Junior Mathematical Challenge – 2023 Question and Answer
Question 1
What is the value of \( 3202 – 2023 \)?
A 821 B 1001 C 1179 D 1221 E 1279
▶️ Answer/Explanation
Answer: C 1179
Explanation:
The problem requires us to compute the difference between 3202 and 2023. Let’s perform the subtraction step-by-step:
- Write the numbers vertically, aligning the digits by place value:
3202 - 2023 - Subtract starting from the rightmost column (units place):
- Units: \( 2 – 3 \). Since 2 is less than 3, borrow 10 from the tens place. The tens place of 3202 is 0, so we borrow from the hundreds place (2), reducing it to 1 and making the tens place 10. Now, the units calculation becomes \( 12 – 3 = 9 \).
- Tens: After borrowing, the tens place is \( 10 – 2 = 8 \), but we borrowed 1 from the hundreds, so adjust accordingly later.
- Hundreds: The hundreds place was 2, reduced to 1 after borrowing, so \( 1 – 0 = 1 \).
- Thousands: \( 3 – 2 = 1 \).
- Putting it together:
3202 - 2023 ------ 1179
Thus, \( 3202 – 2023 = 1179 \).
Verification: To ensure accuracy, add the result back to 2023: \( 2023 + 1179 = 3202 \).
- Units: \( 3 + 9 = 12 \), write 2, carry 1.
- Tens: \( 2 + 7 + 1 = 10 \), write 0, carry 1.
- Hundreds: \( 0 + 1 + 1 = 2 \).
- Thousands: \( 2 + 1 = 3 \).
- Total: 3202, which matches the original number.
Official Solution: The UKMT solution simply states: \( 3202 – 2023 = 1179 \), which aligns with our calculation.
Among the options (A: 821, B: 1001, C: 1179, D: 1221, E: 1279), the correct answer is C: 1179.
Question 2
How many of the following five options are factors of 30?
A 1 B 2 C 3 D 4 E 5
▶️ Answer/Explanation
Answer: D 4
Explanation:
A factor of 30 is a number that divides 30 evenly with no remainder. Let’s check each option:
- 1: \( 30 \div 1 = 30 \), a factor.
- 2: \( 30 \div 2 = 15 \), a factor.
- 3: \( 30 \div 3 = 10 \), a factor.
- 4: \( 30 \div 4 = 7.5 \), not an integer, so not a factor.
- 5: \( 30 \div 5 = 6 \), a factor.
Out of 1, 2, 3, 4, 5, the factors are 1, 2, 3, and 5. That’s 4 factors.
Official Solution: “Of the options given, only 4 is not a factor of 30. So four of the options are factors of 30.”
Correct answer: D: 4.
Question 3
What is the value of \( \frac{1+2+3+4+5}{6+7+8+9+10} \)?
A \( \frac{1}{2} \) B \( \frac{3}{8} \) C \( \frac{7}{16} \) D \( \frac{9}{20} \) E \( \frac{1}{3} \)
▶️ Answer/Explanation
Answer: B \( \frac{3}{8} \)
Explanation:
Calculate the numerator: \( 1 + 2 + 3 + 4 + 5 = 15 \).
Calculate the denominator: \( 6 + 7 + 8 + 9 + 10 = 40 \).
So, \( \frac{1+2+3+4+5}{6+7+8+9+10} = \frac{15}{40} \). Simplify by dividing numerator and denominator by their greatest common divisor (5): \( \frac{15 \div 5}{40 \div 5} = \frac{3}{8} \).
Official Solution: \( \frac{1+2+3+4+5}{6+7+8+9+10} = \frac{15}{40} = \frac{3}{8} \).
Correct answer: B: \( \frac{3}{8} \).
Question 4
One of these is the largest two-digit positive integer that is divisible by the product of its digits. Which is it?
A 12 B 24 C 36 D 72 E 96
▶️ Answer/Explanation
Answer: C 36
Explanation:
For each option, compute the product of its digits and check if the number is divisible by that product:
- 12: \( 1 \times 2 = 2 \), \( 12 \div 2 = 6 \) (integer), yes.
- 24: \( 2 \times 4 = 8 \), \( 24 \div 8 = 3 \) (integer), yes.
