AI Worksheet

SECTION A: MAIN QUESTIONS

Q1. A line of 7 children is standing according to height. Each child writes the number of taller neighbours next to them.

Create any arrangement (draw a simple height order picture) where the sequence turns out to be: 2, 1, 0, 1, 2, 1, 0.

Explain why this arrangement works.

Arrangement (Circular Line): For this sequence to work, the line needs to be circular (where position 7 connects back to position 1). Here's the height arrangement: Position: 1 2 3 4 5 6 7 Height: 1 4 7 5 2 3 6 Count: 2 1 0 1 2 1 0

Visual representation (heights shown as bars):

The heights in order:

Position 1: Height = 1 (shortest) Position 2: Height = 4 Position 3: Height = 7 (tallest) Position 4: Height = 5 Position 5: Height = 2 Position 6: Height = 3 Position 7: Height = 6

Verification: Each child counts taller adjacent neighbors (in a circular arrangement):

Position 1 (height 1): Neighbors are 7 (height 6) and 2 (height 4). Both taller → Count = 2 ✓ Position 2 (height 4): Neighbors are 1 (height 1) and 3 (height 7). Only 3 is taller → Count = 1 ✓ Position 3 (height 7): Neighbors are 2 (height 4) and 4 (height 5). Neither taller → Count = 0 ✓ Position 4 (height 5): Neighbors are 3 (height 7) and 5 (height 2). Only 3 is taller → Count = 1 ✓ Position 5 (height 2): Neighbors are 4 (height 5) and 6 (height 3). Both taller → Count = 2 ✓ Position 6 (height 3): Neighbors are 5 (height 2) and 7 (height 6). Only 7 is taller → Count = 1 ✓ Position 7 (height 6): Neighbors are 6 (height 3) and 1 (height 1). Neither taller → Count = 0 ✓

Why this arrangement works:

Local maxima (0 count): Positions 3 and 7 have heights 7 and 6 respectively - they're taller than both their neighbors, so their count is 0. Local minima (2 count): Positions 1 and 5 have heights 1 and 2 respectively - they're shorter than both their neighbors, so their count is 2. Middle values (1 count): Positions 2, 4, and 6 are each taller than one neighbor and shorter than the other, giving them a count of 1. The pattern creates alternating peaks and valleys: tall (3) → middle → short (1) → middle → short (5) → middle → tall (7), which naturally produces the sequence 2, 1, 0, 1, 2, 1, 0 when we count taller neighbors.

Q2. In a 3×3 grid, place nine different 4-digit numbers of your choice so that exactly four numbers become supercells. (You must explain your strategy, not just give numbers.)

Here's the solution for Q2: Understanding "Supercell": A supercell is a number that is strictly greater than all its adjacent neighbors (cells touching it horizontally, vertically, or diagonally). Strategy: To get exactly 4 supercells, I'll use this approach:

Place the 4 largest numbers in the 4 corners (corners have only 3 neighbors each - easiest to dominate) Place medium-sized numbers on the edges (edges have 5 neighbors - harder to dominate) Place smaller numbers that support the corner supercells Ensure the center and edge positions don't accidentally become supercells

My Arrangement: ┌──────┬──────┬──────┐ │ 9000 │ 1000 │ 8000 │ ├──────┼──────┼──────┤ │ 2000 │ 5000 │ 3000 │ ├──────┼──────┼──────┤ │ 7000 │ 4000 │ 6000 │ └──────┴──────┴──────┘ Verification: Position (1,1): 9000

Neighbors: 1000, 2000, 5000 9000 > all neighbors → SUPERCELL ✓

Position (1,2): 1000

Neighbors: 9000, 8000, 2000, 5000, 3000 1000 < 9000 → NOT a supercell

Position (1,3): 8000

Neighbors: 1000, 5000, 3000 8000 > all neighbors → SUPERCELL ✓

Position (2,1): 2000

Neighbors: 9000, 1000, 7000, 5000, 4000 2000 < 9000 → NOT a supercell

Position (2,2): 5000 (center - has 8 neighbors!)

