Logic and Counting: 11 Applications, Worked Step by Step

Two Savings Boxes of 50 Coins: One Known by Its Mass, the Other by Its Value

heuristicSuppose every coin is a 20-cent coin; each swap for a 50-cent coin adds a fixed mass and a fixed value

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?

Suppose all fifty of Helen’s coins were 20-cent: 135 g.

Suppose all 50 of Helen's coins were 20-cent coins. They would weigh 50 × 2.7 = 135 g.

step 1 of 10

Suppose all of Helen's coins are 20-cent coins and compare with the real mass; each 50-cent coin in place of a 20-cent coin adds the same mass. Ivan has as many coins as Helen, so his extra value comes the same way, one swap at a time.

All steps

Suppose all 50 of Helen's coins were 20-cent coins. They would weigh 50 × 2.7 = 135 g.
Her coins weigh 193 g, which is 193 − 135 = 58 g more.
Swapping one 20-cent coin for a 50-cent coin keeps 50 coins and adds 5.6 − 2.7 = 2.9 g.
The 58 g took 58 ÷ 2.9 = 20 such swaps, so 20 of her coins are 50-cent coins.
(a) Helen had 20 fifty-cent coins and 50 − 20 = 30 twenty-cent coins. Check: 30 × 2.7 + 20 × 5.6 = 81 + 112 = 193 g.
Ivan also has 50 coins, worth $4.50, or 450 cents, more than Helen's.
With the number of coins fixed, his coins can only be worth more through swaps of 20-cent coins for 50-cent coins, and each swap adds 50 − 20 = 30 cents.
Ivan has 450 ÷ 30 = 15 more fifty-cent coins than Helen: 20 + 15 = 35 of them, and 50 − 35 = 15 twenty-cent coins.
His 35 fifty-cent coins weigh 35 × 5.6 = 196 g and his 15 twenty-cent coins 15 × 2.7 = 40.5 g.
(b) Ivan's coins have a total mass of 196 + 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

Dividing the $4.50 by the 2.9 g. Mass and money are separate: 2.9 g a swap answers (a), and 30 cents a swap gives how many more fifty-cent coins Ivan has.
Giving Ivan a different number of coins. He has 50, as Helen does, so his extra value comes only from swaps, 30 cents each.

Drawing Tokens in the Dark: Sure of a Pair, Sure of Six of Every Color

heuristicImagine the worst luck possible, then draw one more

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?

Worst luck for a pair: one of each of the 4 colors.

For a pair, the worst luck is one token of each of the 4 colors: 4 tokens, all different.

step 1 of 6

Play the worst luck: every token that does not help comes out first. The count that forces the goal even then is the answer.

All steps

For a pair, the worst luck is one token of each of the 4 colors: 4 tokens, all different.
The 5th token has only those 4 colors to be, so it matches one of them.
(a) 5 tokens make a pair certain; 4 do not.
For six of every color, the worst luck saves the smallest color for last: every green, blue and red comes out before a single yellow.
That is 25 + 20 + 15 = 60 tokens, and still no yellow.
(b) Only yellow tokens are left, so 6 more draws give six yellows: 60 + 6 = 66 tokens. Saving red for last instead gives only 25 + 20 + 12 + 6 = 63, so yellow, the smallest color, is the worst case.

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

techniqueSample Space

examsO-Level · SAT · GCSE Higher

Common pitfalls

Answering 6 × 4 = 24 for (b). That is the luckiest case, six of each in a row; a certain answer has to survive the unluckiest one.
Stopping at 61, with one yellow. The question asks for six of every color, so six yellows are needed after the 60 others.

Nine Numbers on a Board, Two at a Time Replaced by Their Sum

heuristicFollow how many odd numbers are on the board: no move can make that count even

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?

Nine numbers: five odd, four even.

The board starts with nine numbers: five odd and four even.

step 1 of 7

Count the odd numbers on the board and watch what each kind of move does to that count.