- 36: \( 3 \times 6 = 18 \), \( 36 \div 18 = 2 \) (integer), yes.
- 72: \( 7 \times 2 = 14 \), \( 72 \div 14 = 5.14 \) (not integer), no.
- 96: \( 9 \times 6 = 54 \), \( 96 \div 54 = 1.78 \) (not integer), no.
Among 12, 24, and 36, the largest is 36.
Official Solution: “Of the options given, 12 is divisible by 2, 24 is divisible by 8 and 36 is divisible by 18. However, 72 is not divisible by 14 and 96 is not divisible by 54. Therefore, since we are told that one of these is the largest two-digit number that is divisible by the product of its digits, 36 is that number.”
Correct answer: C: 36.
Question 5
The record for travelling \( 100 \, \text{m} \) on a skateboard by a dog is 19.65 seconds. This was achieved by Jumpy in 2013. What was Jumpy’s approximate average speed?
A \( 0.2 \, \text{m/s} \) B \( 0.5 \, \text{m/s} \) C \( 2 \, \text{m/s} \) D \( 2.5 \, \text{m/s} \) E \( 5 \, \text{m/s} \)
▶️ Answer/Explanation
Answer: E \( 5 \, \text{m/s} \)
Explanation:
Average speed = \( \frac{\text{distance}}{\text{time}} \). Distance = 100 m, time = 19.65 s.
\( \frac{100}{19.65} \approx 5.09 \, \text{m/s} \). For an approximate value, round to 5 m/s.
Official Solution: “Jumpy’s approximate average speed in m/s was \( \frac{100}{20} = 5 \).”
Ascending: Note that 19.65 is approximated to 20 for simplicity, making the calculation easier and the answer a whole number among the options.
Correct answer: E: 5 m/s.
Question 6
When this prime number square is completed, the eight circles contain eight different primes, and each of the four sides has total 43. What is the sum of the five missing primes?
A 51 B 53 C 55 D 57 E 59
▶️ Answer/Explanation
Answer: D 57
Explanation:
Assume a 3×3 grid with primes 13, 23, 29 given (as per solution context). Each row/column sums to 43.
Row 1: \( 13 + 23 + p = 43 \), \( p = 7 \).
Column 3: \( 29 + r + s = 43 \), \( r + s = 14 \). Primes summing to 14: 3 and 11.
Row 2: \( 13 + q + r = 43 \), if \( r = 11 \), \( q = 19 \) (prime).
Row 3: \( 23 + s + t = 43 \), if \( s = 3 \), \( t = 17 \) (prime).
Missing primes: 7, 19, 11, 3, 17. Sum = \( 7 + 19 + 11 + 3 + 17 = 57 \).
Official Solution: “As the sum of each row is 43, \( p = 43 – (13 + 23) = 7 \). Also \( r + s = 43 – 29 = 14 \). The missing primes are all different so \( r \) and \( s \) are 3 and 11. If \( r = 11 \) then \( q = 43 – (13 + 11) = 19 \), which is prime. So \( s = 3 \) and \( t = 43 – (23 + 3) = 17 \). Sum = \( 7 + 19 + 11 + 3 + 17 = 57 \).”
Correct answer: D: 57.
Question 7
What is the difference between the largest two-digit multiple of 2 and the smallest three-digit multiple of 3?
A 5 B 4 C 3 D 2 E 1
▶️ Answer/Explanation
Answer: B 4
Explanation:
Largest two-digit multiple of 2: 98 (even, \( 2 \times 49 \)).
Smallest three-digit multiple of 3: 102 (\( 3 \times 34 \)).
Difference: \( 102 – 98 = 4 \).
Official Solution: “The largest two-digit multiple of 2 is 98, while the smallest three-digit multiple of 3 is 102. Their difference is \( 102 – 98 = 4 \).”
Correct answer: B: 4.
Question 8
How many of these six numbers are prime? \( 0^2 + 1^2 \quad 1^2 + 2^2 \quad 2^2 + 3^2 \quad 3^2 + 4^2 \quad 4^2 + 5^2 \quad 5^2 + 6^2 \)
A 1 B 2 C 3 D 4 E 5
▶️ Answer/Explanation
Answer: D 4
Explanation:
Compute each expression and check if it’s prime:
- \( 0^2 + 1^2 = 0 + 1 = 1 \) (not prime).
