AO3 · HCF and LCM

HCF and LCM Heuristics

10 question types · PSLE Paper 2 · Model Method and algebra, side by side

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) xrightarrow÷ 2 (36, 45).
step 1 of 6 Practice HCF and LCM in the app

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) xrightarrow÷ 2 (36, 45).
  2. Divide both 36 and 45 by 3: (36, 45) xrightarrow÷ 3 (12, 15).
  3. Divide both 12 and 15 by 3: (12, 15) xrightarrow÷ 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.

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 labelled 360 cm by 280 cm.
step 1 of 4 Practice HCF and LCM in the app

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

  1. Draw a rectangle labelled 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.

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 Practice HCF and LCM in the app

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.

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 Practice HCF and LCM in the app

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.

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 Practice HCF and LCM in the app

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.

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 Practice HCF and LCM in the app

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.

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, mathbf24, 30...
step 1 of 6 Practice HCF and LCM in the app

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

  1. List bun counts: 6, 12, 18, mathbf24, 30...
  2. List sausage counts: 8, 16, mathbf24, 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.

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 Practice HCF and LCM in the app

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.

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 wants to pack goodie bags for a school carnival. If she puts 5 badges in each bag, she is short of 3 badges. If she puts 6 badges in each bag, she is short of 3 badges. If she puts 8 badges in each bag, she is also short of 3 badges. 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 Practice HCF and LCM in the app

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).

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 Practice HCF and LCM in the app

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.

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).