Answer: E 111

Explanation:

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. To determine which of the given numbers—101, 103, 107, 109, and 111—is not prime, we need to check each for divisibility by numbers up to its square root. Let’s examine each option step-by-step:

• Option A: 101

  The square root of 101 is approximately 10 (since \(10^2 = 100\) and \(11^2 = 121\)). We test divisibility by primes less than or equal to 10: 2, 3, 5, 7.

  • 2: 101 is odd, so not divisible by 2.
  • 3: Sum of digits = \(1 + 0 + 1 = 2\), not divisible by 3 (since 2 ÷ 3 ≈ 0.67).
  • 5: Last digit is 1, not 0 or 5, so not divisible by 5.
  • 7: \(101 \div 7 \approx 14.43\), not an integer (e.g., \(7 \times 14 = 98\), \(101 – 98 = 3\)).

  No divisors found. Thus, 101 is prime.

• Option B: 103

  Square root ≈ 10. Check primes 2, 3, 5, 7.

  • 2: 103 is odd.
  • 3: Sum = \(1 + 0 + 3 = 4\), not divisible by 3.
  • 5: Last digit is 3.
  • 7: \(103 \div 7 \approx 14.71\) (e.g., \(7 \times 14 = 98\), \(103 – 98 = 5\)).

  No divisors. 103 is prime.

• Option C: 107

  Square root ≈ 10. Check 2, 3, 5, 7.

  • 2: 107 is odd.
  • 3: Sum = \(1 + 0 + 7 = 8\), not divisible by 3.
  • 5: Last digit is 7.
  • 7: \(107 \div 7 \approx 15.29\) (e.g., \(7 \times 15 = 105\), \(107 – 105 = 2\)).

  No divisors. 107 is prime.

• Option D: 109

  Square root ≈ 10. Check 2, 3, 5, 7.

  • 2: 109 is odd.
  • 3: Sum = \(1 + 0 + 9 = 10\), not divisible by 3.
  • 5: Last digit is 9.
  • 7: \(109 \div 7 \approx 15.57\) (e.g., \(7 \times 15 = 105\), \(109 – 105 = 4\)).

  No divisors. 109 is prime.

• Option E: 111

  Square root ≈ 10.5. Check 2, 3, 5, 7.

  • 2: 111 is odd.
  • 3: Sum = \(1 + 1 + 1 = 3\), divisible by 3 (\(111 \div 3 = 37\), since \(3 \times 37 = 111\)).

  Since 111 is divisible by 3, it has divisors 1, 3, 37, and 111. Thus, 111 is not prime.

Conclusion: Among 101, 103, 107, 109, and 111, only 111 is not prime because it is divisible by 3 (and 37), whereas the others have no divisors other than 1 and themselves.

Official Solution: The UKMT solution states: “It is clear that 111 is a multiple of 3 since the sum of its digits is 3. Therefore 111 is not prime.” This confirms our finding using the divisibility rule for 3.

Thus, the correct answer is E: 111.

Answer: D 101

Explanation:

We need to compute \( 2020 \div 20 \). Let’s perform the division step-by-step:

• Write it as a division problem: \( 2020 \div 20 \).
• Perform long division:

     101
  20)2020
     20
     --
      02
       0
      --
       20
       20
      ---
        0
                

  • 20 goes into 20 (first two digits) 1 time: \( 20 \times 1 = 20 \), subtract \( 20 – 20 = 0 \).
  • Bring down 2: 02. 20 goes into 2 zero times: \( 20 \times 0 = 0 \), subtract \( 2 – 0 = 2 \).
  • Bring down 0: 20. 20 goes into 20 one time: \( 20 \times 1 = 20 \), subtract \( 20 – 20 = 0 \).
  • Remainder is 0, so \( 2020 \div 20 = 101 \).
• Alternatively, simplify: \( 2020 \div 20 = \frac{2020}{20} = \frac{2020 \div 2}{20 \div 2} = \frac{1010}{10} = 101 \).

Verification: \( 20 \times 101 = 20 \times (100 + 1) = 2000 + 20 = 2020 \), which matches.

