Logic and counting · applications

Applications: Logic and Counting

10 question types · Model Method and algebra, side by side

O-Level · SAT · GCSE Higher

01

Dual-Constraint Number × Value with Mass (Helen & Ivan Archetype)

heuristicSimultaneous Weight and Value Constraints / Supposition on Mass

Helen and Ivan each had a collection of 20-cent and 50-cent coins in their savings boxes. Each 20-cent coin had a mass of 2.7 g and each 50-cent coin had a mass of 5.6 g. Helen had 50 coins with a total mass of 193.0 g. Ivan also had 50 coins, but the total value of Ivan's coins was $4.50 more than the total value of Helen's coins. (a) How many 50-cent coins did Helen have? (b) What was the total mass of Ivan's coins in grams?

All 20¢50 × 2.7 = 135 g
Suppose all fifty of Helen’s coins were 20-cent: 135 g.
Helen has 50 coins. If all 50 were 20-cent coins: Mass = 50 × 2.7 g = 135.0 g.
step 1 of 10

Assume all 50 of Helen's coins were 20-cent coins to find the mass shortfall. For Ivan, convert the monetary advantage into additional 50-cent coin exchanges.

  1. Helen has 50 coins. If all 50 were 20-cent coins: Mass = 50 × 2.7 g = 135.0 g.
  2. Mass deficit: 193.0 − 135.0 = 58.0 g.
  3. Difference in mass between one 50-cent coin and one 20-cent coin: 5.6 − 2.7 = 2.9 g.
  4. Number of 50-cent coins Helen had: 58.0 ÷ 2.9 = 20 coins.
  5. (a) Helen had 20 50-cent coins.
  6. Ivan has $4.50 = 450 cents more than Helen with the same total coin count.
  7. Replacing 1 20-cent coin with 1 50-cent coin increases value by 50 − 20 = 30 cents.
  8. Ivan has 450 ÷ 30 = 15 more 50-cent coins than Helen.
  9. Ivan has 20 + 15 = 35 50-cent coins and 50 − 35 = 15 20-cent coins.
  10. Mass of Ivan's coins: (35 × 5.6) + (15 × 2.7) = 196.0 + 40.5 = 236.5 g.

answer(a) 20 fifty-cent coins; (b) 236.5 g

techniqueForming Equations · Counting Coins

examsO-Level · SAT · GCSE Higher · H2

Common pitfalls

  • Using the monetary difference ($4.50) to divide by mass difference (2.9g), mixing physical mass with dollar value.
  • Assuming Ivan had a different total number of coins rather than 50 coins.
02

Worst-Case Certainty (Pigeonhole Principle)

heuristicAdversarial Worst-Case Analysis / Pigeonhole Bounds

A container holds 15 red, 20 blue, 25 green, and 12 yellow plastic tokens of identical size and shape. A player draws tokens one at a time in the dark without looking. (a) What is the minimum number of tokens the player must draw to be absolutely certain of having at least one pair of tokens of the same color? (b) What is the minimum number of tokens the player must draw to be absolutely certain of having at least 6 tokens of every single color?

Red15Blue20Green25Yellow12Worst4, no pair yet
Worst luck for a pair: one of each of the 4 colors.
Part (a): 4 colors. Worst luck: 1 Red, 1 Blue, 1 Green, 1 Yellow = 4 tokens.
step 1 of 6

Imagine the worst possible luck: you pick every possible token that doesn't satisfy the goal before being forced to complete it.

  1. Part (a): 4 colors. Worst luck: 1 Red, 1 Blue, 1 Green, 1 Yellow = 4 tokens.
  2. Draw 1 more token: guaranteed a matching pair. 4 + 1 = 5 tokens.
  3. (a) 5 tokens.
  4. Part (b): Goal is 6 of each color. Worst luck: you draw all 25 Greens, all 20 Blues, and all 15 Reds first without getting yellow.
  5. Subtotal = 25 + 20 + 15 = 60 tokens.
  6. Now draw 6 Yellow tokens: 60 + 6 = 66 tokens.

answer(a) 5 tokens; (b) 66 tokens

techniqueSample Space

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding 6 × 4 = 24 tokens for part (b), confusing best-case scenario with worst-case guaranteed certainty.
  • Stopping at 61 tokens, thinking only 1 yellow token is needed instead of 6.
03

Parity Invariance (Odd & Even Deductions)

heuristicModular Invariance / Parity Invariant Sums

Nine integers are written on a whiteboard: five of them are odd numbers and four of them are even numbers. In each move, a student chooses any two numbers on the board, erases them, and writes their sum in their place. This move is repeated until only a single number remains on the board. (a) Is the final remaining number odd or even? Explain your mathematical reasoning clearly. (b) If the rules were changed such that the student replaced the two numbers a and b with their absolute difference |a - b| instead of their sum, would the final remaining number be odd or even?

