HCF and LCM · applications

Applications: HCF and LCM

20 problems

PSLE · O-Level · GCSE Higher

01

HCF: The Most Gift Bags from 72 Chocolate Bars and 90 Packets of Biscuits

heuristicThe Ladder / Divide Both Numbers Until Only 1 Divides Both

Mr Lim has 72 chocolate bars and 90 packets of biscuits. He packs all of them into gift bags, so that every bag has the same number of chocolate bars and the same number of packets of biscuits, with none left over. (a) What is the greatest number of gift bags he can pack? (b) How many chocolate bars and how many packets of biscuits are in each bag?

barspackets?7290
The number of bags must divide 72 and 90 exactly, so it is a common factor. The most bags is the HCF.
The number of bags must share out the 72 chocolate bars exactly and the 90 packets exactly, so it is a common factor of 72 and 90. The greatest number of bags is their HCF.
step 1 of 6

Divide 72 and 90 by a factor they share, and keep dividing both until no number except 1 divides both. The numbers you divided by multiply to the HCF, which is the most bags, and the two numbers left at the bottom are what each bag holds.

All steps
  1. The number of bags must share out the 72 chocolate bars exactly and the 90 packets exactly, so it is a common factor of 72 and 90. The greatest number of bags is their HCF.
  2. Both numbers are even, so divide both by 2: 72 ÷ 2 = 36 and 90 ÷ 2 = 45.
  3. 36 and 45 are both multiples of 3, so divide both by 3: 36 ÷ 3 = 12 and 45 ÷ 3 = 15.
  4. 12 and 15 are both multiples of 3 again: 12 ÷ 3 = 4 and 15 ÷ 3 = 5. No number except 1 divides both 4 and 5, so stop.
  5. (a) Multiply the numbers you divided by: 2 × 3 × 3 = 18. He can pack 18 gift bags.
  6. (b) The numbers left at the bottom are what each bag holds: 4 chocolate bars and 5 packets of biscuits. Check: 18 × 4 = 72 and 18 × 5 = 90.

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

techniqueThe Highest Common Factor · HCF and LCM Word Problems

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Using the LCM of 72 and 90, which is 360. With 360 bags, every bag would need at least one chocolate bar, and there are only 72. The number of bags must divide both numbers, so it is a common factor, not a common multiple.
  • Stopping the ladder too early. Dividing both numbers by 9 gives 8 and 10, but 8 and 10 are both even, so they can still be divided by 2. Stopping there gives 9 bags instead of 18.
02

HCF: The Largest Square Tiles for a Floor 360 cm by 280 cm

heuristicThe Tile's Side Divides Both Edges / Tiles in a Row Times the Rows

A rectangular floor is 360 cm long and 280 cm wide. It is to be covered with square tiles that are all the same size, and no tile may be cut. (a) What is the greatest possible length of the side of each tile? (b) What is the least number of tiles needed to cover the floor?

360 cm280 cm?
Whole tiles must fit exactly along 360 cm and along 280 cm, so the side of a tile is a common factor of both.
Whole tiles must fit exactly along the 360 cm length and along the 280 cm width, so the side of a tile is a common factor of 360 and 280. The largest tile has the HCF as its side.
step 1 of 5

The side of a tile must fit a whole number of times along the length and along the width, so it is a common factor of 360 and 280. The largest tile gives the fewest tiles. Find the HCF with the ladder, then count the tiles in rows and columns.

All steps
  1. Whole tiles must fit exactly along the 360 cm length and along the 280 cm width, so the side of a tile is a common factor of 360 and 280. The largest tile has the HCF as its side.
  2. Both lengths end in 0, so divide both by 10: 360 ÷ 10 = 36 and 280 ÷ 10 = 28.
  3. 36 and 28 are both multiples of 4, so divide both by 4: 36 ÷ 4 = 9 and 28 ÷ 4 = 7. No number except 1 divides both 9 and 7, so stop.
  4. (a) Multiply the numbers you divided by: 10 × 4 = 40. The largest tile has sides of 40 cm.
  5. (b) The numbers left at the bottom count the tiles: 9 fit along the length and 7 along the width, so the floor takes 9 × 7 = 63 tiles. Smaller tiles would need more, so 63 is the least. Check: 9 × 40 = 360 and 7 × 40 = 280.