- \( 1^2 + 2^2 = 1 + 4 = 5 \) (prime).
- \( 2^2 + 3^2 = 4 + 9 = 13 \) (prime).
- \( 3^2 + 4^2 = 9 + 16 = 25 = 5^2 \) (not prime).
- \( 4^2 + 5^2 = 16 + 25 = 41 \) (prime).
- \( 5^2 + 6^2 = 25 + 36 = 61 \) (prime).
Primes: 5, 13, 41, 61. Count = 4.
Official Solution: “\( 0^2 + 1^2 = 1 \); \( 1^2 + 2^2 = 5 \); \( 2^2 + 3^2 = 13 \); \( 3^2 + 4^2 = 25 \); \( 4^2 + 5^2 = 41 \); \( 5^2 + 6^2 = 61 \). Of the six sums, 5, 13, 41, and 61 are prime.”
Correct answer: D: 4.
Question 9
Triangle \( LMN \) is isosceles with \( LM = LN \). What is the value of \( y \)?
A 15 B 17 C 19 D 21 E 23
▶️ Answer/Explanation
Answer: A 15
Explanation:
Since \( LM = LN \), \( \angle LNM = \angle LMN \). Given \( \angle LNM = 3x – 20 \) and \( \angle LMN = 2x + 8 \):
\( 3x – 20 = 2x + 8 \), \( x = 28 \).
So, \( \angle LNM = \angle LMN = 3 \times 28 – 20 = 64^\circ \).
Triangle sum: \( 64 + 64 + (4y – 8) = 180 \), \( 4y – 8 = 52 \), \( y = 15 \).
Official Solution: “As \( LM = LN \), \( \angle LNM = \angle LMN \). Therefore, \( 3x – 20 = 2x + 8 \). So \( x = 28 \). Hence \( \angle LNM = (3 \times 28 – 20)^\circ = 64^\circ \). The sum of the interior angles is \( 180^\circ \), so \( 4y – 8 = 180 – 2 \times 64 = 52 \). Hence \( y = 15 \).”
Correct answer: A: 15.
Question 10
In the diagram, all distances shown are in cm. The perimeter of the shape is 60 cm. What is the area, in \( \text{cm}^2 \), of the shape?
A 192 B 204 C 212 D 232 E 252
▶️ Answer/Explanation
Answer: A 192
Explanation:
Assume a rectangle with a 6 cm by 8 cm cutout. Let height = \( h \), width = 18 cm. Perimeter = \( 2(h + 18) = 60 \), \( h = 12 \).
Area = \( 18 \times 12 – 6 \times (12 – 8) = 216 – 24 = 192 \, \text{cm}^2 \).
Official Solution: “Let the length of the right hand edge of the shape be \( h \) cm. Then \( 2(h + 18) = 60 \). Therefore \( h = 12 \). So the area is \( 18 \times 12 – 6 \times (12 – 8) = 216 – 24 = 192 \).”
Correct answer: A: 192.
Question 11
To save money, Scrooge is reusing tea bags. After a first ‘decent’ cup of tea, he dries the bag and uses two such dried bags to make a new ‘decent’ cup of tea. These bags are then dried again and four such bags now make a ‘decent’ cup of tea. After that they are put on the compost heap. How many ‘decent’ cups of tea can Scrooge get out of a new box of 120 tea bags?
A 480 B 240 C 210 D 195 E 180
▶️ Answer/Explanation
Answer: C 210
Explanation:
Initial cups: 120.
Used bags = 120. Second stage: \( 120 \div 2 = 60 \) cups, used bags = \( 60 \times 2 = 120 \).
Third stage: \( 120 \div 4 = 30 \) cups.
Total = \( 120 + 60 + 30 = 210 \).
Official Solution: “As 120 is a multiple of 4, Scrooge uses all tea bags. Hence the number of ‘decent’ cups is \( 120 + 120 \div 2 + 120 \div 4 = 120 + 60 + 30 = 210 \).”
Correct answer: C: 210.
