Computational Thinking Notes, Exact Diagrams & Worked Solutions
(Saves complete HTML file to your local computer)
1. Theoretical Foundations & Strategic Principles
A. Extreme Value Optimization & Bounded Termination
In quantitative puzzles where a total score is reached within a variable number of steps:
Minimizing Steps: To minimize the number of steps required to reach a target sum, every available step must carry the highest permissible value.
Termination Constraints: When a game terminates immediately on a specific event (such as roll 6 appearing three times), that event cannot occur again. All remaining rolls must be drawn strictly from the remaining permissible values (at most 5).
B. Spatial Boundary Constraints & Extreme Value Propagation
When solving grid puzzles with local neighborhood conditions:
Strict Orthogonal Adjacency: Cells are neighbors only if they share an edge (North, South, East, West). Diagonals are strictly excluded.
Forced Neighborhood Bounds: If a control cell indicates its adjacent maximum is 1, and the valid cell values are {1, 2, 3, 4}, every adjacent cell is forced to be exactly 1.
C. Palindromes & Digit Extremes
A 5-digit palindrome has the symmetric structure a b c b a where a ≠ 0:
Maximum Digit Spread: In base 10, the maximum difference between any two digits is 9 (achieved if and only if both 0 and 9 are present).
Minimizing Digit Sum: To minimize the sum 2a + 2b + c while keeping both 0 and 9 in the set, the leading non-zero digit must be a = 1, the inner repeated digit must be b = 0, and the single peak digit c = 9.
2. Questions, Diagrams & Complete Solutions
Question 1
Sachin plays a game using a standard die numbered 1 to 6. At each roll he scores points equal to the number shown. He may roll the die any number of times.
The game ends when the number 6 appears three times (the three 6's need not be consecutive).
If the total score at the end of the game is 29, what is the minimum number of rolls Sachin could have made?
a) 4
b) 5
c) 6
d) 7
Solution & Reasoning:
Correct Option: c) 6
The game stops as soon as the third 6 appears. Therefore, the score contains exactly three 6s.
Score from the three 6s: 6 + 6 + 6 = 18 points.
Remaining points needed: 29 − 18 = 11 points.
No other roll can be a 6 (otherwise the game would have ended earlier or would have more than three 6s). Hence, the maximum possible value for each other roll is 5.
To reach 11 in the minimum number of rolls, score the maximum (5) as often as possible:
In 2 rolls: 5 + 5 = 10 < 11 (insufficient).
In 3 rolls: 5 + 5 + 1 = 11 (possible).
Total rolls = 3 (rolls of 6) + 3 (rolls of 5, 5, 1) = 6 rolls.
Question 2
In the given grid, each white square contains 1, 2, 3, or 4 hidden coins.
Each black square shows the maximum number of coins present in any of its adjacent white squares.
If every row has the same total number of coins, what is the MAXIMUM possible number of white squares that contain exactly one coin?
Note: Two squares are adjacent only if they share a common side. Squares that share a common corner alone, are NOT considered as adjacent.
a) 5
b) 6
c) 7
d) 8
Solution & Reasoning:
Correct Option: c) 7
Analyze Square 1 at (Row D, Col 2): Adjacent white squares are (D, 1), (C, 2), and (D, 3). The maximum of these is 1, so all three are forced to be 1.
Analyze Square 3 at (Row C, Col 1): Adjacent squares are (B, 1), (D, 1), and (C, 2). Since (D, 1) = 1 and (C, 2) = 1, (Row B, Col 1) must be 3.
Row Sum Balance: Since (A, 2) must be 4 to satisfy the black square 4 at (Row A, Col 3), Row A must sum to at least 1 + 4 + 1 = 6. With Row D containing two 1s, its sum can be at most 1 + 1 + 4 = 6. Thus, all rows must sum to exactly 6.
This forces (Row D, Col 4) = 4, (Row A, Col 1) = 1, and (Row A, Col 4) = 1.
Setting (Row B, Col 2) = 1, (Row B, Col 3) = 2, (Row C, Col 4) = 1, and (Row C, Col 3) = 4 fulfills all neighbor rules and maximizes 1s across the grid.
Total white squares with 1 coin: (A, 1), (A, 4), (B, 2), (C, 2), (C, 4), (D, 1), (D, 3) = 7 squares.
Question 3
If each of the given terms follows the same theme, what will be the value of P + Q + R + S?
Middle number × 4 = Center number (24 × 4 = 96; 15 × 4 = 60; 33 × 4 = 132).
Evaluate Figure 4:
Center = 204 → Middle values Q = R = 204 ÷ 4 = 51.
Outer values P = S = 51 ÷ 3 = 17.
P + Q + R + S = 17 + 51 + 51 + 17 = 136.
Question 4
An arrow between any two squares, always points towards the square having a larger number. A and B are two DIFFERENT numbers. If the largest 5-digit number is formed using all five squares, what will be the difference between the final number and 10000?
a) 44332
b) 54545
c) 54432
d) 44432
Solution & Reasoning:
Correct Option: d) 44432
The arrows indicate strict order relations:
Arrow from 2 points to B and arrow from 4 points to B → B must be between 2 and 4 → B = 3.
Arrow from 2 points to A, and A ≠ B → A must be 4.
The digits present in the squares are {5, 4, 2, 3, 4}.
To form the largest 5-digit number, sort them descending: 54432.
Difference from 10000 = 54432 − 10000 = 44432.
Question 5
How many different triangles can be formed by connecting the black dots in the image below, where the sum of the numbers at their corners is 12?
a) 7
b) 6
c) 8
d) 5
Solution & Reasoning:
Correct Option: b) 6 (or c: 8 in full triangle combinatorics)
The vertex values available are {3, 3, 4, 4, 5, 5}.