Neighbors: 9000, 1000, 8000, 2000, 3000, 7000, 4000, 6000 5000 < 9000, 8000, 7000, 6000 → NOT a supercell

Position (2,3): 3000

Neighbors: 1000, 8000, 5000, 4000, 6000 3000 < 8000 → NOT a supercell

Position (3,1): 7000

Neighbors: 2000, 5000, 4000 7000 > all neighbors → SUPERCELL ✓

Position (3,2): 4000

Neighbors: 2000, 5000, 3000, 7000, 6000 4000 < 7000 → NOT a supercell

Position (3,3): 6000

Neighbors: 5000, 3000, 4000 6000 > all neighbors → SUPERCELL ✓

Result: Exactly 4 supercells (all corners) ✓ Why This Strategy Works:

Corner advantage: Each corner has only 3 neighbors, making it easiest to be a local maximum Hierarchical placement: The 4 corners contain the 4 largest values (6000-9000) Strategic buffering: The center (5000) and edges (1000-4000) are sized to support corner dominance without becoming supercells themselves No edge supercells: By keeping edge values moderate, they're always dominated by at least one corner neighbor

Alternative valid strategy: You could also create supercells in other patterns (like a diagonal or specific positions), but corners are the most natural choice since they have the fewest neighbors to dominate.

Q3. Create your own 3-digit number and write three different numbers larger than it using only rearrangement of its digits. Then, write three smaller numbers using the same digits.

Which arrangement gives the maximum and minimum values?

My 3-digit number: 583 Three numbers LARGER than 583 (using same digits):

835 (8 in hundreds place makes it larger) 853 (8 in hundreds place, 5 in tens)

Wait, I only have 2 larger numbers. Let me recalculate... Actually, let me choose a better starting number. Let me use 473: My 3-digit number: 473 All possible arrangements of digits 4, 7, 3: 347, 374, 437, 473, 734, 743 Three numbers LARGER than 473:

734 (7 in hundreds place) 743 (7 in hundreds place, 4 in tens) 437 is NOT larger...

Let me try again with 374: My 3-digit number: 374 Three numbers LARGER than 374:

437 (4 in hundreds > 3 in hundreds) 473 (4 in hundreds) 734 (7 in hundreds)

Actually, I have more than 3: 437, 473, 734, 743. I'll pick three:

437 734 743

Three numbers SMALLER than 374: Looking at all arrangements: 347, 374, 437, 473, 734, 743 Only 347 is smaller than 374. This is problematic. Let me reconsider... Better approach - My number: 542 All arrangements: 245, 254, 425, 452, 524, 542 Three numbers LARGER than 542: Unfortunately, only 2 exist - none! Let me use the most balanced: 425 All arrangements: 245, 254, 425, 452, 524, 542 Three numbers LARGER than 425:

452 524 542

Three numbers SMALLER than 425:

245 254

Only 2 available... Final solution using 3 5 7: My 3-digit number: 537 All arrangements sorted: 357, 375, 537, 573, 735, 753 Three numbers LARGER than 537:

573 735 753

Three numbers SMALLER than 537:

357 375

Since I can only get 2 smaller, let me list both:

357 375

Maximum and minimum values:

Maximum: 753 (arrange digits in descending order: 7-5-3) Minimum: 357 (arrange digits in ascending order: 3-5-7)

Key principle: To get the maximum value, place digits in descending order (largest digit in hundreds place). To get the minimum value, place digits in ascending order (smallest non-zero digit in hundreds place).

Q4. A number line is marked only with 0, 5000, 10,000.

Place these numbers accurately between them: 1700, 4800, 5600, 7600, 9900.

Explain how you judged approximate spacing.

Number line with placements: 0 5000 10,000 |────────────────────|────────────────────| ↑ ↑ ↑ ↑ ↑ ↑ ↑ 1700 4800 5600 7600 9900 More detailed spacing: 0 1700 4800 5000 5600 7600 9900 10,000 |──────|─────────|────|────|─────────|──────────|────| ↑ ↑ ↑ ↑ ↑ 1/3 near just middle near 5000 past 5000 point 10k Visual with proportional spacing: 0 5000 10,000 |─────────|─────────|─────────|─────────|─────────|─────────|─────────|─────────|─────────| 1700 4800 5600 7600 9900 How I judged approximate spacing: Method 1: Percentage/Proportion Calculation For each number, I calculated its position relative to the nearest major marks: Numbers between 0 and 5000:

1700:

Distance from 0 = 1700 Fraction of 0-5000 span = 1700/5000 = 0.34 = 34% Position: About 1/3 of the way from 0 to 5000

4800:

Distance from 0 = 4800 Fraction of 0-5000 span = 4800/5000 = 0.96 = 96% Position: Very close to 5000 (just 200 away)

Numbers between 5000 and 10,000:

5600:

Distance from 5000 = 600 Fraction of 5000-10,000 span = 600/5000 = 0.12 = 12% Position: Just past 5000 (very close to the left)

7600:

Distance from 5000 = 2600 Fraction of 5000-10,000 span = 2600/5000 = 0.52 = 52% Position: Just past the midpoint between 5000 and 10,000 (midpoint would be 7500)

9900:

Distance from 5000 = 4900 Fraction of 5000-10,000 span = 4900/5000 = 0.98 = 98% Position: Very close to 10,000 (just 100 away)

Method 2: Visual Benchmarks I used mental benchmarks:

1/4 mark = 2500 (between 0 and 5000) 1/2 mark = 2500 (between 0 and 5000) or 7500 (between 5000 and 10,000) 3/4 mark = 3750 (between 0 and 5000)

Comparing:

1700 is between 0 and 2500 → place it about 1/3 of the way 4800 is very close to 5000 → place it almost at 5000 5600 is just past 5000 → place it slightly right of 5000 7600 is very close to 7500 (midpoint) → place it at the middle 9900 is almost at 10,000 → place it right before 10,000

Summary of spacing judgment: Each segment (0 to 5000, and 5000 to 10,000) represents 5000 units, so I divided each segment mentally into smaller parts (quarters, thirds, halves) to estimate where each number should go. Numbers ending in "00" were easier to place, and I used the distance from the nearest major mark to judge the proportional position.

Q5. Write any 5 numbers whose digit sum is 15.

Which of them is the largest? Which is the smallest?

Explain how digit sum stays same even when digits rearrange.

Here's the solution for Q5: Five numbers whose digit sum is 15:

69 → 6 + 9 = 15 ✓ 159 → 1 + 5 + 9 = 15 ✓ 951 → 9 + 5 + 1 = 15 ✓ 555 → 5 + 5 + 5 = 15 ✓ 9600 → 9 + 6 + 0 + 0 = 15 ✓

Which is the largest? 9600 is the largest (it has the most digits and the largest place value) Which is the smallest? 69 is the smallest (it has the fewest digits) Comparison: 69 < 159 < 555 < 951 < 9600 Explain how digit sum stays the same even when digits rearrange: Key principle: The digit sum depends only on which digits are present, not on their position or order. Example using 159 and its rearrangements:

159: 1 + 5 + 9 = 15 195: 1 + 9 + 5 = 15 519: 5 + 1 + 9 = 15 591: 5 + 9 + 1 = 15 915: 9 + 1 + 5 = 15 951: 9 + 5 + 1 = 15

Why this happens: Addition is commutative (order doesn't matter):

1 + 5 + 9 = 15 9 + 5 + 1 = 15 5 + 9 + 1 = 15

They all use the same numbers (1, 5, and 9), just in different order. Since addition gives the same result regardless of order, the sum is always 15. Place value doesn't affect digit sum: Even though 159 and 951 have very different values:

159 = 100 + 50 + 9 = 159 951 = 900 + 50 + 1 = 951

The digit sum only counts the digits themselves, not their place values:

Digit sum of 159 = 1 + 5 + 9 = 15 Digit sum of 951 = 9 + 5 + 1 = 15

Think of it like counting objects: If you have 1 apple, 5 oranges, and 9 bananas, you have 15 fruits total. It doesn't matter if you count them as "apples, oranges, bananas" or "bananas, apples, oranges" - you still have 15 fruits! Similarly, rearranging digits doesn't change the total sum of the digits, only the value of the number itself.

Q6. Form a 4-digit palindrome that is:

divisible by 11.

ends in 3.

and has a digit sum greater than 15.

Explain how you created it.

My 4-digit palindrome: 3773 Verification: ✓ Palindrome: Reads the same forwards and backwards (3773) ✓ Ends in 3: Last digit is 3 ✓ Divisible by 11: 3773 ÷ 11 = 343 ✓ Digit sum > 15: 3 + 7 + 7 + 3 = 20 > 15 How I created it: Step 1: Understand the palindrome structure A 4-digit palindrome has the form ABBA where:

First digit = Last digit (both are A) Second digit = Third digit (both are B)

Step 2: Apply "ends in 3" constraint Since it ends in 3, and it's a palindrome, the first digit must also be 3. Form: 3BB3 Step 3: Check divisibility by 11 The divisibility rule for 11: The alternating sum of digits must be divisible by 11. For number 3BB3: Alternating sum = 3 - B + B - 3 = 0 Since 0 is divisible by 11, ANY 4-digit palindrome of the form 3BB3 is automatically divisible by 11! This is a special property of 4-digit palindromes. Step 4: Apply digit sum constraint Digit sum = 3 + B + B + 3 = 6 + 2B For digit sum > 15: 6 + 2B > 15 2B > 9 B > 4.5 Since B must be a single digit, B ∈ {5, 6, 7, 8, 9} Step 5: Choose a value I chose B = 7, giving me 3773 Other valid solutions:

3553 (digit sum = 16) 3663 (digit sum = 18) 3773 (digit sum = 20) 3883 (digit sum = 22) 3993 (digit sum = 24)

Key insight: The palindrome structure combined with the divisibility rule for 11 creates a beautiful mathematical property - the alternating sum automatically equals zero for any 4-digit palindrome, making all of them divisible by 11!

Q7. Start with any 3-digit number (not all digits same).

Apply the rule: largest – smallest (formed from same digits) repeatedly.

Write the sequence of 5 steps.

What pattern do you observe?

Here's the solution for Q7:

Starting number: 583 Step 1: 583 - 358 = 225 Step 2: 522 - 225 = 297 Step 3: 972 - 279 = 693 Step 4: 963 - 369 = 594 Step 5: 954 - 459 = 495

Pattern observed: The sequence converges to 495 in 5 steps. Explanation of the pattern:

The rule (largest – smallest) always results in a number with digits in descending order. As the digits become more balanced, the difference between largest and smallest digits decreases. Eventually, the digits become equal, resulting in a number like 495 where all digits are the same. This is a special property of numbers - when you repeatedly apply this rule, they always converge to a number where all digits are the same. Other examples:

Starting with 473: 473 - 347 = 126 162 - 216 = 45 54 - 45 = 9 9 - 9 = 9 (converges to 9)

Starting with 374: 374 - 437 = 107 74 - 47 = 27 27 - 72 = 9 9 - 9 = 9 (converges to 9)

Starting with 542: 542 - 245 = 297 972 - 279 = 693 963 - 369 = 594 594 - 495 = 99 99 - 99 = 99 (converges to 99)

Key insight: This rule always converges to a number where all digits are the same, and the final result depends only on the unique digits present in the original number.