All steps

The board starts with nine numbers: five odd and four even.
Two even numbers add to an even number, and an odd and an even number to an odd one: the count of odd numbers stays the same. Two odd numbers add to an even number: the count falls by 2.
So the count of odd numbers falls by 0 or 2 at each move. It starts at 5, an odd number, and stays odd: 5, 3 or 1, never 0.
After eight moves one number is left. The count of odd numbers is still odd, and with one number on the board it can only be 1.
(a) The last number is odd, whichever numbers are chosen and in whatever order.
With the difference in place of the sum, the results are the same kind: odd − odd and even − even are even, and odd − even is odd. Each move again changes the count of odd numbers by 0 or 2.
(b) The count of odd numbers again stays odd from 5, so the last number is odd.

answer(a) odd: the count of odd numbers starts at 5 and stays odd, so with one number left the count is 1 and that number is odd; (b) odd, for the same reason

techniqueOdd and Even Numbers · Divisibility Tests

examsPSLE

Common pitfalls

Trying particular numbers and orders of moves. The answer cannot depend on them: every move keeps the count of odd numbers odd.
Expecting differences to behave unlike sums. The sum and the difference of two numbers are both odd or both even, so the count of odd numbers changes in the same way.

Four Teams, Six Matches, and a Table of 7, 4, 3 and 1 Points

heuristicA match with a winner gives out 3 points and a draw only 2, so the points missing from 18 count the draws

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?

Six matches; all decisive would give 18 points.

Each of the 4 teams plays the other 3, and each match has two teams: 4 × 3 ÷ 2 = 6 matches. If every match had a winner, the table would add to 6 × 3 = 18 points.

step 1 of 11

Compare the table's total with the 18 points that six matches with winners would give out. Then read each team's points as wins, draws and losses, the fixed ones first.

All steps

Each of the 4 teams plays the other 3, and each match has two teams: 4 × 3 ÷ 2 = 6 matches. If every match had a winner, the table would add to 6 × 3 = 18 points.
The table adds to 7 + 4 + 3 + 1 = 15 points.
A draw gives out 1 + 1 = 2 points instead of 3, so each draw leaves the total 1 point short of 18: 18 − 15 = 3 draws.
(a) 6 matches were played, and 3 of them were draws.
Team A's 7 points from 3 matches can only be 2 wins and a draw: 3 + 3 + 1.
Team D's 1 point can only be a draw and 2 losses.
Team B's 4 points can only be a win, a draw and a loss: 3 + 1 + 0.
Each draw is a drawn result for both of its teams, so 3 draws make 6 drawn results. A, B and D have one each, 3 in all.
The other 3 drawn results are Team C's, so C drew all three of its matches: 0 wins, 3 draws, 0 losses. Check: C has 3 points, and the wins, 2 for A and 1 for B, make the 3 matches that had a winner.
Team A's draw was with C, so A's 2 wins were against B and D.
(b) Team A beat Team B; Team C had 0 wins, 3 draws and 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

Counting 3 draws as 3 drawn results. Each draw is a drawn result for both teams, so 3 draws make 6, and C's share is 3 of them.
Giving C 1 win and 2 losses for its 3 points. Then A, B and C would have 4 wins between them, but only 3 matches had a winner.

Four-Digit Numbers from 0, 2, 3, 5 and 8: All of Them, and the Even Ones Over 5000

heuristicFill the most restricted place first, and split into cases when two restrictions want the same digit

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?

Thousands: not 0, so 4 choices; then 4, 3, 2 for the rest.

Fill the thousands first, because it cannot be 0: 4 choices. The hundreds can then be any of the 4 digits left, the tens any of 3, and the ones any of 2.

step 1 of 7

Draw four boxes for thousands, hundreds, tens and ones, and fill the box with the strictest rule first. When the first and last boxes could both want the 8, count each start separately.

All steps

Fill the thousands first, because it cannot be 0: 4 choices. The hundreds can then be any of the 4 digits left, the tens any of 3, and the ones any of 2.
(a) 4 × 4 × 3 × 2 = 96 numbers.
For (b), a number over 5000 starts with 5 or 8, and an even number ends in 0, 2 or 8. The 8 could be wanted in both places, so take the two starting digits separately.
Starting with 5: the ones digit is 0, 2 or 8, 3 choices, and the middle two places take any of the 3 digits left and then any of 2: 3 × 3 × 2 = 18 numbers.
Starting with 8: the 8 is used, so the ones digit is 0 or 2, 2 choices, and the middle two places again have 3 × 2: 2 × 3 × 2 = 12 numbers.
The two cases share no number, so together they make 18 + 12 = 30.
(b) 30 even numbers greater than 5000.

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

techniqueSample Space · Rows and Columns

examsO-Level · SAT · GCSE Higher

Common pitfalls

Letting 0 lead in (a), which gives 5 × 4 × 3 × 2 = 120. A number written 0235 is the three-digit number 235.
Taking 2 choices for the thousands and 3 for the ones at once: 2 × 3 × 3 × 2 = 36. That includes six numbers that start and end with 8, which repeat a digit.

Four Boys, One Broken Window, Exactly One Liar

heuristicFind two statements that cannot both be true: the one liar is among them

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?

Carl and Dan contradict each other: exactly one of them lies.

Carl says Dan broke the window, and Dan says Carl is lying. If Carl tells the truth, Dan's statement is false; if Carl lies, Dan's is true. Exactly one of the two is lying.

step 1 of 7

Look for two statements that contradict each other. One of them must be the lie, which clears everyone else at once.

All steps

Carl says Dan broke the window, and Dan says Carl is lying. If Carl tells the truth, Dan's statement is false; if Carl lies, Dan's is true. Exactly one of the two is lying.
There is only one liar, and it is Carl or Dan, so Alex and Ben are both telling the truth.
Alex is truthful, so Carl broke the window. Ben's 'I did not break the window' agrees.
Then Carl's statement, that Dan broke it, is false: Carl is the liar.
Dan, who says Carl is lying, is telling the truth, so exactly one boy lied, as the question says.
(a) Carl is lying.
(b) Carl broke the window.

answer(a) Carl: trying each of the 4 boys as the liar, only Carl leaves exactly 1 liar; (b) Carl broke the window

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

Common pitfalls

Trying the boys one by one without a plan. Carl's and Dan's statements contradict each other, so the liar is one of them, and Alex and Ben can be believed at once.
Assuming the boy who broke the window is the liar. Here it is Carl, but that follows from Alex's true statement, not from a rule that the culprit lies.

A Purse Worth 32 Dollars, and 38 with the Coin Counts Swapped

heuristicPut the purse beside its swapped copy: every coin pairs a 20-cent coin with a 50-cent coin

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?

The purse and its swapped version together: $70.

Put the purse and its swapped version side by side. Together they are worth $32 + $38 = $70, or 7000 cents.

step 1 of 8

Put the purse and its swapped copy together. Each coin in the purse pairs with its swapped coin, a 20-cent coin with a 50-cent coin, so the pairs give the number of coins; the change in value gives the difference between the two kinds.

All steps

Put the purse and its swapped version side by side. Together they are worth $32 + $38 = $70, or 7000 cents.
Every 20-cent coin in the purse is a 50-cent coin in the swapped version, and every 50-cent coin a 20-cent coin. So each coin of the purse pairs with one in the copy, 20 cents with 50 cents: 70 cents a pair.
There are as many pairs as coins in the purse: 7000 ÷ 70 = 100.
(a) Rachel had 100 coins.
Swapping raised the value by $6, or 600 cents. It turned every 20-cent coin into a 50-cent coin and every 50-cent coin into a 20-cent coin, so the purse held more 20-cent coins.
Each 20-cent coin that becomes a 50-cent coin gains 30 cents, and each 50-cent coin that becomes a 20-cent coin loses 30 cents.
The gains outweigh the losses by 600 cents, so there were 600 ÷ 30 = 20 more 20-cent coins than 50-cent coins.
(b) The 50-cent coins were (100 − 20) ÷ 2 = 40, and the 20-cent coins 60. Check: 60 × 20 + 40 × 50 = 1200 + 2000 = 3200 cents.

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

techniqueSimultaneous by Elimination · Counting Coins

examsO-Level · SAT · GCSE Higher

Common pitfalls

Giving the 60 to the 50-cent coins. The value went up when the counts were swapped, so the purse held fewer 50-cent coins: 40.
Dividing the 600 cents by 70. The 70 cents is the value of a pair; each swapped coin changes the value by the difference, 30 cents.

The Ones Digit of a Power of 3, and of a Power of 3 Plus a Power of 7

heuristicThe ones digits of the powers repeat every 4, so only the remainder of the power after dividing by 4 matters

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

The units digits cycle: 3, 9, 7, 1 and 7, 9, 3, 1.

The ones digit of a power depends only on the ones digit of the power before it. Powers of 3 go 3, 9, 27, 81, 243, so their ones digits run 3, 9, 7, 1 and then repeat.

step 1 of 9

Write the ones digits of the first few powers until they repeat. A whole number of cycles brings the digit back to where it was, so only the steps after the last whole cycle need counting.

All steps

The ones digit of a power depends only on the ones digit of the power before it. Powers of 3 go 3, 9, 27, 81, 243, so their ones digits run 3, 9, 7, 1 and then repeat.
Powers of 7 go 7, 49, 343, 2401, so their ones digits run 7, 9, 3, 1 and repeat. Both cycles are 4 long.
2024 is 4 × 506, a whole number of cycles.
So the 2024th power of each ends a cycle: its ones digit is 1.
The 2025th power starts a new cycle: ones digit 3 for the powers of 3, and 7 for the powers of 7.
The 2026th power is second in the cycle: 9 for the powers of 3, and 9 for the powers of 7.
(a) The ones digit of 3²⁰²⁶ is 9.
The ones digit of a sum depends only on the ones digits added: 9 + 9 = 18, which ends in 8.
(b) The ones digit of 3²⁰²⁶ + 7²⁰²⁶ is 8.

answer(a) 9; (b) 8

techniqueModular Arithmetic on a Clock Face · Divisibility Tests

examsPSLE

Common pitfalls

Reading a remainder of 0 as the start of the cycle. A power divisible by 4, such as the 2024th, takes the 4th place, the last: it ends in 1.
Adding the bases, as if 3²⁰²⁶ + 7²⁰²⁶ were 10²⁰²⁶, which ends in 0. Powers do not add that way: 3² + 7² = 9 + 49 = 58, not 10² = 100.

120 Pupils, Badminton and Basketball, and 15 Who Play Neither

heuristicAdding the two sports counts the pupils who play both twice; the excess over those who play at least one is the overlap

Of the 120 pupils in a grade, 75 play badminton, 60 play basketball, and 15 play neither sport. (a) How many pupils play both badminton and basketball? (b) What percentage of the pupils who play badminton do not play basketball?

15 play neither, so 105 are inside the two circles.

Draw a circle for each sport inside a box of 120 pupils. The 15 who play neither stand outside both circles, so 120 − 15 = 105 pupils are inside.

step 1 of 7

Draw a circle for each sport inside a box for the 120 pupils. The two circles hold fewer pupils than their counts add to; the difference is the overlap.

All steps

Draw a circle for each sport inside a box of 120 pupils. The 15 who play neither stand outside both circles, so 120 − 15 = 105 pupils are inside.
Adding the two sports gives 75 + 60 = 135, more than 105, because a pupil who plays both is counted once in each circle.
The extra is the overlap, counted twice: 135 − 105 = 30 pupils.
(a) 30 pupils play both sports.
Badminton only: 75 − 30 = 45 pupils. Check: 45 + 30 + (60 − 30) + 15 = 120.
Of the 75 badminton players, 45 do not play basketball: 4575 = 35.
(b) 35 = 60%.

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

techniqueTwo-Way Tables · Sample Space

examsSAT · GCSE Higher

Common pitfalls

Taking 120 from 135. The 15 who play neither are in neither circle; the overlap is 135 − 105 = 30, not 15.
Dividing 45 by all 120 pupils, which gives 37.5%. The question asks about the badminton players, so divide by 75.

Shortest Walks Across a 4-by-3 Grid of Streets, with One Crossing Closed

heuristicThe ways to reach a crossing are the ways to the crossing on its left plus the ways to the crossing below it

A town's streets form a grid 4 blocks wide and 3 blocks high. Sam walks from the southwest corner, (0, 0), to the northeast corner, (4, 3), going only east or north at every crossing. (a) How many different shortest routes are there? (b) The crossing at (2, 2) is closed for roadwork. How many shortest routes avoid it?

Only east or north: count the ways to reach each corner.

Walking only east or north, a route reaches each crossing from the crossing to its left or the crossing below it. So count, for every crossing, the ways to reach it.

step 1 of 13

Write at each crossing the number of ways to reach it. A route arrives from the left or from below, so each number is the sum of those two; a closed crossing gets 0.