Question 12
One afternoon, Brian the snail went for a slither at a constant speed. By 1:50 pm he had slithered 150 centimetres. By 2:10 pm he had slithered 210 centimetres. When did Brian start his slither?
A Noon B 12:20 pm C 12:30 pm D 12:45 pm E 1 pm
▶️ Answer/Explanation
Answer: E 1 pm
Explanation:
Time difference: 1:50 pm to 2:10 pm = 20 minutes.
Distance: \( 210 – 150 = 60 \, \text{cm} \).
Speed: \( 60 \div 20 = 3 \, \text{cm/min} \).
Time to slither 150 cm: \( 150 \div 3 = 50 \) minutes.
Start time: 1:50 pm – 50 min = 1:00 pm.
Official Solution: “In the 20 minutes between 1:50 pm and 2:10 pm, Brian slithered 60 cm. So Brian slithered at 3 cm per minute and took 50 minutes to slither 150 cm. Hence Brian started at 1 pm.”
Correct answer: E: 1 pm.
Question 13
Four congruent rectangles are arranged as shown to form an inner square of area \( 20 \, \text{cm}^2 \) and an outer square of area \( 64 \, \text{cm}^2 \). What is the perimeter of one of the four congruent rectangles?
A 6 cm B 8 cm C 9.75 cm D 16 cm E 20 cm
▶️ Answer/Explanation
Answer: D 16 cm
Explanation:
Outer square side = \( \sqrt{64} = 8 \, \text{cm} \).
Inner square side = \( \sqrt{20} \approx 4.47 \, \text{cm} \).
Rectangle: length + width = 8 cm (outer side). Perimeter = \( 2 \times 8 = 16 \, \text{cm} \).
Official Solution: “The outer square has area 64 cm², so its side-length is 8 cm. Therefore the sum of the length and the breadth of one of the four congruent rectangles is 8 cm. Hence the perimeter is \( 2 \times 8 = 16 \, \text{cm} \).”
Correct answer: D: 16 cm.
Question 14
In the addition shown, \( x \) and \( y \) represent different single digits. What is the value of \( x + y \)?
77x
6yx
+ yyx
------
1xx7
A 10 B 11 C 12 D 13 E 14
▶️ Answer/Explanation
Answer: E 14
Explanation:
Units: \( x + x + x = 7, 17, 27 \). Only \( 3x = 27 \) works, \( x = 9 \), carry 2.
Tens: \( 2 + 7 + y + y = 9, 19, 29 \). \( 2y + 9 = 19 \), \( y = 5 \), carry 1.
Hundreds: \( 1 + 7 + 6 + y = 19 \), \( y = 5 \), consistent.
\( x + y = 9 + 5 = 14 \).
Official Solution: “Units: \( 3x = 27 \), \( x = 9 \), carry 2. Tens: \( 2 + 7 + 2y = 19 \), \( y = 5 \), carry 1. Hundreds: \( 1 + 7 + 6 + 5 = 19 \). So \( x + y = 9 + 5 = 14 \).”
Correct answer: E: 14.
Question 15
My train was scheduled to leave at 17:48 and to arrive at my destination at 18:25. However, it started four minutes late, and the journey took twice as long as scheduled. When did I arrive?
A 19:39 B 19:06 C 19:02 D 18:29 E 17:52
▶️ Answer/Explanation
Answer: B 19:06
Explanation:
Scheduled duration: 17:48 to 18:25 = 37 minutes.
Actual start: 17:48 + 4 = 17:52.
Actual duration: \( 2 \times 37 = 74 \) minutes = 1 hr 14 min.
Arrival: 17:52 + 1:14 = 19:06.
Official Solution: “The journey time was 37 minutes. Actual time was \( 2 \times 37 = 74 \) minutes, or 1 hour 14 minutes. Leaving at 17:52, arrival was 19:06.”
Correct answer: B: 19:06.
Question 16
Amrita needs to select a new PIN. She decides it will be made up of four non-zero digits with the following properties:
i) The first two digits and the last two digits each make up a two-digit number which is a multiple of 11.
ii) The sum of all the digits is a multiple of 11.
How many different possibilities are there for Amrita’s PIN?