Official Solution: “The value of \( 2020 \div 20 \) is equal to the value of \( 202 \div 2 = 101 \).” This suggests a quick method: divide numerator and denominator by 10 first (\( \frac{2020}{20} = \frac{202}{2} = 101 \)).

Among the options (A: 10, B: 11, C: 100, D: 101, E: 111), the correct answer is D: 101.

Answer: C 3

Explanation:

Rotational symmetry of order two means a figure looks identical after a 180° rotation around its center. Since the problem references “these figures” but no diagrams are provided, we assume five figures based on a rectangle, each with a marked center, as typical in JMC problems. A rectangle itself has rotational symmetry of order 2 (180° rotation about its center maps it onto itself). We need to determine how many of five hypothetical figures retain this property.

Official Solution: “The only rotational symmetry of the rectangle is rotation through 180° and so has order two. The diagrams below show the effect of this rotation on each of the five figures. Only the first, fourth and fifth of these figures remain unchanged and so just three figures have rotational symmetry of order two.”

Since the figures aren’t provided, let’s hypothesize based on common patterns:

• A plain rectangle: Rotates 180° to itself (order 2).
• Rectangle with an asymmetric mark (e.g., one corner shaded): After 180°, the mark moves, no symmetry.
• Rectangle with symmetric marks (e.g., opposite corners shaded): 180° rotation swaps marks symmetrically (order 2).
• Rectangle with a central symmetric feature: Retains appearance (order 2).
• Rectangle with an off-center asymmetric feature: Changes appearance, no symmetry.

Without specific figures, we trust the official solution: three figures (e.g., 1st, 4th, 5th) remain unchanged. Thus, C: 3 is correct.

Answer: B 6660

Explanation:

To convert meters to centimeters, use the fact that 1 meter = 100 centimeters. For 66.6 meters:

• \( 66.6 \times 100 = 6660 \) centimeters.
• Break it down: \( 66 \times 100 = 6600 \), \( 0.6 \times 100 = 60 \), \( 6600 + 60 = 6660 \).

Official Solution: “There are 100 cm in 1 m. So the number of centimetres in 66.6 m is \( 100 \times 66.6 = 6660 \).”

Verification: \( 6660 \div 100 = 66.6 \), which matches 66.6 meters.

Options: A: 66600 (too large, implies 666 m), B: 6660 (correct), C: 666 (implies 6.66 m), D: 66.6 (meters, not cm), E: 66 (too small). Thus, B: 6660.

Answer: E 6

Explanation:

Let Amrita’s number be \( n \). Follow her operations: double it (\( 2n \)), add 9 (\( 2n + 9 \)), divide by 3 (\( \frac{2n + 9}{3} \)), subtract 1 (\( \frac{2n + 9}{3} – 1 \)), and this equals \( n \). Set up the equation:

• \( \frac{2n + 9}{3} – 1 = n \)
• Multiply through by 3: \( 2n + 9 – 3 = 3n \)
• Simplify: \( 2n + 6 = 3n \)
• Subtract \( 2n \): \( 6 = 3n – 2n \), so \( 6 = n \).

Verification: \( n = 6 \): \( 2 \times 6 = 12 \), \( 12 + 9 = 21 \), \( 21 \div 3 = 7 \), \( 7 – 1 = 6 \). It works.

Official Solution: “Let Amrita’s number be \( n \). Then \( (2n + 9) \div 3 – 1 = n \). Therefore \( 2n + 9 = 3(n + 1) = 3n + 3 \). Hence \( n = 9 – 3 = 6 \).”

Options: A: 1 (yields 0), B: 2 (yields 1), C: 3 (yields 2), D: 4 (yields 3), E: 6 (yields 6). Thus, E: 6.

Answer: C \( \frac{1}{4} \)

Explanation:

All fractions have a common denominator of 12, so combine the numerators:

• \( \frac{6 – 5 + 4 – 3 + 2 – 1}{12} \)
• Step-by-step: \( 6 – 5 = 1 \), \( 1 + 4 = 5 \), \( 5 – 3 = 2 \), \( 2 + 2 = 4 \), \( 4 – 1 = 3 \).
• So, \( \frac{3}{12} = \frac{1}{4} \).