On the boardoddoddoddoddoddeveneveneveneven5 odd, 4 even
Nine numbers: five odd, four even.
Pair up the 5 odd numbers: (O + O) + (O + O) + O = E + E + O = O.
step 1 of 7

Group odd numbers into pairs. An odd plus an odd becomes even. 5 odds produce 2 pairs (even) and 1 leftover odd that can never be eliminated.

  1. Pair up the 5 odd numbers: (O + O) + (O + O) + O = E + E + O = O.
  2. No matter how you combine numbers, odd parity only changes when an odd pairs with an odd (reducing the odd count by 2).
  3. Starting with 5 odd numbers (an odd count), subtracting 2 leaves 3, then 1.
  4. You will always end with exactly 1 odd parity.
  5. (a) Strictly Odd.
  6. Replacing with difference |a − b|: O − O = E, E − E = E, O − E = O. The count of odd numbers still decreases only in pairs of 2.
  7. (b) Strictly Odd.

answer(a) odd: the count of odd numbers goes 5 → 3 → 1 and never reaches 0; (b) odd, for the same reason

techniqueOdd and Even Numbers · Divisibility Tests

examsPSLE

Common pitfalls

  • Assuming the result depends on which specific numbers are chosen or in what order they are erased.
  • Believing that taking differences can turn an odd initial parity into an even final number.
04

Tournament / Round-Robin Scoring Matrix

heuristicCombinatorial Score Balancing / Constraint Elimination

Four football teams, Team A, Team B, Team C, and Team D, competed in a round-robin tournament where every team played every other team exactly once. Teams received 3 points for a win, 1 point for a draw, and 0 points for a loss. The final standings showed: Team A had 7 points, Team B had 4 points, Team C had 3 points, and Team D had 1 point. (a) How many matches were played in total, and how many matches ended in a draw? (b) What was the result of the match between Team A and Team B, and how many wins, draws, and losses did Team C have?

A 7WWDB 4WDC 3?D 1D
Six matches; all decisive would give 18 points.
6 matches: if all were decisive wins, points would be 6 × 3 = 18 points.
step 1 of 11

A decisive match yields 3 total points; a draw yields only 2 total points. The gap between maximum points (18) and actual points (15) reveals the number of draws.

  1. 6 matches: if all were decisive wins, points would be 6 × 3 = 18 points.
  2. Actual total points = 7 + 4 + 3 + 1 = 15 points.
  3. Every draw loses 1 point: 18 − 15 = 3 draws.
  4. (a) 6 matches, 3 draws.
  5. Team A (7 pts) = 2 wins + 1 draw.
  6. Team D (1 pt) = 1 draw + 2 losses.
  7. Team B (4 pts) = 1 win + 1 draw + 1 loss.
  8. Total draws allocated: A (1) + B (1) + D (1) = 3. But each draw involves 2 teams (6 team-draw slots total).
  9. Team C must take the remaining 3 team-draw slots: Team C drew all 3 matches.
  10. Since Team A drew with C, Team A's remaining matches (against B and D) were wins.
  11. (b) Team A beat Team B; Team C had 0 wins, 3 draws, 0 losses.

answer(a) 6 matches, 3 draws; (b) A beat B; C had 0 wins, 3 draws and 0 losses

techniqueForming Equations · Two-Way Tables

examsO-Level · SAT · GCSE Higher · H2

Common pitfalls

  • Thinking that 3 draws means 3 team-draw slots, forgetting that every draw gives 1 draw point to BOTH participating teams.
  • Assuming Team C scored 3 points from 1 win and 2 losses, which would leave an odd number of draw slots.
05

Arrangements with Positional Restrictions

heuristicCasework Permutations / Restricted Digit Slots

The five digits 0, 2, 3, 5, 8 are used to form 4-digit whole numbers without repeating any digit. (a) How many distinct 4-digit whole numbers can be formed in total? (b) How many distinct 4-digit EVEN whole numbers greater than 5000 can be formed?

Th H T U4432choices per slot
Thousands: not 0, so 4 choices; then 4, 3, 2 for the rest.
Part (a): [Th: 4 options] × [H: 4 options] × [T: 3 options] × [U: 2 options] = 96.
step 1 of 7

Set up 4 slot boxes: [Th][H][T][U]. Count options slot by slot, separating cases to prevent double-counting digit 8.

  1. Part (a): [Th: 4 options] × [H: 4 options] × [T: 3 options] × [U: 2 options] = 96.
  2. (a) 96.
  3. Part (b): Number must start with 5 or 8, and end in 0, 2, or 8.
  4. Branch 1 (Starts with 5): Units can be 0, 2, or 8 (3 options). Middle two slots have 3 × 2 = 6 options. 3 × 6 = 18 numbers.
  5. Branch 2 (Starts with 8): Units can be 0 or 2 (2 options). Middle two slots have 3 × 2 = 6 options. 2 × 6 = 12 numbers.
  6. Total: 18 + 12 = 30 numbers.
  7. (b) 30 numbers.

answer(a) 96 numbers; (b) 30 numbers

techniqueSample Space · Rows and Columns

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Allowing 0 in the thousands place in part (a), yielding 5 × 4 × 3 × 2 = 120.
  • Lumping 5 and 8 together as 2 choices for thousands and 3 choices for units without separating cases, falsely getting 2 × 3 × 3 × 2 = 36 (which double-counts digit 8 when it is used in both slots).
06

Truth and Lie (Knight and Knave) Casework

heuristicContradiction Pairs / Deductive Hypothesis Elimination

Four school boys, Alex, Ben, Carl, and Dan, were questioned about who broke a classroom window. Exactly one of the four boys broke the window. Exactly three of the boys are telling the truth, while exactly one boy is lying. The boys made the following statements: Alex said, 'Carl broke the window.' Ben said, 'I did not break the window.' Carl said, 'Dan broke the window.' Dan said, 'Carl is lying.' (a) Which boy is telling a lie? (b) Which boy broke the window?

AlexCarl broke itBennot meCarlDan broke itDanCarl is lying
Carl and Dan contradict each other: exactly one of them lies.
Notice Carl and Dan point at each other: one must be True and one must be False.
step 1 of 7

Set up a truth table across the 4 suspects and check for the scenario where exactly 3 are truthful and 1 is a liar.

  1. Notice Carl and Dan point at each other: one must be True and one must be False.
  2. That uses up the 1 allowed Lie. So Alex and Ben must be telling the truth.
  3. Alex says: 'Carl broke the window' → True! So Carl is the culprit.
  4. Carl says: 'Dan broke the window' → Lie! (Carl is lying).
  5. Dan says: 'Carl is lying' → True! (Dan is telling the truth).
  6. (a) Carl is the liar.
  7. (b) Carl broke the window.

answer(a) Carl, the 1 liar among the 4; (b) Carl broke the window

techniqueFinding the Flaw in an Argument · From a Special Case to Every Case

Common pitfalls

  • Testing all four possibilities randomly without first identifying the contradictory pair (Carl and Dan).
  • Assuming the person who broke the window must be the one who lied (while often true in puzzles, it must be proved through deduction).
07

Coin Denomination Exchanges with Invariant Value

heuristicSwapped Quantities System / Sum and Difference Elimination

Rachel had a purse containing only 20-cent coins and 50-cent coins with a total value of $32.00. If the number of 20-cent coins and 50-cent coins were swapped, the total value of the coins would increase by $6.00 to $38.00. (a) How many coins did Rachel have in her purse altogether? (b) How many 50-cent coins did Rachel have at first?

Purse$32Swapped$38
The purse and its swapped version together: $70.
Combine original purse and swapped purse: Total value = $32 + $38 = $70.00.
step 1 of 8

Combine 1 original set and 1 swapped set. Every coin becomes a pair of 1 twenty-cent and 1 fifty-cent coin (70¢ per pair).

  1. Combine original purse and swapped purse: Total value = $32 + $38 = $70.00.
  2. Each coin now has a counterpart: 1 20¢ coin + 1 50¢ coin = 70¢.
  3. Total number of coins in one purse: 7000¢ ÷ 70¢ = 100 coins.
  4. (a) 100 coins altogether.
  5. Swapping increased the value by $6.00 = 600¢.
  6. Since 50¢ is 30¢ more than 20¢, the increase means there were more 20¢ coins than 50¢ coins initially.
  7. Difference in coin count: 600¢ ÷ 30¢ = 20 more 20¢ coins than 50¢ coins.
  8. Number of 50-cent coins: (100 − 20) ÷ 2 = 40 coins.

answer(a) 100 coins; (b) 40 fifty-cent coins

techniqueSimultaneous by Elimination · Counting Coins

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Assigning 60 to the 50-cent coins and 40 to the 20-cent coins, forgetting that the value increased upon swapping, meaning there were fewer 50-cent coins initially.
  • Dividing $6.00 by 70¢ instead of the unit value difference of 30¢.
08

Divisibility & Last Digit Modular Mechanics

heuristicPeriodic Remainder Cycles / Modular Arithmetic of Powers

(a) What is the ones (units) digit of 32026? (b) What is the ones (units) digit of the sum 32026 + 72026?

Powers of 33971repeats every 4Powers of 77931repeats every 4
The units digits cycle: 3, 9, 7, 1 and 7, 9, 3, 1.
Cycle for base 3: [3] -> [9] -> [7] -> [1] (repeats every 4 steps).
step 1 of 9

Map the repeating units digit cycle onto a 4-position wheel dial.

  1. Cycle for base 3: [3] -> [9] -> [7] -> [1] (repeats every 4 steps).
  2. Cycle for base 7: [7] -> [9] -> [3] -> [1] (repeats every 4 steps).
  3. 2026 steps: 2024 is a multiple of 4 (4 × 506).
  4. So step 2024 lands on the last item [1].
  5. Step 2025 lands on [3] for base 3, and [7] for base 7.
  6. Step 2026 lands on [9] for base 3, and [9] for base 7.
  7. (a) Ones digit of 32026 is 9.
  8. Sum of ones digits: 9 + 9 = 18. The units digit is 8.
  9. (b) 8.

answer(a) 9; (b) 8

techniqueModular Arithmetic on a Clock Face · Divisibility Tests

examsPSLE

Common pitfalls

  • Using remainder 0 as position 0 instead of the 4th (last) number in the cycle.
  • Adding exponents directly (32026 + 72026 = 102026), which is mathematically false.
09

Inclusion-Exclusion Principle (Venn Overlaps)

heuristicTwo-Set Venn Overlap / Complementary Set Deduction

In a cohort of 120 primary school pupils: 75 pupils play badminton, 60 pupils play basketball, and 15 pupils play neither of these two sports. (a) How many pupils play both badminton and basketball? (b) What percentage of the pupils who play badminton do not play basketball?

?badminton 75basketball 60neither 15
15 play neither, so 105 are inside the two circles.
Total inside the sports circles: 120 − 15 = 105 pupils.
step 1 of 7

Draw two overlapping circles in a 120-pupil universe box. 15 sit outside. The remaining 105 must fit 75 and 60, with the overlap absorbing the excess.

  1. Total inside the sports circles: 120 − 15 = 105 pupils.
  2. Sum of both sports counts: 75 + 60 = 135.
  3. The excess count represents the double-counted overlap: 135 − 105 = 30 pupils.
  4. (a) 30 pupils play both.
  5. Badminton only (without basketball): 75 − 30 = 45 pupils.
  6. Fraction of badminton players: 4575 = 35 = 60%.
  7. (b) 60%.

answer(a) 30 pupils; (b) 60%

techniqueTwo-Way Tables · Sample Space

examsSAT · GCSE Higher

Common pitfalls

  • Subtracting 135 from 120 directly, forgetting to exclude the 15 pupils who play neither sport.
  • Dividing 45 by the total cohort of 120 (37.5%) instead of by the badminton group of 75 (60%).
10

Shortest Grid Path with Blocked Intersection

heuristicCombinatorial Grid Walking / Complementary Obstacle Avoidance

A student navigates a street network forming a 4 by 3 rectangular grid of city blocks (4 blocks East and 3 blocks North). The student starts at the South-West corner (0, 0) and walks to the North-East corner (4, 3), moving only East (Right) or North (Up) at every intersection. (a) How many unique shortest paths are there in total from (0, 0) to (4, 3)? (b) If road construction completely blocks the intersection at (2, 2), how many valid shortest paths can the student take without passing through the blocked intersection (2, 2)?

startend
Only east or north: count the ways to reach each corner.
Label (0, 0) = 1. The bottom row and leftmost column are all 1s.
step 1 of 13

Use Pascal's Triangle addition principle on the grid vertices, labeling each intersection with the number of routes leading into it.

  1. Label (0, 0) = 1. The bottom row and leftmost column are all 1s.
  2. Fill by adding (Left + Down):
  3. Row 1: (1, 1)=2, (2, 1)=3, (3, 1)=4, (4, 1)=5.
  4. Row 2: (1, 2)=3, (2, 2)=6, (3, 2)=10, (4, 2)=15.
  5. Row 3: (1, 3)=4, (2, 3)=10, (3, 3)=20, (4, 3)=35.
  6. (a) 35 paths.
  7. With (2, 2) blocked, set the value at (2, 2) to 0:
  8. At (3, 2): comes from (2, 2) [0] + (3, 1) [4] = 4.
  9. At (4, 2): comes from (3, 2) [4] + (4, 1) [5] = 9.
  10. At (2, 3): comes from (1, 3) [4] + (2, 2) [0] = 4.
  11. At (3, 3): comes from (2, 3) [4] + (3, 2) [4] = 8.
  12. At (4, 3): comes from (3, 3) [8] + (4, 2) [9] = 17.
  13. (b) 17 paths.

answer(a) 35 paths; (b) 17 paths

techniqueSequences Worth Knowing · The nth Term

examsPSLE · GCSE Higher

Common pitfalls

  • Adding the stage paths (6 + 3 = 9) instead of multiplying them (6 × 3 = 18) by the fundamental counting principle.
  • Moving West or South, violating the shortest-path condition.
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