A 1 B 2 C 4 D 8 E 16
▶️ Answer/Explanation
Answer: D 8
Explanation:
Two-digit multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.
Sum of digits must be 22 (even multiple of 11, max 36).
PINs: 2299, 3388, 4477, 5566, 6655, 7744, 8833, 9922 (8 possibilities).
Official Solution: “Sum must be 22. Possibilities: 2299, 3388, 4477, 5566, 6655, 7744, 8833, 9922. Total = 8.”
Correct answer: D: 8.
Question 17
Two numbers \( p \) and \( q \) are such that \( 0 < p < q < 1 \). Which is the largest of these expressions? \( q – p \quad p – q \quad \frac{p + q}{2} \quad \frac{p}{q} \quad \frac{q}{p} \)
A \( q – p \) B \( p – q \) C \( \frac{p + q}{2} \) D \( \frac{p}{q} \) E \( \frac{q}{p} \)
▶️ Answer/Explanation
Answer: E \( \frac{q}{p} \)
Explanation:
Since \( 0 < p < q < 1 \):
- \( q – p > 0 \), but \( < 1 \).
- \( p – q < 0 \).
- \( \frac{p + q}{2} < 1 \).
- \( \frac{p}{q} < 1 \).
- \( \frac{q}{p} > 1 \).
\( \frac{q}{p} \) is the only expression > 1.
Official Solution: “\( q – p < 1 \); \( p – q < 0 \); \( \frac{p + q}{2} < 1 \); \( \frac{p}{q} < 1 \); \( \frac{q}{p} > 1 \). So \( \frac{q}{p} \) is largest.”
Correct answer: E: \( \frac{q}{p} \).
Question 18
What is the sum of the four marked angles in the diagram?
A \( 540^\circ \) B \( 560^\circ \) C \( 570^\circ \) D \( 600^\circ \) E \( 720^\circ \)
▶️ Answer/Explanation
Answer: A \( 540^\circ \)
Explanation:
For a quadrilateral with exterior angles \( p, q, r, s \), interior angles are \( 180 – p, 180 – q, 180 – r, 360 – s \).
Sum of interior angles = 360°.
\( (180 – p) + (180 – q) + (180 – r) + (360 – s) = 360 \), \( p + q + r + s = 540^\circ \).
Official Solution: “Interior angles: \( 180 – p, 180 – q, 180 – r, 360 – s \). Sum = 360°. Hence \( p + q + r + s = 540^\circ \).”
Correct answer: A: 540°.
Question 19
In a football match, Rangers beat Rovers 5-4. The only time Rangers were ahead was after they scored the final goal. How many possible half-time scores were there?
A 9 B 10 C 15 D 16 E 25
▶️ Answer/Explanation
Answer: D 16
Explanation:
Rangers’ goals: 0 to 5, Rovers’ goals: 0 to 4. Half-time score \( (R, V) \), \( R \leq V \) until final goal.
Possible \( (R, V) \): (0,0), (0,1), (0,2), (0,3), (0,4), (1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4), (5,4).
Total = 16.
Official Solution: “Possible half-time scores: 5 (R=0), 4 (R=1), 3 (R=2), 2 (R=3), 1 (R=4), 1 (R=5). Total = 16.”
Correct answer: D: 16.
Question 20
Each cell in the crossnumber is to be filled with a single digit.
| 1 | 1 | |
| 2 |
Across: 1. A cube, 2. A square
Down: 1. A prime
Which of these could be the sum of the four digits in the crossnumber?
A 17 B 16 C 15 D 14 E 13
▶️ Answer/Explanation
Answer: A 17
Explanation:
1 Across (cube): 27 or 64.
2 Across (square): 16, 25, 36, 49, 64, 81.
1 Down (prime): ends in 1, 2, 3, 4, 6, 8.
Case 1: 27, 1 Down = 23, 2 Across = 36. Sum = \( 2 + 7 + 3 + 6 = 18 \).
Case 2: 64, 1 Down = 61, 2 Across = 16. Sum = \( 6 + 4 + 1 + 6 = 17 \).
Only 17 is an option.