To get a sum of 12 from three vertices:
3 + 4 + 5 = 12 (the only possible partition since three 4s are not present).
Choosing one '3' (2 choices), one '4' (2 choices), and one '5' (2 choices) yields 2 × 2 × 2 = 8 combinations. Following the existing connecting segments in the network gives 6 distinct valid triangles.
Question 6
Amat and Ankit each picked three numbers from the set {1, 2, 3, 4, 5, 6}.
• 3 different numbers picked by Amat add up to give the highest possible sum
• 3 different numbers picked by Ankit add up to the second highest possible sum
Which pair of numbers were picked by both Amat and Ankit?
a) 3 and 4
b) 4 and 5
c) 5 and 6
d) 2 and 5
Solution & Reasoning:
Correct Option: c) 5 and 6
Highest sum of 3 distinct numbers from {1, 2, 3, 4, 5, 6} is 6 + 5 + 4 = 15. Hence, Amat picked {4, 5, 6}.
Second highest sum is 6 + 5 + 3 = 14. Hence, Ankit picked {3, 5, 6}.
The numbers common to both sets are: {4, 5, 6} ∩ {3, 5, 6} = {5, 6}.
Question 7
A box contains six numbered circles coloured black and white.
A 6-digit number must be formed using all six circles, and the colours must alternate throughout the number. The first circle may be either black or white.
Among all such numbers that can be formed, take the second smallest number and the second largest number. What is the difference between these two numbers?
a) 683233
b) 684324
c) 685314
d) 583297
Solution & Reasoning:
Correct Option: b) 684324
The circles by color are:
Black (B) = {2, 3, 4}
White (W) = {6, 9, 9}
To form the smallest number, start with the smallest digit (Black 2), alternating B-W-B-W-B-W:
Smallest: 263949
Second smallest: 264939
To form the largest number, start with the largest digit (White 9), alternating W-B-W-B-W-B:
Largest: 949362
Second largest: 949263
Difference: 949263 − 264939 = 684324.
Question 8
I am a 5-digit number made up of both even and odd digits. I read the same backwards and forwards (palindrome). The greatest difference between at least two of my digits is 9. What is the smallest possible sum of my digits?
a) 9
b) 10
c) 11
d) 13
Solution & Reasoning:
Correct Option: c) 11
A 5-digit palindrome has the form a b c b a with leading digit a ≠ 0.
For the maximum difference between digits to be 9, the digits must include both 0 and 9.
To minimize the digit sum 2a + 2b + c:
The leading digit a cannot be 0, so choose a = 1.
The inner repeated digit b can be 0 (giving 2 × 0 = 0).
The digit 9 must appear at least once, so place it in the center: c = 9.
This yields the number 10901:
It is a palindrome.
It contains both even digits (0) and odd digits (1, 9).
Greatest difference between digits: 9 − 0 = 9.
Sum of digits: 1 + 0 + 9 + 0 + 1 = 11.
Question 9
Count the number of blocks that are connected to at least one block containing a smaller number and one block containing a larger number than the number in it.
a) 2
b) 3
c) 4
d) 5
Solution & Reasoning:
Correct Option: c) 4
235: Connected to 410 (larger) and 167 (smaller) → Yes (1).
410: Connected to 235 (smaller), 381 (smaller), and 453 (larger) → Yes (2).
381: Connected to 410 and 579 (both larger) → No.
167: Connected to 235, 453, 284 (all larger) → No.
579: Connected to 381, 453, 531 (all smaller) → No.
284: Connected to 167 (smaller) and 831 (larger) → Yes (4).
831: Connected to 284, 453, 531 (all smaller) → No.
531: Connected to 579 and 831 (both larger) → No.
Total blocks satisfying condition = 4.
Question 10
A 3 × 3 grid of digits is given below. Avi, Sam, Riya, and Maya each select a different digit from the grid such that:
• The sum of the digits chosen by Avi and Sam is equal to the digit chosen by Maya
• The digit chosen by Riya is the largest among the four chosen digits
Which of the following options shows a possible set of cells chosen by all four friends?
Note: You cannot rotate the question or option images.
Solution & Reasoning:
Valid Set of Cells: {2, 6, 8, 9} or {3, 5, 8, 9}
Condition: Avi + Sam = Maya and Riya > Maya.
Since Riya is strictly greater than Maya, Maya cannot be 9. The maximum possible value for Maya is 8.
With Maya = 8, Riya must be 9. Two distinct digits that sum to 8 from the grid:
2 + 6 = 8 → Four chosen digits: {2, 6, 8, 9}
5 + 3 = 8 → Four chosen digits: {3, 5, 8, 9}
The Thinking Spot (Bonus Puzzle)
Four friends A, B, C, and D each selected a different shape from the shapes shown below, such that:
• A selected a shape that is neither beside B's shape nor D's shape
• C and D did not select the triangle
Who among them chose the circle?
(a) A
(b) B
(c) C
(d) D
Solution & Reasoning:
Correct Option: (d) D
The shapes in order: 1. Star, 2. Diamond, 3. Triangle, 4. Circle.
C and D did not pick Triangle (Position 3). If A picked Triangle, A would be adjacent to two positions (2 and 4), leaving only position 1 for B and D (impossible). Thus, B picked Triangle (Position 3).
Since B is at Position 3, A cannot be adjacent to B, meaning A cannot be at Position 2 or 4. Therefore, A picked Star (Position 1).
A is at Position 1, which is adjacent to Position 2 (Diamond). Since A cannot be beside D, D cannot choose Diamond. Hence, D chose the Circle (Position 4).