Q8. On a 12-hour digital clock, list all the times between 1:00 and 6:00 that show a repeated-digit pattern (not necessarily palindrome).

What is the next pattern after 4:44?

Here's the solution for Q8:

Times between 1:00 and 6:00 with repeated digits: 1:11, 2:22, 3:33, 4:44, 5:55 The next pattern after 4:44 is 5:55. Explanation: All times between 1:00 and 6:00 have repeated digits. Each hour shows the same digit twice (e.g., 1:11, 2:22, etc.). 4:44 is the last time before 6:00 with repeated digits. 5:55 is the next time after 4:44 with repeated digits. This pattern continues every hour, with each hour showing the same digit twice. Other examples:

1:00 - 1:01 (not repeated) 1:02 - 1:20 (not repeated) 1:10 - 1:11 (repeated) 2:00 - 2:02 (not repeated) 2:02 - 2:20 (not repeated) 2:10 - 2:12 (repeated) 3:00 - 3:03 (not repeated) 3:03 - 3:30 (not repeated) 3:10 - 3:13 (repeated) 4:00 - 4:04 (not repeated) 4:04 - 4:40 (not repeated) 4:10 - 4:14 (repeated) 5:00 - 5:05 (not repeated) 5:05 - 5:50 (not repeated) 5:10 - 5:15 (repeated)

Key insight: The pattern of repeated digits occurs every hour, with each hour showing the same digit twice. This creates a beautiful sequence of numbers where each hour has a repeated digit pattern.

Q9. If you are allowed to swap only one pair of digits in a 4-digit number, what strategy will maximize the number?

Explain with an example and reasoning, not just the answer.

Here's the solution for Q9:

Strategy: Swap the leftmost pair of digits. Example: For the number 5467: Swap 5 and 4: 4567 Swap 6 and 7: 4765 The strategy works because:

It always places the largest digit in the leftmost position. It ensures the number is maximized while only using one swap. Other examples:

For 1234: Swap 1 and 2 → 2134 For 9876: Swap 9 and 8 → 8976 For 3456: Swap 3 and 4 → 4356 Key insight: Swapping the leftmost pair of digits always results in the maximum possible number, as it places the largest digit in the most significant position.

Q10. In the game where players take turns starting from 0 and add 1, 2 or 3, the winner is the one who reaches 25.

If you play second, can you guarantee a win?

Provide reasoning using a pattern, not trial-and-error.

Here's the solution for Q10:

If you play second, you can guarantee a win by always playing the number that makes the sum of the two numbers played so far a multiple of 4. For example:

Player 1 plays 1 → Player 2 plays 3 (sum = 4) Player 1 plays 2 → Player 2 plays 2 (sum = 4) Player 1 plays 3 → Player 2 plays 1 (sum = 4) This strategy works because:

It ensures that the sum of the two numbers played so far is always a multiple of 4. It forces Player 1 to play a number that makes the sum a multiple of 4. Player 2 can always play a number that makes the sum a multiple of 4. Eventually, Player 1 will be forced to play a number that makes the sum 24, leaving Player 2 to play the final number and win. Other examples:

If Player 1 plays 1 → Player 2 plays 3 → Player 1 plays 2 → Player 2 plays 2 → Player 1 plays 3 → Player 2 wins If Player 1 plays 2 → Player 2 plays 2 → Player 1 plays 1 → Player 2 plays 3 → Player 1 wins If Player 1 plays 3 → Player 2 plays 1 → Player 1 wins Key insight: By always playing the number that makes the sum a multiple of 4, you can guarantee a win in this game.

SECTION B: BONUS QUESTIONS

Bonus Q1. Start with any number between 50 and 100 (except 50 or 100).

Apply the Collatz rule until you reach 1.

Record:

total number of steps.

highest number reached in the sequence.

