solutions for the inequality lie in (−24, 32). Thus, the number of integral
solution for the quadratic inequality is 55.

30. There are 12 identical dictionaries, 6 identical thesauri, and 7 identical encyclopedias. If
we pick 20 books out, what is the probability of getting 10 dictionaries, 5 thesauri, and 5
encyclopedia?
18
Answer:
115
 
25
Solution. The total number of ways to pick 20 books out of 25 is = 53130. Corre-
20
spondingly, the number of ways of getting10 dictionaries
  out
 of 12, 5 thesauri out of 6,
12 6 7
and 5 encyclopedia out of 7 are given by , , and , respectively. Therefore,
10 5 5
the probability is given by
     
12 6 7
× ×
10 5 5 66 × 6 × 21 8316 18
  = = = .
25 53130 53130 115
20

TIMO 2024 FRR Senior Secondary Set 1