Official Solution: “The value of \( \frac{6}{12} – \frac{5}{12} + \frac{4}{12} – \frac{3}{12} + \frac{2}{12} – \frac{1}{12} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \).” (Note: This seems to misgroup terms; the direct sum is clearer.)

Options: A: \( \frac{1}{2} \) (0.5), B: \( \frac{1}{3} \) (≈0.333), C: \( \frac{1}{4} \) (0.25), D: \( \frac{1}{5} \) (0.2), E: \( \frac{1}{6} \) (≈0.167). \( \frac{3}{12} = 0.25 \), so C: \( \frac{1}{4} \).

Answer: A 19

Explanation:

Find four different positive integers whose product is 110. Factorize 110:

• \( 110 = 2 \times 5 \times 11 \), but that’s three numbers.
• Introduce 1: \( 1 \times 2 \times 5 \times 11 = 110 \).
• Check: all distinct (1, 2, 5, 11), product = 110.
• Sum: \( 1 + 2 + 5 + 11 = 19 \).

Other sets (e.g., 1, 1, 5, 22) repeat numbers, violating “different.”

Official Solution: “Note that \( 110 = 10 \times 11 = 2 \times 5 \times 11 \), which is the prime factorisation of 110. Therefore the four different positive integers whose product is 110 are 1, 2, 5 and 11. Their sum is 19.”

Options: A: 19 (correct), B: 22, C: 24, D: 25, E: 28. Thus, A: 19.

Answer: D 5

Explanation:

Consider the grid from the official solution: a 2×3 grid labeled 1, 2, 3 (top row), 4, 5, 6 (bottom row). Cells share a vertex (corner) if diagonally adjacent, not a side if not horizontal/vertical neighbors.

1 2
3 4 5
  6
          

Pairs sharing a vertex but not a side:

• 1 and 4 (diagonal, no side).
• 2 and 3 (diagonal).
• 2 and 5 (diagonal).
• 3 and 6 (diagonal).
• 5 and 6 (diagonal).

Total: 5 ways.

Official Solution: “The pairs of cells which Wesley can colour are 1 and 4; 2 and 3; 2 and 5; 3 and 6; 5 and 6. Therefore the diagram can be coloured in five ways.”

Options: A: 2, B: 3, C: 4, D: 5 (correct), E: 6. Thus, D: 5.

Answer: A 240

Explanation:

Let the number be \( n \). Then: \( \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times n = 2 \).

• Combine fractions: \( \frac{1}{2 \times 3 \times 4 \times 5} = \frac{1}{120} \).
• So, \( \frac{n}{120} = 2 \).
• Solve: \( n = 2 \times 120 = 240 \).

Verification: \( \frac{240}{5} = 48 \), \( \frac{48}{4} = 12 \), \( \frac{12}{3} = 4 \), \( \frac{4}{2} = 2 \). Correct.

Official Solution: “Let the required number be \( n \). Then, \( \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times n = 2 \). Therefore \( \frac{1}{120} \times n = 2 \). Hence \( n = 120 \times 2 = 240 \).”

Options: A: 240 (correct), B: 120 (yields 1), C: 60 (yields 0.5), D: \( \frac{1}{120} \) (too small), E: \( \frac{1}{240} \) (smaller). Thus, A: 240.

Answer: D 1

Explanation:

Solve each equation:

• \( x – 2 = 10 \): \( x = 10 + 2 = 12 \). True.
• \( \frac{x}{2} = 24 \): \( x = 24 \times 2 = 48 \). False.
• \( 10 – x = 2 \): \( 10 – 2 = x \), \( x = 8 \). False.
• \( 2x – 1 = 25 \): \( 2x = 26 \), \( x = 13 \). False.

Only \( x – 2 = 10 \) has \( x = 12 \).

Official Solution: “The solutions of the four equations are, from left to right, \( x = 12 \); \( x = 48 \); \( x = 8 \); \( x = 13 \). Therefore exactly one of the equations has solution \( x = 12 \).”