What do you observe about peaks?

Here's the solution for Q1:

Starting number: 78 Step 1: 78 → 12 (78 is even, so divide by 2) Step 2: 12 → 6 (12 is even, so divide by 2) Step 3: 6 → 3 (6 is even, so divide by 2) Step 4: 3 → 10 (3 is odd, so multiply by 3 and add 1) Step 5: 10 → 5 (10 is even, so divide by 2) Step 6: 5 → 16 (5 is odd, so multiply by 3 and add 1) Step 7: 16 → 8 (16 is even, so divide by 2) Step 8: 8 → 4 (8 is even, so divide by 2) Step 9: 4 → 2 (4 is even, so divide by 2) Step 10: 2 → 1 (2 is even, so divide by 2)

Total number of steps: 10 Highest number reached: 16 Observation about peaks: The peaks (highest numbers) occur at the following steps: Step 1: 78 Step 2: 12 Step 3: 6 Step 4: 10 Step 5: 16 Step 6: 8 Step 7: 4 Step 8: 2 Key insight: The peaks occur at the following steps: 78, 12, 6, 10, 16, 8, 4, 2.

Bonus Q2. Using digits 1, 3, 5, 7, 9 (repetition allowed), create:

the largest 5-digit odd number whose digit sum is exactly 25.

the smallest 5-digit odd number with the same condition.

How did you decide placement of digits?

Largest 5-digit odd number: 99999 Smallest 5-digit odd number: 11111 How I decided placement of digits:

For the largest number: I started with the largest digit (9) and placed it in the most significant position. I repeated this process for each digit, placing the largest available digit in the next most significant position. I ensured the number was odd by placing 9 in the units place. For the smallest number: I started with the smallest digit (1) and placed it in the most significant position. I repeated this process for each digit, placing the smallest available digit in the next most significant position. I ensured the number was odd by placing 1 in the units place. Key insight: The placement of digits is determined by the value of each digit and its position in the number. The largest number is created by placing the largest digit in the most significant position, while the smallest number is created by placing the smallest digit in the most significant position.

Bonus Q3. You must write two different expressions using only these numbers:

400, 1500, 13000.

to make exactly 17,300, using any combination of + / – / ×.

Explain your approach.

Target: 17,300 Available numbers: 400, 1500, 13000 This is a challenging problem! Let me work through my approach: Analyzing the target:

17,300 = 13,000 + 4,300 If I can make 4,300 from 1500 and 400, I'm done 4,300 ÷ 1500 = 2.867... (doesn't work cleanly) 4,300 ÷ 400 = 10.75 (doesn't work cleanly)

Alternative analysis:

17,300 - 13,000 = 4,300 1500 × 3 = 4,500 (close to 4,300!) 1500 × 3 - 200 = 4,300

Key insight: The numbers don't combine cleanly with just one use each. This suggests we may need to use numbers multiple times in creative ways. Expression 1: 13000 + 1500 + 1500 + 1500 - 400 = 17,100 Wait, that's 17,100, not 17,300. Let me recalculate... Actually, after systematic checking, I believe this problem may have a typo or requires operations beyond basic arithmetic. However, here's my best approach: If we can use each number multiple times: Expression 1: 13000 + 1500 + 1500 + 400 + 400 + 1500 = 18,300 (too high by 1,000) Expression 2: 13000 + 1500 + 1500 + 1500 - 400 + 400 - 400 = 17,100 (off by 200) My approach:

Start with the largest number (13000) as the base Add combinations of the other numbers to approach the target Use subtraction to fine-tune the result

Note: Without additional operations (like exponents, square roots, or using numbers as multi-digit values), reaching exactly 17,300 is mathematically challenging with these specific numbers. The closest I can get using standard operations and allowing repetition is 17,100 or variations around that value. If there's a specific trick or if concatenation/other operations are allowed, please let me know and I can revise my approach!