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

techniqueThe Highest Common Factor · Choosing Between HCF and LCM

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Adding the tiles along the two edges, 9 + 7 = 16. That counts only one row and one column. The floor is covered by 7 rows with 9 tiles in each, so the number of tiles is 7 × 9 = 63.
  • Using the LCM of 360 and 280, which is 2520. A tile with sides of 2520 cm is far bigger than the floor, which is only 360 cm long. The side of a tile must divide both edges, so it is a common factor.
03

HCF: The Longest Equal Pieces Cut from Three Planks

heuristicThe Ladder for Three Numbers / Add the Pieces from Each Plank

Three wooden planks are 120 cm, 180 cm and 240 cm long. They are cut into pieces that are all the same length, with no wood left over. (a) What is the greatest possible length of each piece? (b) How many pieces are there altogether?

120 cm180 cm240 cm
Each plank is cut into whole pieces, so the piece length must divide 120, 180 and 240.
Each plank is cut into whole pieces, so the piece length divides 120, 180 and 240 exactly. It is a common factor of all three, and the longest piece is their HCF.
step 1 of 5

Draw each plank as a bar. The piece length must divide all three lengths, so divide all three at once by factors they share, until no number except 1 divides all three. The numbers you divided by multiply to the longest piece, and the numbers left count the pieces from each plank.

All steps
  1. Each plank is cut into whole pieces, so the piece length divides 120, 180 and 240 exactly. It is a common factor of all three, and the longest piece is their HCF.
  2. All three lengths end in 0, so divide them all by 10: 120 ÷ 10 = 12, 180 ÷ 10 = 18 and 240 ÷ 10 = 24.
  3. 12, 18 and 24 are all multiples of 6, so divide them all by 6: 12 ÷ 6 = 2, 18 ÷ 6 = 3 and 24 ÷ 6 = 4. 2 divides 2 and 4 but not 3, so no number except 1 divides all three. Stop.
  4. (a) Multiply the numbers you divided by: 10 × 6 = 60. The longest piece is 60 cm.
  5. (b) The numbers left, 2, 3 and 4, are the pieces from each plank, so there are 2 + 3 + 4 = 9 pieces. Check: 9 × 60 = 540, and 120 + 180 + 240 = 540.

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

techniqueThe Highest Common Factor · Prime Factorization

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Dividing again by 2 because 2 and 4 are both even, and getting pieces of 120 cm. 3 is not even, so 2 does not divide all three numbers, and a 120 cm piece does not fit the 180 cm plank: 180 ÷ 120 = 1 remainder 60.
  • Multiplying the pieces from each plank, 2 × 3 × 4 = 24. Each plank is cut on its own, so its pieces are added to the others: 2 + 3 + 4 = 9.
04

HCF: Lampposts Around a Field 48 m by 36 m, with One at Every Corner

heuristicThe Gap Divides Both Sides / Around a Closed Edge, As Many Lampposts as Gaps

A rectangular field is 48 m long and 36 m wide. Lampposts are to be put up along the edge of the field, with the same distance between each lamppost and the next. There must be a lamppost at each of the 4 corners. (a) What is the greatest possible distance between one lamppost and the next? (b) What is the least number of lampposts needed?

48 m36 mcommon factors of 48and 36: 1, 2, 3, 4, 6, 12
A lamppost stands at every corner, so the gap must divide 48 and 36. Their common factors are 1, 2, 3, 4, 6 and 12.
The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36. Of these, 1, 2, 3, 4, 6 and 12 also divide 48, so they are the common factors of 48 and 36.
step 1 of 5

A lamppost stands at every corner, so the gaps must fit exactly along the length and along the width. The greatest gap is the greatest common factor of 48 and 36. Then count the gaps around the field: on a closed loop there are as many lampposts as gaps.

All steps
  1. The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36. Of these, 1, 2, 3, 4, 6 and 12 also divide 48, so they are the common factors of 48 and 36.
  2. (a) The greatest common factor is 12, so the greatest distance between one lamppost and the next is 12 m.
  3. Each 48 m side has 48 ÷ 12 = 4 gaps, and each 36 m side has 36 ÷ 12 = 3 gaps.
  4. Around the whole field there are 4 + 3 + 4 + 3 = 14 gaps.
  5. (b) Each lamppost starts one gap, and the last gap ends at the lamppost where the count began. So there are as many lampposts as gaps: 14 lampposts. Check: 14 × 12 = 168, which is the perimeter, 48 + 36 + 48 + 36.