Options: A: 4, B: 3, C: 2, D: 1 (correct), E: 0. Thus, D: 1.

Answer: E 840 cm

Explanation:

A 3×3 grid has 9 squares, with 4 horizontal and 4 vertical wires (each 3 cm), totaling \( 4 \times 3 + 4 \times 3 = 24 \) cm. For a 20×20 grid:

• 21 horizontal wires (0 to 20), each 20 cm: \( 21 \times 20 = 420 \) cm.
• 21 vertical wires, each 20 cm: \( 21 \times 20 = 420 \) cm.
• Total: \( 420 + 420 = 840 \) cm.

Official Solution: “In a 20 by 20 grid there are 21 horizontal wires each of length 20 cm and 21 vertical wires each of length 20 cm. Therefore the total length of wire required for a 20 by 20 grid is \( 42 \times 20 \, \text{cm} = 840 \, \text{cm} \).”

Options: A: 400, B: 420, C: 441, D: 800, E: 840 (correct). Thus, E: 840 cm.

Answer: B \( \frac{3}{16} \)

Explanation:

Large triangle split into 4 equal triangles, each \( \frac{1}{4} \) of the area. One is subdivided into 4 smaller triangles, each \( \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \) of the large triangle. Three are shaded:

• Shaded area = \( 3 \times \frac{1}{16} = \frac{3}{16} \).

Official Solution: “Each of these triangles has an area equal to one quarter of the area of the large triangle. The equilateral triangle which is partly shaded has been divided into four congruent equilateral triangles, three of which are shaded. Therefore the fraction of the area of the large triangle which has been shaded is \( \frac{3}{4} \times \frac{1}{4} = \frac{3}{16} \).”

Options: A: \( \frac{1}{8} \), B: \( \frac{3}{16} \) (correct), C: \( \frac{1}{4} \), D: \( \frac{5}{16} \), E: \( \frac{3}{8} \). Thus, B: \( \frac{3}{16} \).

Answer: B 4

Explanation:

Four integers, mean = 5, so sum = \( 4 \times 5 = 20 \). Median = 6, so ordered list \( a, b, c, d \) has \( \frac{b + c}{2} = 6 \), \( b + c = 12 \). Then \( a + d = 20 – 12 = 8 \). Mean of smallest (\( a \)) and largest (\( d \)) = \( \frac{a + d}{2} = \frac{8}{2} = 4 \).

Example: 1, 6, 6, 7 (sum = 20, median = \( \frac{6+6}{2} = 6 \), mean of 1 and 7 = 4).

Official Solution: “Sum = 20. Median is 6, so sum of two middle integers is 12. Hence sum of smallest and largest is \( 20 – 12 = 8 \). Mean = \( 8 \div 2 = 4 \).”

Options: A: 3, B: 4 (correct), C: 5, D: 6, E: 8. Thus, B: 4.

Answer: D 112°

Explanation:

Without the diagram, use the official solution’s logic. Let \( \angle PON = x \), so \( \angle OLM = 2x \). Vertically opposite \( \angle PON \) is \( \angle KOL = x \). \( \angle LKO = 180° – 124° = 56° \) (straight line). In triangle \( KOL \), exterior angle \( \angle OLM = 2x \) equals \( \angle KOL + \angle LKO = x + 56 \). Solve: \( 2x = x + 56 \), \( x = 56 \), \( 2x = 112° \).

Official Solution: “\( \angle OLM = 2x \), \( \angle PON = x \), \( \angle KOL = x \). \( \angle LKO = 180° – 124° = 56° \). In triangle \( KOL \), \( 2x = x + 56 \). So \( x = 56 \), \( \angle OLM = 112° \).”

Options: A: 102°, B: 106°, C: 108°, D: 112° (correct), E: 124°. Thus, D: 112°.

Answer: E 35

Explanation:

Let \( f \) = just football, \( t \) = just tennis, \( b \) = both. Total = 42: \( f + t + b = 42 \). Tennis total = just football: \( t + b = f \). Twice as many play both as just tennis: \( b = 2t \). Solve:

• \( t + b = f \), \( b = 2t \), so \( t + 2t = f \), \( f = 3t \).
• \( f + t + b = 42 \), substitute \( f = 3t \), \( b = 2t \): \( 3t + t + 2t = 42 \), \( 6t = 42 \), \( t = 7 \).
• \( b = 2 \times 7 = 14 \), \( f = 3 \times 7 = 21 \).
• Football players = \( f + b = 21 + 14 = 35 \).