All steps

Walking only east or north, a route reaches each crossing from the crossing to its left or the crossing below it. So count, for every crossing, the ways to reach it.
Along the bottom street and up the left street there is only one way to each crossing: straight along. Write 1 at each.
Every other crossing gets the sum of the crossing to its left and the one below it. The first row up: (1, 1) = 2, (2, 1) = 3, (3, 1) = 4, (4, 1) = 5.
The second row: (1, 2) = 3, (2, 2) = 6, (3, 2) = 10, (4, 2) = 15.
The third row: (1, 3) = 4, (2, 3) = 10, (3, 3) = 20, (4, 3) = 35.
(a) There are 35 shortest routes.
Now close (2, 2). No route may pass through it, so its count becomes 0, and every crossing reached through it must be counted again.
(3, 2) is reached from (2, 2), now 0, and from (3, 1), 4: 0 + 4 = 4.
(4, 2) is reached from (3, 2), 4, and from (4, 1), 5: 4 + 5 = 9.
(2, 3) is reached from (1, 3), 4, and from (2, 2), 0: 4 + 0 = 4.
(3, 3) is reached from (2, 3), 4, and from (3, 2), 4: 4 + 4 = 8.
(4, 3) is reached from (3, 3), 8, and from (4, 2), 9: 8 + 9 = 17.
(b) 17 shortest routes avoid the closed crossing. Check: the 6 ways to (2, 2) times the 3 ways on from it are 18 routes through it, and 35 − 18 = 17.

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

techniqueSequences Worth Knowing · The nth Term

examsPSLE · GCSE Higher

Common pitfalls

Adding the ways to (2, 2) and the ways on from it, 6 + 3 = 9. Each way there can go on by each way onward, so they multiply: 18.
Counting routes that go west or south. A shortest route never turns back: every step is east or north.

A Move That Lowers the Total by 1, Repeated Until One Number Is Left

heuristicTrack the Parity of a Count / Count the Moves

A board shows 25 whole numbers: 13 of them are odd and 12 are even. A computer repeats one move. It chooses any two numbers x and y on the board, rubs both out, and writes the single number x + y − 1 in their place. Each move leaves one number fewer on the board, and the computer stops when only one number is left. (a) Is x + y − 1 odd or even when (i) x and y are both odd, (ii) one of them is odd and the other is even, (iii) x and y are both even? (b) Is the last number on the board odd or even? Explain why. (c) Suppose instead that each move writes the difference between the two numbers, the larger minus the smaller. Is the last number odd or even?

(a) Two odd or two even numbers add to an even number, and 1 less is odd. An odd and an even number add to an odd number, and 1 less is even.

(a) Add the two numbers, then take away 1. (i) Two odd numbers add up to an even number, and 1 less is odd. (ii) An odd and an even number add up to an odd number, and 1 less is even. (iii) Two even numbers add up to an even number, and 1 less is odd.

step 1 of 6

Count the odd numbers on the board. Work out how each kind of move changes that count, then use the number of moves to see whether the count ends odd or even.

All steps

(a) Add the two numbers, then take away 1. (i) Two odd numbers add up to an even number, and 1 less is odd. (ii) An odd and an even number add up to an odd number, and 1 less is even. (iii) Two even numbers add up to an even number, and 1 less is odd.
Now count the odd numbers on the board. In (i) two odd numbers go and an odd number comes back: one odd number fewer. In (ii) one odd number goes and an even number comes back: one fewer. In (iii) two even numbers go and an odd number comes back: one more. Every move changes the count of odd numbers by exactly 1.
Each move leaves one number fewer, so going from 25 numbers to 1 takes 25 − 1 = 24 moves. Every move switches the count of odd numbers between odd and even, and 24 switches, an even number, bring it back to the kind it started as.
(b) The count starts at 13, which is odd, so it is odd at the end too. With one number left, the count of odd numbers is 0 or 1, so it is 1: the last number is odd.
With the difference instead, two odd numbers give an even number, so the count of odd numbers falls by 2. An odd and an even number give an odd number, and two even numbers give an even number, so in those moves the count does not change. The count changes by 0 or 2 each time, so it stays odd all the way from 13.
(c) The last number is odd again.

answer(a) (i) odd, (ii) even, (iii) odd; (b) odd, because the count of odd numbers changes by 1 on each of the 24 moves and starts at 13; (c) odd

techniqueOdd and Even Numbers · From a Special Case to Every Case

examsPSLE

Common pitfalls

Thinking the answer depends on which numbers the computer chooses. The moves are of three kinds, but each one changes the count of odd numbers by exactly 1, so after 24 moves the result is the same whatever the choices.
Counting 25 moves instead of 24. Each move removes one number, and going from 25 numbers to 1 removes 24; with 25 moves the count of odd numbers would switch an odd number of times and the answer would come out even.