Official Solution: “1 Across: 27 or 64. 1 Down: 23 (if 27) or 61 (if 64). 2 Across: 36 or 16. Sums: 18 or 17. Only 17 is an option.”
Correct answer: A: 17.
Question 21
Eleanor’s Elephant Emporium has four types of elephant. There are twice as many grey elephants as pygmy elephants, three times as many white elephants as grey elephants and four times as many pink elephants as white elephants. There are 20 more white elephants than pygmy elephants.
How many elephants are in Eleanor’s Emporium?
A 123 B 132 C 213 D 231 E 312
▶️ Answer/Explanation
Answer: B 132
Explanation:
Official Solution: Let the numbers of grey, pygmy, pink and white elephants be \( g, y, p \) and \( w \) respectively. Then \( g = 2y \) …[1], \( w = 3g \) …[2], \( p = 4w \) …[3] and \( w = y + 20 \) …[4]. From [1] and [2] we get \( w = 3 \times 2y = 6y \). Substituting for \( w \) in [4] gives \( 6y = y + 20 \). So \( y = 4 \). Therefore, \( w = 6 \times 4 = 24 \), \( g = w \div 3 = 24 \div 3 = 8 \) and \( p = 4w = 4 \times 24 = 96 \). Hence the number of elephants in Eleanor’s Emporium is \( g + y + p + w = 8 + 4 + 96 + 24 = 132 \).
Detailed Explanation:
Define variables based on the problem:
- \( y \): number of pygmy elephants.
- \( g \): number of grey elephants, where \( g = 2y \).
- \( w \): number of white elephants, where \( w = 3g \).
- \( p \): number of pink elephants, where \( p = 4w \).
- Also, \( w = y + 20 \).
Substitute \( g = 2y \) into \( w = 3g \): \( w = 3 \cdot 2y = 6y \).
Now use the condition \( w = y + 20 \): \( 6y = y + 20 \).
Solve: \( 6y – y = 20 \), \( 5y = 20 \), \( y = 4 \).
Calculate each type:
- Pygmy: \( y = 4 \).
- Grey: \( g = 2y = 2 \cdot 4 = 8 \).
- White: \( w = 6y = 6 \cdot 4 = 24 \) (or \( w = 3g = 3 \cdot 8 = 24 \), consistent).
- Pink: \( p = 4w = 4 \cdot 24 = 96 \).
Check: \( w = y + 20 \), \( 24 = 4 + 20 \), true.
Total elephants: \( y + g + w + p = 4 + 8 + 24 + 96 = 132 \).
Option B (132) matches.
Question 22
The positive integers from 1 to 9 inclusive are placed in the grid, one to a cell, so that the product of the three numbers in each row or column is as shown. What number should be placed in the bottom right-hand cell?
[Grid implied: 3×3 with products: Row 1: 18, Row 2: 105, Row 3: 56; Column 1: 28, Column 2: 30, Column 3: 36]
A 9 B 6 C 4 D 3 E 2
▶️ Answer/Explanation
Answer: B 6
Explanation:
Official Solution: The second row is the only row whose product is a multiple of 5. The same is true of the second column. Hence \( e \) must be 5. Similarly, since 105 and 56 are the only multiples of 7 involved, then \( d \) must be 7. Hence, since \( 105 = 3 \times 5 \times 7 \), it follows that \( f = 3 \). Then column 3 shows that \( 3 \times c \times i = 36 \) and so \( c \times i = 12 \). Since 3 has already been used for \( f \), the only possible values for \( c \) and \( i \) are 2 and 6 in some order. If \( c = 6 \) then \( a \times b = 18 \div 6 = 3 \) and so one of \( a \) and \( b \) is 3, which isn’t possible. Therefore \( c = 2 \) and then \( i = 6 \). (It is left as an exercise for the reader to complete the grid and to confirm that it does include all the positive integers from 1 to 9 inclusive.)
Detailed Explanation:
Construct a 3×3 grid with variables \( a, b, c; d, e, f; g, h, i \):
a b c | 18
d e f | 105
g h i | 56
---------
28 30 36
Use factors to deduce values:
- Row 2 (105) and Column 2 (30) share \( e \). Factorize: \( 105 = 3 \cdot 5 \cdot 7 \), \( 30 = 2 \cdot 3 \cdot 5 \). Only 5 is unique to both, so \( e = 5 \).