Official Solution: “\( f + t + b = 42 \), \( t + b = f \), \( b = 2t \). Then \( f = t + 2t = 3t \), \( 3t + t + 2t = 42 \), \( t = 7 \), football = \( 42 – t = 35 \).”

Options: A: 7, B: 14, C: 21, D: 28, E: 35 (correct). Thus, E: 35.

Answer: A \( \frac{6}{25} \)

Explanation:

Possible numbers: .0625, 0.625, 6.25, 62.5, 625.

• \( .0625 = \frac{625}{10000} = \frac{1}{16} \) (C).
• \( 0.625 = \frac{625}{1000} = \frac{5}{8} \) (B).
• \( 6.25 = \frac{625}{100} = \frac{25}{4} \) (D).
• \( 62.5 = \frac{625}{10} \) (not listed).
• \( 625 = 25^2 \) (E).
• \( \frac{6}{25} = 0.24 \), not achievable.

Official Solution: “\( \frac{6}{25} = 0.24 \). Others: \( \frac{5}{8} = 0.625 \), \( \frac{1}{16} = .0625 \), \( \frac{25}{4} = 6.25 \), \( 25^2 = 625 \).”

Options: A: \( \frac{6}{25} \) (correct), B: \( \frac{5}{8} \), C: \( \frac{1}{16} \), D: \( \frac{25}{4} \), E: \( 25^2 \). Thus, A: \( \frac{6}{25} \).

Answer: E 1056

Explanation:

Total area = sum of square areas: \( 1^2 + 4^2 + 7^2 + 8^2 + 9^2 + 10^2 + 14^2 + 15^2 + 18^2 = 1 + 16 + 49 + 64 + 81 + 100 + 196 + 225 + 324 = 1056 \). Rectangle dimensions must be 32×33 (as per official solution), and \( 32 \times 33 = 1056 \).

Official Solution: “Side-lengths deduced as 32 and 33. Area = \( 32 \times 33 = 1056 \).” (Or \( 18 + 15 = 33 \), \( 18 + 14 = 32 \)).

Options: A: 144, B: 225, C: 900, D: 1024, E: 1056 (correct). Thus, E: 1056.

Answer: B 5

Explanation:

Non-prime digits: 0, 1, 4, 6, 8, 9. Primes can’t end in 0, 4, 6, 8. Check units 1, 9:

• 11, 19, 41, 61, 89 (all prime, digits non-prime).

Total: 5.

Official Solution: “Units digit 1 or 9. Primes: 11, 19, 41, 61, 89. Hence five.”

Options: A: 6, B: 5 (correct), C: 4, D: 3, E: 2. Thus, B: 5.

Answer: E 7

Explanation:

Hey there! Let’s figure out what \( L \) is in this fun table puzzle. The table has letters \( J \), \( K \), and \( L \), and we need to use the sums to find \( L \). Let’s go step by step—it’s like a treasure hunt!

Step 1: Write down what we know from the table.

• Row 1: \( J + K + J = 5 \) (that’s \( 2J + K = 5 \))
• Row 2: \( K + K + L = 13 \) (that’s \( 2K + L = 13 \))
• Row 3: \( L + J + L = 15 \) (that’s \( J + 2L = 15 \))
• Column 1: \( J + K + L = 11 \)
• Column 2: \( K + K + J = 7 \) (that’s \( 2K + J = 7 \))

Step 2: Find a clue by adding two sums together.

Let’s add Row 1 and Column 2 because they both have \( J \) and \( K \):

• Row 1: \( 2J + K = 5 \)
• Column 2: \( 2K + J = 7 \)
• Add them: \( (2J + K) + (2K + J) = 5 + 7 \)
• That’s \( 2J + K + 2K + J = 12 \), or \( 3J + 3K = 12 \).

Now, divide both sides by 3: \( J + K = 4 \). Cool, we’ve got something!

Step 3: Use our clue to find \( L \).

Look at Column 1: \( J + K + L = 11 \). We know \( J + K = 4 \), so:

• \( 4 + L = 11 \)
• Subtract 4 from 11: \( L = 7 \)

Step 4: Check if it works!

If \( L = 7 \), let’s find \( J \) and \( K \):

• Row 3: \( J + 2 \times 7 = 15 \), so \( J + 14 = 15 \), \( J = 1 \)
• Row 1: \( 2 \times 1 + K = 5 \), so \( 2 + K = 5 \), \( K = 3 \)
• Check Column 1: \( 1 + 3 + 7 = 11 \) (yes!)
• Row 2: \( 3 + 3 + 7 = 13 \) (yes!)
• Column 2: \( 3 + 3 + 1 = 7 \) (yes!)

Everything fits perfectly, like a puzzle clicking into place!

Official Solution (grown-up version): They say, “Add the top row and middle column: \( 2J + K + 2K + J = 12 \), so \( 3J + 3K = 12 \), and \( J + K = 4 \). Then from the first column, \( J + K + L = 11 \), so \( L = 11 – 4 = 7 \).” Same answer, just faster!

So, \( L = 7 \), and the answer is E: 7. Great job, detective!

Answer: D 184

Explanation:

Hi, friends! Imagine building with little toy blocks. Edmund makes a cube with 8 blocks, and Samuel makes a bigger shape. Let’s figure out how many extra blocks Samuel uses!

Step 1: Picture Edmund’s cube.

A cube is like a box that’s the same size all around. Since it’s made of 8 small cubes, think of it as 2 cubes long, 2 cubes wide, and 2 cubes high (because \( 2 \times 2 \times 2 = 8 \)). Easy, right?

Step 2: Figure out Samuel’s big cuboid.

Samuel’s shape is bigger:

• Twice as long: \( 2 \times 2 = 4 \) cubes long
• Three times as wide: \( 3 \times 2 = 6 \) cubes wide
• Four times as high: \( 4 \times 2 = 8 \) cubes high

Now, how many blocks does Samuel need? Multiply those numbers:

• \( 4 \times 6 = 24 \)
• \( 24 \times 8 = 192 \) (do it step-by-step: \( 24 \times 8 = 24 \times (2 + 6) = 48 + 144 = 192 \))

So, Samuel uses 192 little cubes!

Step 3: Find out how many more.

Edmund uses 8, Samuel uses 192. Subtract:

• \( 192 – 8 = 184 \)

Samuel uses 184 more cubes than Edmund!

Step 4: Double-check.

If Samuel uses 184 more than 8: \( 8 + 184 = 192 \). That matches his total, so we’re good!

Official Solution: They say, “Edmund’s cube is \( 2 \times 2 \times 2 = 8 \). Samuel’s cuboid is \( 4 \times 6 \times 8 = 192 \). So, \( 192 – 8 = 184 \).” Same answer, just quicker!

The answer is D: 184. Awesome building skills!

Answer: D \( 42 \times 48 \)

Explanation:

Hello! This is like a magic trick with numbers. We need to find a pair of numbers where if you flip their digits around, the answer stays the same when you multiply them. Let’s try each one!

Step 1: Understand the example.

\( 62 \times 13 = 806 \), and if you flip them to \( 26 \times 31 \), it’s still 806. Cool, huh? Let’s test the options.

Step 2: Check each pair.

• • A: \( 25 \times 36 \)

\( 25 \times 36 = 900 \) (25 × 30 = 750, 25 × 6 = 150, 750 + 150 = 900)

Flip: \( 52 \times 63 = 3276 \) (52 × 60 = 3120, 52 × 3 = 156, 3120 + 156 = 3276)

900 ≠ 3276, so nope!

• • B: \( 34 \times 42 \)

\( 34 \times 42 = 1428 \) (34 × 40 = 1360, 34 × 2 = 68, 1360 + 68 = 1428)

Flip: \( 43 \times 24 = 1032 \) (43 × 20 = 860, 43 × 4 = 172, 860 + 172 = 1032)

1428 ≠ 1032, no match.

• • C: \( 54 \times 56 \)

\( 54 \times 56 = 3024 \) (54 × 50 = 2700, 54 × 6 = 324, 2700 + 324 = 3024)

Flip: \( 45 \times 65 = 2925 \) (45 × 60 = 2700, 45 × 5 = 225, 2700 + 225 = 2925)

3024 ≠ 2925, not this one.

• • D: \( 42 \times 48 \)

\( 42 \times 48 = 2016 \) (42 × 40 = 1680, 42 × 8 = 336, 1680 + 336 = 2016)

Flip: \( 24 \times 84 = 2016 \) (24 × 80 = 1920, 24 × 4 = 96, 1920 + 96 = 2016)

2016 = 2016, yay, it works!

• • E: \( 32 \times 43 \)

\( 32 \times 43 = 1376 \) (32 × 40 = 1280, 32 × 3 = 96, 1280 + 96 = 1376)

Flip: \( 23 \times 34 = 782 \) (23 × 30 = 690, 23 × 4 = 92, 690 + 92 = 782)

1376 ≠ 782, no good.

Step 3: Pick the winner.

Only \( 42 \times 48 \) and \( 24 \times 84 \) both equal 2016. That’s our magic pair!

Official Solution: They check the last digits first and say, “\( 42 \times 48 = 2016 \), and \( 24 \times 84 = 2016 \), so D works.” We did the full math, but same answer!

The answer is D: \( 42 \times 48 \). You’re a number wizard now!

Answer: E 144 cm²

Explanation:

Hey, let’s fold some paper with Harriet! She starts with a square, folds it twice, and the final shape’s perimeter is 30 cm. We need to find the area of her original square. Let’s do it together!

Step 1: Imagine the square.

Let’s call the side of the square \( s \) cm. A square has all sides equal, so it’s \( s \) by \( s \).

Step 2: Fold it in half.

Fold the square in half one way (say, top to bottom). Now it’s a rectangle that’s \( s \) long and \( \frac{s}{2} \) wide.

Step 3: Fold it again.

Fold it in half the other way (side to side). The new rectangle is \( s \) long and \( \frac{s}{4} \) wide (because \( \frac{s}{2} \) divided by 2 is \( \frac{s}{4} \)). It’s not a square because \( s \) isn’t the same as \( \frac{s}{4} \).

Step 4: Use the perimeter.

The perimeter is all the way around: 2 lengths plus 2 widths.

• Length = \( s \), Width = \( \frac{s}{4} \)
• Perimeter = \( 2 \times s + 2 \times \frac{s}{4} = 30 \)
• Simplify: \( 2s + \frac{s}{2} = 30 \)
• Get rid of the fraction—multiply everything by 2: \( 4s + s = 60 \)
• That’s \( 5s = 60 \), so \( s = 12 \) cm

Step 5: Find the area.

Area of a square is side times side: \( 12 \times 12 = 144 \) cm².

Step 6: Check it.

First fold: \( 12 \times 6 \). Second fold: \( 12 \times 3 \). Perimeter = \( 2 \times 12 + 2 \times 3 = 24 + 6 = 30 \) cm. Perfect!

Official Solution: They use \( 4x \) for the side, get \( 10x = 30 \), \( x = 3 \), so side = 12 cm, area = 144 cm². Same as us!

The answer is E: 144 cm². Folding is fun!

Answer: B 210

Explanation:

Hi, explorers! We’re looking for special numbers that leave a remainder of 1 when divided by the four smallest prime numbers. Let’s find the two smallest ones and see how far apart they are!

Step 1: What are the four smallest primes?

Primes are numbers only divisible by 1 and themselves. The smallest ones are: 2, 3, 5, and 7.

Step 2: Find a number that works.

We need a number that, when divided by 2, 3, 5, and 7, leaves 1 left over. Like:

• \( 2 \div 2 = 1 \) remainder 0 (nope)
• \( 3 \div 2 = 1 \) remainder 1, \( 3 \div 3 = 1 \) remainder 0 (nope)

We need it to work for all four. Let’s try a trick: multiply them together (2 × 3 × 5 × 7 = 210) and add 1.

• \( 210 + 1 = 211 \)
• \( 211 \div 2 = 105 \) remainder 1
• \( 211 \div 3 = 70 \) remainder 1
• \( 211 \div 5 = 42 \) remainder 1
• \( 211 \div 7 = 30 \) remainder 1

It works! 211 is our first number.

Step 3: Find the next one.

Add 210 again: \( 211 + 210 = 421 \).

• \( 421 \div 2 = 210 \) remainder 1
• \( 421 \div 3 = 140 \) remainder 1
• \( 421 \div 5 = 84 \) remainder 1
• \( 421 \div 7 = 60 \) remainder 1

Yes! 421 works too.

Step 4: Find the difference.

\( 421 – 211 = 210 \). That’s how far apart they are!

Official Solution: They say, “Multiply 2 × 3 × 5 × 7 = 210. First number = 211, next = 421, difference = 210.” Same as our adventure!

The answer is B: 210. You’re a prime number hero!

Answer: B 4

Explanation:

Hi, friends! Let’s imagine a big room with chairs for Susan’s talk. We need to find the most empty chairs that can be next to each other in one row. Let’s play with the numbers!

Step 1: Count all the chairs.

8 rows with 10 chairs each: \( 8 \times 10 = 80 \) chairs total.

Step 2: See how many are empty.

54 parents are sitting, so: \( 80 – 54 = 26 \) chairs are empty.

Step 3: Understand the rule.

Every parent is alone or has just one neighbor. In a row of 10:

• Best way: P-P-P-P-P-P-P-_-_, 7 parents (3 alone, 4 paired), 3 empty.

Step 4: Pack most parents in 7 rows.

Put 7 parents in 7 rows: \( 7 \times 7 = 49 \) parents, 3 empty per row.

Step 5: Fill the last row.

54 – 49 = 5 parents left for the 8th row. Try it:

• P-P-P-P-P-_-_-_, 5 parents, 5 empty, but 4 are together!

Step 6: Check the empties.

7 rows × 3 empty = 21, plus 5 in the last row = 26 empty total. The most together in one row is 4!

Official Solution: They say, “Max 7 per row, 49 total, 8th row has 5, max adjacent empty = 4.” We got it too!

The answer is B: 4. Great seating plan!

Answer: E 38°

Explanation:

Hey, angle hunters! We don’t have the picture, but we can still find \( \angle JKL \) using some clever clues from the grown-ups’ solution. Let’s go on an angle adventure!

Step 1: Look at the straight lines.

On line \( PQRS \), there are angles we’ll call \( x \), \( 2y \), and \( a \) that add to 180°: \( x + 2y + a = 180 \). Another part has \( y + 2x + b = 180 \).

Step 2: Add them up.

Put those together:

• \( (x + 2y + a) + (y + 2x + b) = 180 + 180 \)
• That’s \( x + 2y + a + y + 2x + b = 360 \)
• Simplify: \( 3x + 3y + a + b = 360 \)

Step 3: Check a triangle.

In triangle \( QMR \), angles \( a + b + 33 = 180 \), so \( a + b = 147 \).

Step 4: Find \( x + y \).

Use that in our big equation:

• \( 3x + 3y + 147 = 360 \)
• Subtract 147: \( 3x + 3y = 213 \)
• Divide by 3: \( x + y = 71 \)

Step 5: More angles!

Some angles match because they’re opposite: \( d = 2y \), \( e = 2x \). In triangle \( QRK \), \( d + e + c = 180 \), so:

• \( 2x + 2y + c = 180 \)
• Since \( x + y = 71 \), \( 2x + 2y = 2 \times 71 = 142 \)
• \( 142 + c = 180 \), \( c = 38° \)

Step 6: That’s our angle!

\( \angle JKL = c = 38° \). We found it!

Official Solution: They say, “Add equations, use the triangle, get \( x + y = 71 \), then \( c = 38° \).” We took the scenic route, but same spot!

The answer is E: 38°. You’re an angle superstar!