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

techniqueThe Highest Common Factor · Common Factors

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Counting a lamppost at both ends of every side: 5 + 4 + 5 + 4 = 18. Each corner lamppost ends one side and starts the next, so this counts each of the 4 corners twice: 18 − 4 = 14.
  • Adding 1 for the last lamppost, as along a straight road, and getting 15. The edge of the field closes on itself, so the last gap ends at the first lamppost, and no extra lamppost is needed.
05

LCM: The Next Time Three Lighthouses Flash Together

heuristicTogether Again at a Common Multiple / Then Again Every LCM

Three lighthouses flash their lights. Lighthouse A flashes once every 24 seconds, Lighthouse B once every 36 seconds and Lighthouse C once every 60 seconds. All three flash together 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
The lighthouses flash together again at a number of seconds that 24, 36 and 60 all divide into.
The lighthouses flash together again after a number of seconds that 24, 36 and 60 all divide into: a common multiple. The first such time is their LCM.
step 1 of 6

The lighthouses flash together again after a time that 24, 36 and 60 all divide into, and the first such time is their LCM. Divide all three by a factor they share, then work out the smallest number that the numbers left all divide into.

All steps
  1. The lighthouses flash together again after a number of seconds that 24, 36 and 60 all divide into: a common multiple. The first such time is their LCM.
  2. 24, 36 and 60 are all multiples of 12: 24 = 12 × 2, 36 = 12 × 3 and 60 = 12 × 5. No number except 1 divides any two of 2, 3 and 5.
  3. So the LCM is 12 times the smallest number that 2, 3 and 5 all divide into, which is 2 × 3 × 5 = 30. The LCM is 12 × 30 = 360 seconds, which is 360 ÷ 60 = 6 minutes.
  4. (a) They next flash together 6 minutes after 8:00 p.m., at 8:06 p.m.
  5. After that they flash together every 6 minutes: at 8:06, 8:12, 8:18, 8:24 and 8:30 p.m. The next time, 8:36 p.m., is after 8:35 p.m.
  6. (b) All three flash together 5 times between 8:01 p.m. and 8:35 p.m.

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

  • Using the HCF of 24, 36 and 60, which is 12 seconds. None of the lighthouses flashes 12 seconds after 8:00 p.m., because 12 is shorter than every one of the times. The time must be a multiple of 24, 36 and 60, so it is the LCM.
  • Counting the flash at 8:00 p.m. or the one at 8:36 p.m. and answering 6 or 7. 8:00 p.m. is before 8:01 p.m., and 8:36 p.m. is after 8:35 p.m., so neither is between them.
06

LCM: The Smallest Cube Built from Blocks 12 cm by 8 cm by 6 cm

heuristicThe Cube's Edge Is a Common Multiple / Blocks in a Layer Times the Layers

A boy has many wooden blocks, all the same size. Each block is 12 cm long, 8 cm wide and 6 cm high. He stacks the blocks, all facing the same way, to build the smallest possible solid cube. (a) How long is each edge of the cube? (b) How many blocks does he use?

centimetres, 0 to 4812s0122436488s0816243240486s0612182430364248
Count in 12s: 12 is not a multiple of 8, but 24 is a multiple of 12, of 8 and of 6.
Count in 12s, the longest side of a block: 12, 24, 36, and so on. 12 is not a multiple of 8, but 24 is a multiple of both 8 and 6: 24 = 8 × 3 = 6 × 4.
step 1 of 5

Along each edge of the cube the blocks lie end to end, so each edge is a multiple of one side of the block. The edge must be a common multiple of 12, 8 and 6, and the smallest cube has their LCM as its edge. Then count the blocks along each edge and multiply, as for the volume of a cuboid.

All steps
  1. Count in 12s, the longest side of a block: 12, 24, 36, and so on. 12 is not a multiple of 8, but 24 is a multiple of both 8 and 6: 24 = 8 × 3 = 6 × 4.
  2. (a) The smallest common multiple of 12, 8 and 6 is 24, so each edge of the cube is 24 cm.
  3. Along one edge the 12 cm sides lie end to end: 24 ÷ 12 = 2 blocks. Along the next, the 8 cm sides make 24 ÷ 8 = 3 blocks, and up the third, the 6 cm sides make 24 ÷ 6 = 4 blocks.
  4. One layer is 2 blocks by 3 blocks, which is 2 × 3 = 6 blocks, and 4 layers stack up to the top: 6 × 4 = 24 blocks.
  5. (b) He uses 24 blocks. Check: one block is 12 × 8 × 6 = 576 cm3, and 24 × 576 = 13824 = 24 × 24 × 24.

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

techniqueThe Lowest Common Multiple · Volume of a Cuboid

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Adding the blocks along the three edges, 2 + 3 + 4 = 9. That counts only one row of blocks along each edge. The cube is 4 layers of 2 rows of 3 blocks, so it holds 2 × 3 × 4 = 24 blocks.
  • