HCF and LCM · applications

Applications: HCF and LCM

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

PSLE · O-Level · GCSE Higher

01

HCF: Equal Distribution into Maximum Sets (No Remainder)

heuristicHighest Common Factor / Maximum Common Divisor

Mr. Lim has 72 chocolate bars and 90 packets of biscuits. He wants to pack all of them into identical gift bags such that every bag contains the exact same number of chocolate bars and biscuit packets, with none left over. (a) What is the greatest number of gift bags he can pack? (b) How many chocolate bars and biscuit packets will be in each gift bag?

272903645chocolate · biscuits
Divide both counts by a common factor, 2. Each division is one more way to split every bag the same.
Divide both 72 and 90 by 2: (72, 90) ÷ 2⟶ (36, 45).
step 1 of 6

Use the ladder method (repeated division by common prime factors) to extract common groups simultaneously.

  1. Divide both 72 and 90 by 2: (72, 90) ÷ 2⟶ (36, 45).
  2. Divide both 36 and 45 by 3: (36, 45) ÷ 3⟶ (12, 15).
  3. Divide both 12 and 15 by 3: (12, 15) ÷ 3⟶ (4, 5).
  4. Since 4 and 5 share no common factors other than 1, stop.
  5. Multiply the left-hand divisors: 2 × 3 × 3 = 18 bags.
  6. The remaining quotients directly give the items per bag: 4 chocolate bars and 5 biscuit packets.

answer(a) 18 gift bags; (b) 4 chocolate bars and 5 biscuit packets in each bag

techniqueThe Highest Common Factor · HCF and LCM Word Problems

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Confusing HCF with LCM and calculating LCM(72, 90) = 360, resulting in an impossible bag count greater than the items available.
  • Stopping the ladder division early (e.g., dividing by 9 once and obtaining 8 and 10, forgetting that 8 and 10 still share a common factor of 2).
  • Mixing up the answers to parts (a) and (b), writing 4 or 5 as the total number of bags.
02

HCF: Paving a 2D Surface with Largest Identical Square Tiles

heuristicCommon Dimensional Divisor / Area Conservation

A rectangular floor measures 360 cm in length and 280 cm in breadth. A tiler wants to cover the entire floor completely with identical square ceramic tiles without cutting any of them. (a) What is the largest possible length of the side of each square tile? (b) What is the minimum number of tiles needed to cover the entire floor?

360 cm280 cm
The floor: 360 cm by 280 cm.
Draw a rectangle labeled 360 cm by 280 cm.
step 1 of 4

Partition the 2D rectangular grid into unit squares by finding the largest common measurement that divides both borders.

  1. Draw a rectangle labeled 360 cm by 280 cm.
  2. Find common divisors for both edges using short division: divide by 10 → (36, 28), then divide by 4 → (9, 7).
  3. Multiply common scale factors: 10 × 4 = 40 cm per tile edge.
  4. Total tile count is the product of row and column intervals: 9 columns × 7 rows = 63 tiles.

answer(a) 40 cm; (b) 63 tiles

techniqueThe Highest Common Factor · Choosing Between HCF and LCM

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Dividing the total floor area by the side length instead of the tile area: (360 × 280) ÷ 40 = 2520 instead of ÷ 402 = 63.
  • Adding the row and column counts (9 + 7 = 16) instead of multiplying them (9 × 7 = 63).
  • Finding the LCM of 360 and 280, producing an enormous tile size larger than the floor itself.
03

HCF: Cutting Unequal Lengths into Longest Equal Pieces (1D Segmentation)

heuristicCommon Linear Segmenting / Zero Wastage

Three wooden planks of lengths 120 cm, 180 cm, and 240 cm are to be cut into shorter pieces of identical length without any wood left over. (a) What is the greatest possible length of each cut piece? (b) How many such cut pieces will there be altogether?

120 cm180 cm240 cm
Three planks, one piece length for all, nothing wasted: the length must divide 120, 180 and 240.
Perform simultaneous short division on 120, 180, and 240:
step 1 of 5

Represent each plank as a bar and partition each bar into identical blocks using repeated division.

  1. Perform simultaneous short division on 120, 180, and 240:
  2. Divide by 10: (12, 18, 24).
  3. Divide by 6: (2, 3, 4).
  4. Common factor: 10 × 6 = 60 cm (length of 1 block).
  5. Sum the remaining unit blocks: 2 + 3 + 4 = 9 pieces.

answer(a) 60 cm; (b) 9 pieces

techniqueThe Highest Common Factor · Prime Factorization

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Multiplying the remaining pieces (2 × 3 × 4 = 24) instead of adding them, confusing 1D piece counting with 2D/3D grid arrays.
  • Finding the HCF of only the first two numbers (120 and 180) and forgetting to check divisibility against 240.
04

HCF: Equidistant Spacing Along Perimeter with Corner Anchors

heuristicBoundary Segmentation / Common Interval Divisor

A rectangular field measures 48 m by 36 m. Lampposts are to be erected along the boundary of the field at equal intervals. A lamppost must be placed at each of the 4 corners of the field. (a) What is the maximum distance between any two consecutive lampposts? (b) What is the least number of lampposts needed?

48 m36 m4 intervals of 12 m
A post at every corner and equal gaps: the gap must divide 48 and 36. The largest is 12 m. Along a length: 48 ÷ 12 = 4 intervals.
Intervals along each length: 48 ÷ 12 = 4 intervals.
step 1 of 4

Analyze the intervals along each side using the closed loop perimeter principle (Number of posts = Number of intervals).

  1. Intervals along each length: 48 ÷ 12 = 4 intervals.
  2. Intervals along each breadth: 36 ÷ 12 = 3 intervals.
  3. Total intervals around the closed perimeter: 4 + 3 + 4 + 3 = 14 intervals.
  4. In a closed loop, the number of posts equals the number of intervals: 14 lampposts.

answer(a) 12 m; (b) 14 lampposts

techniqueThe Highest Common Factor · Common Factors

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Double-counting the corner posts: calculating (4 + 1) + (3 + 1) + (4 + 1) + (3 + 1) = 18 posts instead of recognizing that corners are shared.
  • Dividing the area of the rectangle by 12 rather than using the perimeter.
05

LCM: Periodic Synchronization (Simultaneous Events)

heuristicLowest Common Multiple / Cycle Alignment

Three lighthouses flash their lights at regular intervals. Lighthouse A flashes every 24 seconds, Lighthouse B flashes every 36 seconds, and Lighthouse C flashes every 60 seconds. All three lighthouses flash simultaneously at 8:00 p.m. (a) At what time will they next flash together? (b) How many times will all three flash together between 8:01 p.m. and 8:35 p.m.?

seconds after 8:00 p.m.A · 24 s0360B · 36 s03672108144180216252288324360C · 60 s060120180240300360
Three rhythms on one clock. They coincide again at the first common multiple.
Ladder method for LCM of (24, 36, 60):
step 1 of 5

Create an interval timeline bar or ladder chart to identify the earliest shared milestone on the time axis.

  1. Ladder method for LCM of (24, 36, 60):
  2. Divide by 12 → (2, 3, 5).
  3. Since 2, 3, 5 are pairwise coprime, multiply: 12 × 2 × 3 × 5 = 360 seconds = 6 minutes.
  4. Step through intervals: 8:00, 8:06, 8:12, 8:18, 8:24, 8:30, 8:36 (exceeds 8:35).
  5. Count matches strictly between 8:01 and 8:35: 5 occurrences.

answer(a) 8:06 p.m.; (b) 5 times

techniqueThe Lowest Common Multiple · HCF and LCM Word Problems

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Selecting the HCF instead of LCM (giving HCF = 12 seconds, which is far too short to align with a 36- or 60-second cycle).
  • Including the 8:00 p.m. event when the question specifies 'between 8:01 p.m. and 8:35 p.m.'.
  • Forgetting to convert 360 seconds into minutes when stating the clock time.
06

LCM: Building the Smallest Solid Cube from Rectangular Blocks

heuristicSpatial Lowest Common Multiple / 3D Volumetric Matching

A student has a collection of identical rectangular wooden blocks. Each block measures 12 cm by 8 cm by 6 cm. He stacks these blocks together to form the smallest possible solid cube. (a) What is the length of one edge of the cube? (b) How many wooden blocks does he need to build this cube?

Edge 2412122 blocks longEdge 248883 blocks wideEdge 2466664 blocks high
(a) The edge must be a multiple of 12, 8 and 6 at once. 24 is the first: two 12s, three 8s, four 6s.
Find smallest common side length: multiples of 12 are 12, 24, 36...; check divisibility by 8 and 6 → 24 cm.
step 1 of 4

Calculate volume scaling by comparing the target cube volume to 1 individual block volume.

  1. Find smallest common side length: multiples of 12 are 12, 24, 36...; check divisibility by 8 and 6 → 24 cm.
  2. Volume of 1 small block = 12 × 8 × 6 = 576 cm3.
  3. Volume of the target solid cube = 24 × 24 × 24 = 13824 cm3.
  4. Total blocks = 13824 ÷ 576 = 24 blocks.

answer(a) 24 cm; (b) 24 blocks

techniqueThe Lowest Common Multiple · Volume of a Cuboid

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Adding the edge unit counts (2 + 3 + 4 = 9) instead of multiplying them across all 3 dimensions (2 × 3 × 4 = 24).
  • Calculating the HCF ( HCF(12, 8, 6) = 2 cm , which describes cutting a block into smaller cubes, not stacking blocks to form a larger cube).
07

LCM: Equal Quantity Matching Across Different Pack Sizes

heuristicCommon Multiple Quantity / Group Cost Evaluation

At a bakery, hotdog buns are sold in packs of 6 for $4, and sausages are sold in packs of 8 for $5. Melissa needs to buy the exact same number of hotdog buns and sausages to make hotdogs, without having any buns or sausages left over. What is the least amount of money she can spend in total?

Buns66666packs of 6
Buns come in sixes: 6, 12, 18, 24, 30.
List bun counts: 6, 12, 18, 24, 30...
step 1 of 6

Set up a comparison table listing multiples of each item until equal item quantities are matched.

  1. List bun counts: 6, 12, 18, 24, 30...
  2. List sausage counts: 8, 16, 24, 32...
  3. First common quantity match is 24 items.
  4. Draw 4 unit packs for buns: 4 × $4 = $16.
  5. Draw 3 unit packs for sausages: 3 × $5 = $15.
  6. Combine costs: $16 + $15 = $31.

answer$31

techniqueThe Lowest Common Multiple · Common Multiples

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Finding the LCM of the prices ( LCM(4, 5) = 20 ) instead of the pack quantities.
  • Assuming that buying 1 pack of each yields equal items (6 ≠ 8).
  • Multiplying 24 items directly by the pack price rather than dividing by pack size first.
08

LCM with Constant Remainder (Equal Excess)

heuristicCommon Multiple with Constant Positive Offset

A farmer has a basket of eggs. When he packs them into cartons of 6, cartons of 8, or cartons of 9, there are always 2 eggs left over each time. What is the smallest possible number of eggs the farmer has in the basket (assuming he has more than 2 eggs)?

BaselineLCM(6, 8, 9) = 72
Full cartons of 6, 8 or 9 with nothing over: the smallest such count is the LCM, 72.
Find the common baseline where cartons leave zero remainder: LCM(6, 8, 9) = 72.
step 1 of 4

Visualize the quantity as a baseline LCM block plus a fixed excess block of 2 units.

  1. Find the common baseline where cartons leave zero remainder: LCM(6, 8, 9) = 72.
  2. Every scenario describes an excess of 2 eggs above a full pack: [Complete Packs] + 2.
  3. Smallest non-trivial total = 72 + 2 = 74 eggs.
  4. Verification: 74 ÷ 6 = 12 R 2, 74 ÷ 8 = 9 R 2, 74 ÷ 9 = 8 R 2.

answer74 eggs

techniqueThe Lowest Common Multiple · Remainders

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Subtracting the remainder from the LCM (72 − 2 = 70) instead of adding it back.
  • Adding 2 to each divisor before finding the LCM (e.g., finding the LCM of 8, 10, 11).
09

LCM with Constant Shortage (Equal Deficit / Negative Offset)

heuristicCommon Multiple with Constant Shortage

A teacher packs all her badges into goodie bags for a school carnival, filling one bag at a time. If she puts 5 badges in each bag, the last bag is 3 badges short of full. The same happens with 6 badges in each bag, and with 8 badges in each bag. What is the smallest possible number of badges the teacher has?

TargetLCM(5, 6, 8) = 120
Bags of 5, 6 or 8 would come out exact at the LCM, 120.
Determine the complete round number benchmark: LCM(5, 6, 8) = 120.
step 1 of 4

Draw the target multiple bar and deduct the missing gap to find the actual quantity.

  1. Determine the complete round number benchmark: LCM(5, 6, 8) = 120.
  2. Notice that in every case, the teacher is 3 units away from filling the final bag: [Actual Count] + [3 missing] = 120.
  3. Deduct the deficit from the LCM: 120 − 3 = 117 badges.
  4. Verification: 117 ÷ 5 = 23 R 2 (short of 3 to make 24); 117 ÷ 6 = 19 R 3 (short of 3 to make 20); 117 ÷ 8 = 14 R 5 (short of 3 to make 15).

answer117 badges

techniqueThe Lowest Common Multiple · Remainders

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Treating a 'shortage of 3' as an excess and adding 3 (120 + 3 = 123), which would leave a remainder of 3 rather than a shortage.
  • Failing to recognize that 'remainder of 2 when divided by 5' is structurally identical to a 'shortage of 3' (5 − 2 = 3).
10

LCM with Shifting Non-Uniform Remainders

heuristicSystematic Listing / Congruence Stepping

A florist has a collection of roses numbering fewer than 100. When she arranges them in bunches of 5, she has 3 roses left over. When she arranges them in bunches of 7, she has 4 roses left over. What are all the possible total numbers of roses she could have?

roses, 0 to 1007s + 4
Bunches of 7 with 4 over: 4, 11, 18, 25, 32, ...
Column 1 (Groups of 7 plus 4): 4, 11, 18, 25, 32, 39...
step 1 of 6

Construct a two-column listing table to identify the first common overlapping value, then step forward by the LCM interval.

  1. Column 1 (Groups of 7 plus 4): 4, 11, 18, 25, 32, 39...
  2. Column 2 (Groups of 5 plus 3): 3, 8, 13, 18, 23, 28...
  3. Identify the first intersection in both lists: 18.
  4. Calculate repeating block: 5 × 7 = 35.
  5. Step forward by 35: 18, (18 + 35 = 53), (53 + 35 = 88).
  6. Stop since the next number 88 + 35 = 123 is greater than 100.

answer18, 53 or 88 roses

techniqueCommon Multiples · Modular Arithmetic on a Clock Face

examsPSLE

Common pitfalls

  • Listing only the first value (18) and missing the subsequent values (53, 88) requested by 'all possible numbers'.
  • Adding the remainders together (3 + 4 = 7) and trying to apply a constant offset approach.
  • Stepping forward by the sum of divisors (5 + 7 = 12) instead of their LCM (35).
11

Divisibility: Packing 2,346 Cans into Boxes of 4, 6 or 9, and the Fewest Extra Cans Needed

heuristicDivisibility Tests / Last Two Digits and Digit Sum

A school food drive has collected 2,346 cans. The cans are to be packed into boxes that are all the same size, with every box full and no cans left over. The boxes come in three sizes: 4 cans, 6 cans and 9 cans. (a) Without dividing, use the divisibility tests to find which box size or sizes the school can use. (b) The school would rather use the boxes of 9. What is the smallest number of extra cans, 0 or more, it must collect so that boxes of 9 can be used, and how many boxes of 9 will it then fill?

2,346 cansBoxes of 4last two digits: 46 = 4 × 11 + 2no
Boxes of 4: only the last two digits decide, because 100 is a multiple of 4. 46 = 4 × 11 + 2, so 2 cans would be left over.
Test for 4: look only at the last two digits, 46. Since 46 = 4 × 11 + 2, 46 is not a multiple of 4, so boxes of 4 would leave 2 cans over.
step 1 of 6

Test each box size with its divisibility test instead of dividing: the last two digits for 4, the digit sum for 9, and the tests for 2 and 3 together for 6. Then use the remainder to find how many cans complete one more box of 9.

  1. Test for 4: look only at the last two digits, 46. Since 46 = 4 × 11 + 2, 46 is not a multiple of 4, so boxes of 4 would leave 2 cans over.
  2. Test for 9: add the digits. 2 + 3 + 4 + 6 = 15, and 15 is not a multiple of 9, so boxes of 9 cannot be used as things stand.
  3. Test for 6: the number must pass the tests for 2 and for 3. It ends in 6, so it is even, and its digit sum, 15, is a multiple of 3. So 2,346 is a multiple of 6.
  4. (a) Only boxes of 6 can be used. Check: 6 × 391 = 2346.
  5. The digit sum 15 is 6 more than 9, so boxes of 9 leave 6 cans over. Another 9 − 6 = 3 cans complete one more box, making 2,349 cans, whose digit sum is 2 + 3 + 4 + 9 = 18.
  6. (b) The school collects 3 extra cans, and 2349 ÷ 9 = 261, so it fills 261 boxes of 9. Check: 9 × 261 = 2349.

answer(a) Boxes of 6 only; (b) 3 extra cans, filling 261 boxes of 9

techniqueDividing Exactly · Divisibility Tests

examsPSLE

Common pitfalls

  • Deciding that 2,346 is a multiple of 4 because it is even. Being even only shows that it divides by 2; the test for 4 looks at the last two digits, and 46 is not a multiple of 4.
  • Answering 6 extra cans because boxes of 9 leave 6 over. The 6 cans are what is left over; the extra cans needed are what completes the next box, 9 − 6 = 3.
  • Thinking that a number which divides by 3 must also divide by 9. The digit sum 15 is a multiple of 3 but not of 9.
12

Factors: Every Equal Team Size for a School Camp, Then the Sizes the Rules Allow

heuristicListing Factors in Pairs / Applying Each Rule in Turn

At a school camp, 60 students are to be split into teams for the games, with every team the same size and nobody left out. (a) Teams of 1 and a single team of all 60 count too. List every team size that splits the 60 students equally. (b) The camp rules say that each team must have at least 4 students and that there must be at least 3 teams. Which team sizes are still possible?

factor pairs of 60factorpartner1×602×303×204×155×126×10
Every team size comes with a number of teams, so find the factors of 60 in pairs.
Find the factors of 60 in pairs, starting from 1: 1 × 60, 2 × 30, 3 × 20, 4 × 15, 5 × 12 and 6 × 10.
step 1 of 6

List the factors of 60 in pairs, because every team size comes with a number of teams. Then apply each rule to the list and cross out the sizes it forbids.

  1. Find the factors of 60 in pairs, starting from 1: 1 × 60, 2 × 30, 3 × 20, 4 × 15, 5 × 12 and 6 × 10.
  2. Try 7: 60 ÷ 7 = 8 remainder 4, so 7 is not a factor. The next number, 8, has 8 × 8 = 64, which is more than 60, so any partner of 8 would be smaller than 8 and already found. The list is complete.
  3. (a) The team sizes are the 12 factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
  4. Apply the first rule, at least 4 students in a team. That crosses out team sizes 1, 2 and 3.
  5. Apply the second rule, at least 3 teams. The number of teams is the partner in each pair, so team sizes 30 and 60, which give only 2 teams and 1 team, are crossed out.
  6. (b) The team sizes still possible are 4, 5, 6, 10, 12, 15 and 20, which is 7 sizes. Check: teams of 20 make 60 ÷ 20 = 3 teams, just enough, and teams of 4 make 15 teams.

answer(a) 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60 (12 sizes); (b) 4, 5, 6, 10, 12, 15 and 20 (7 sizes)

techniqueFactors · Factor Pairs

examsPSLE · GCSE Higher

Common pitfalls

  • Leaving out 1 and 60, or stopping the list at 6. Every factor has a partner, so listing the factors in pairs finds 10, 12, 15, 20, 30 and 60 along with 1 to 6.
  • Applying the rule of at least 3 teams to the team size instead of to the number of teams. A team of 30 students is large, but it gives only 60 ÷ 30 = 2 teams, so it is the one that breaks the rule.
13

Factor Pairs: 96 Chairs in a Rectangle, and the Arrangement the Hall Has Room For

heuristicFactor Pairs / Testing Each Arrangement Against the Limits

A caretaker is setting out 96 chairs in a school hall in a rectangle: every row has the same number of chairs, and there are no gaps. (a) Counting 8 rows of 12 and 12 rows of 8 as two different arrangements, and counting a single row of 96 as an arrangement, how many different arrangements are there? (b) The hall has room for no more than 14 chairs in a row and no more than 10 rows. Which arrangement must the caretaker use?

factor pairs of 961 × 962 × 483 × 324 × 246 × 168 × 125, 7, 9 leave remainders; 10 × 10 is past 96
The factor pairs of 96. 5, 7 and 9 leave remainders, and 10 × 10 = 100 is more than 96, so the list is complete.
List the factor pairs of 96, testing 1, 2, 3, and so on: 1 × 96, 2 × 48, 3 × 32, 4 × 24, 6 × 16 and 8 × 12. The numbers 5, 7 and 9 leave remainders, and 10 × 10 = 100 is more than 96, so the list is complete.
step 1 of 5

List the factor pairs of 96. Each pair gives two arrangements, one for each way round. Then test every arrangement against the two limits of the hall.

  1. List the factor pairs of 96, testing 1, 2, 3, and so on: 1 × 96, 2 × 48, 3 × 32, 4 × 24, 6 × 16 and 8 × 12. The numbers 5, 7 and 9 leave remainders, and 10 × 10 = 100 is more than 96, so the list is complete.
  2. (a) Each pair gives two arrangements, because either number can be the number of rows. So there are 6 × 2 = 12 arrangements.
  3. Apply the first limit, no more than 14 chairs in a row. That rules out rows of 96, 48, 32, 24 and 16 chairs, and leaves 8 rows of 12, 12 rows of 8, 16 rows of 6, 24 rows of 4, 32 rows of 3, 48 rows of 2 and 96 rows of 1.
  4. Apply the second limit, no more than 10 rows. Of those seven arrangements, only 8 rows of 12 has 10 rows or fewer.
  5. (b) The caretaker must use 8 rows of 12 chairs. Check: 8 × 12 = 96, 12 chairs is within the limit of 14, and 8 rows is within the limit of 10.

answer(a) 12 arrangements; (b) 8 rows of 12 chairs

techniqueFactor Pairs · Factors

examsPSLE · GCSE Higher

Common pitfalls

  • Counting 6 arrangements, one for each factor pair. 8 rows of 12 and 12 rows of 8 look different in the hall, and the question counts them separately, so each pair gives two.
  • Choosing 12 rows of 8 because 8 chairs is within the limit of 14. It has 12 rows, and the hall has room for only 10, so both limits must be checked for every arrangement.
14

Primes: 60 Numbered Lockers Sorted by the Sieve, and the Rest Shared Equally

heuristicSieve of Eratosthenes / Counting What Is Crossed Out

The 60 lockers in a school corridor are numbered 1 to 60. For Math Week, every locker with a prime number goes to a Primary 6 student. Locker 1 holds the first-aid kit. All the other lockers are shared equally among Primary 3, Primary 4 and Primary 5. (a) Use the Sieve of Eratosthenes to find how many lockers go to Primary 6 students. (b) How many lockers does each of the other three levels get?

1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859601 is neither prime nor composite: the first-aid locker
Lockers 1 to 60 in rows of 10. Cross out 1: it is neither prime nor composite, and it holds the first-aid kit.
Write 1 to 60 in rows of 10. Cross out 1: it is neither prime nor composite, and it is the first-aid locker.
step 1 of 6

Write the numbers 1 to 60 in rows of 10. Keep each prime and cross out its other multiples, starting with 2, until the next prime times itself is more than 60. The numbers left are the primes.

  1. Write 1 to 60 in rows of 10. Cross out 1: it is neither prime nor composite, and it is the first-aid locker.
  2. Keep 2 and cross out every other multiple of 2: 4, 6, 8, and so on up to 60. That crosses out 29 numbers.
  3. Keep 3 and cross out the multiples of 3 that are still there: 9, 15, 21, 27, 33, 39, 45, 51 and 57. That is 9 more.
  4. Keep 5 and cross out 25, 35 and 55; keep 7 and cross out 49. The next prime is 11, and 11 × 11 = 121 is more than 60, so every number left is prime.
  5. (a) The numbers left are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53 and 59, so 17 lockers go to Primary 6 students.
  6. (b) The composite lockers are the ones crossed out after locker 1: 29 + 9 + 3 + 1 = 42, which is also 60 − 1 − 17. Shared among three levels, each gets 42 ÷ 3 = 14 lockers.

answer(a) 17 lockers; (b) 14 lockers each

techniqueThe Sieve of Eratosthenes · Multiples

examsO-Level · GCSE Higher

Common pitfalls

  • Counting 1 as a prime. A prime has exactly two factors, 1 and itself, and 1 has only one factor, so it is not prime; here it is also the first-aid locker.
  • Stopping the sieve after 5 and leaving 49 in the list. 49 = 7 × 7, so the multiples of 7 must be crossed out too; only once the next prime times itself passes 60 can the sieve stop.
Mr. Chalk Drill factors and multiples