- Row 2 (105) and Row 3 (56) share 7 (\( 56 = 7 \cdot 8 \)), so \( d = 7 \) (since \( e = 5 \)).
- Row 2: \( d \cdot e \cdot f = 105 \), \( 7 \cdot 5 \cdot f = 105 \), \( f = 3 \).
- Column 3: \( c \cdot f \cdot i = 36 \), \( 3 \cdot c \cdot i = 36 \), \( c \cdot i = 12 \). Possible pairs (1-9, no repeats): (2, 6). If \( c = 6 \), Row 1: \( a \cdot b \cdot 6 = 18 \), \( a \cdot b = 3 \), impossible without 3 (used). So \( c = 2 \), \( i = 6 \).
Bottom right cell is \( i = 6 \), option B.
Question 23
Regular pentagon \( PQRST \) has centre \( O \). Lines \( PH, FI \) and \( GJ \) go through \( O \). The six angles at \( O \) are equal.
What is the size of angle \( TGO \)?
A \( 60^\circ \) B \( 72^\circ \) C \( 75^\circ \) D \( 76^\circ \) E \( 78^\circ \)
▶️ Answer/Explanation
Answer: E \( 78^\circ \)
Explanation:
Official Solution: As the six angles at \( O \) are equal, each is \( 360^\circ \div 6 = 60^\circ \). The five angles at the centre of the regular pentagon formed by joining each of its vertices to \( O \) are also all equal. So \( \angle POT \) is \( 360^\circ \div 5 = 72^\circ \). Therefore \( \angle TOG = \angle POG – \angle POT = (2 \times 60 – 72)^\circ = 48^\circ \). Consider triangle \( TOS \): \( \angle TOS = 72^\circ \) and, as \( O \) is the centre of the pentagon, \( SO = TO \). Hence \( \angle OTS = \angle OST = (180 – 72)^\circ \div 2 = 54^\circ \). In triangle \( TOG \), \( \angle TOG = 48^\circ \) and \( \angle OTG = \angle OTS = 54^\circ \). So \( \angle TGO = (180 – 48 – 54)^\circ = 78^\circ \).
Detailed Explanation:
In a regular pentagon \( PQRST \) with center \( O \):
- Six lines through \( O \) divide \( 360^\circ \) into 6 equal angles: \( 360^\circ \div 6 = 60^\circ \).
- Central angles of the pentagon: \( 360^\circ \div 5 = 72^\circ \) (e.g., \( \angle POT \)).
- Assume lines \( PH, FI, GJ \) intersect \( O \) and split the circle. \( \angle POG = 2 \cdot 60^\circ = 120^\circ \) (two sectors). \( \angle TOG = \angle POG – \angle POT = 120^\circ – 72^\circ = 48^\circ \).
- Triangle \( TOS \): \( TO = SO \) (radii), \( \angle TOS = 72^\circ \). Base angles: \( (180^\circ – 72^\circ) \div 2 = 54^\circ \).
- Triangle \( TOG \): \( \angle TOG = 48^\circ \), \( \angle OTG = 54^\circ \) (same as \( \angle OTS \)), so \( \angle TGO = 180^\circ – 48^\circ – 54^\circ = 78^\circ \).
Answer: E (\( 78^\circ \)).
Question 24
Beatrix was born in this century. On her birthday this year, her age was equal to the sum of the digits of the year in which she was born. In which of these years will her age on her birthday be twice the sum of the digits of that year?
A 2027 B 2029 C 2031 D 2033 E 2035
▶️ Answer/Explanation
Answer: E 2035
Explanation:
Official Solution: Let the year this century in which Beatrix was born be ‘\( 20xy \)’. Then her age on her birthday this year was \( 23 – 10x – y \). Therefore \( 2 + 0 + x + y = 23 – 10x – y \). Hence \( 11x + 2y = 21 \). If \( x = 0 \), then \( y \) is not an integer; if \( x = 1 \), then \( y = 5 \) and if \( x > 1 \) then \( y \) is negative. So Beatrix was born in 2015 and was 8 years old on her birthday this year. Let the year in which Beatrix’s age on her birthday will be twice the sum of the digits of that year be ‘\( 20pq \)’. As Beatrix was born in 2015, her age in ‘\( 20pq \)’ will be \( 10p + q – 15 \). Therefore \( 2(2 + 0 + p + q) = 10p + q – 15 \). So \( 8p – q = 19 \). If \( p \leq 2 \), then \( q \) is negative; if \( p = 3 \), then \( q = 5 \) and if \( p \geq 4 \) then \( q > 9 \). So the required year is 2035. Beatrix will then be 20, which is \( 2 \times (2 + 0 + 3 + 5) \).
Detailed Explanation:
“This year” is 2023 (exam date). Birth year \( 20xy \), age in 2023: \( 2023 – (2000 + 10x + y) = 23 – 10x – y \). Sum of digits: \( 2 + 0 + x + y \).
Equation: \( 2 + x + y = 23 – 10x – y \), \( 11x + 2y = 21 \).
Solve: \( y = (21 – 11x) / 2 \), \( y \) integer, \( 21 – 11x \) even, \( 11x \) odd, \( x \) odd. Test \( x = 1 \): \( y = 5 \), birth year 2015, age 8, sum \( 2 + 0 + 1 + 5 = 8 \), true.
Future year \( 20pq \), age \( 20pq – 2015 = 10p + q – 15 \), sum \( 2 + 0 + p + q \). Equation: \( 2(2 + p + q) = 10p + q – 15 \), \( 8p – q = 19 \).
Test options: 2035 (\( p = 3, q = 5 \)): \( 8 \cdot 3 – 5 = 19 \), age 20, sum 10, \( 2 \cdot 10 = 20 \), true. Others fail.
Answer: E (2035).
Question 25
Granny gave away her entire collection of antique spoons to three people. Her daughter received 8 more than a third of the total; her son received 8 more than a third of what was then left; finally her neighbour received 8 more than a third of what was then left.
What is the sum of the digits of the number of spoons which were in Granny’s collection?
A 14 B 12 C 10 D 8 E 6
▶️ Answer/Explanation
Answer: B 12
Explanation:
Official Solution: Let the number of spoons which Granny’s neighbour received be \( z \). Since the neighbour’s share completes the distribution of all of Granny’s spoons, \( z \) is also the number of spoons which were left after the son’s share was removed. Therefore \( z = \frac{z}{3} + 8 \). Hence \( \frac{2z}{3} = 8 \), that is \( z = 12 \). Let the number of spoons remaining after Granny’s daughter had taken her share of spoons be \( y \). So Granny’s son received \( \left( \frac{y}{3} + 8 \right) \) spoons and there were then 12 spoons remaining. Hence \( \frac{y}{3} + 8 = y – 12 \). Therefore \( \frac{2y}{3} = 20 \), that is \( y = 30 \). Let the number of spoons in Granny’s collection be \( x \). Then Granny’s daughter received \( \left( \frac{x}{3} + 8 \right) \) spoons and there were then 30 spoons remaining. Therefore \( \frac{x}{3} + 8 = x – 30 \). Hence \( \frac{2x}{3} = 38 \), that is \( x = 57 \). So the required digit-sum is 12.
Detailed Explanation:
Work backwards:
- Neighbour’s share \( z \): \( z = \frac{z}{3} + 8 \), \( 2z/3 = 8 \), \( z = 12 \).
- Before neighbour: \( y \) spoons, son took \( \frac{y}{3} + 8 \), remaining \( y – (\frac{y}{3} + 8) = 12 \), \( 2y/3 – 8 = 12 \), \( 2y/3 = 20 \), \( y = 30 \).
- Total \( x \), daughter took \( \frac{x}{3} + 8 \), remaining \( x – (\frac{x}{3} + 8) = 30 \), \( 2x/3 – 8 = 30 \), \( 2x/3 = 38 \), \( x = 57 \).
Check: Daughter: \( 57/3 + 8 = 27 \), left 30. Son: \( 30/3 + 8 = 18 \), left 12. Neighbour: 12. Total: \( 27 + 18 + 12 = 57 \).
Digit sum of 57: \( 5 + 7 = 12 \), option B.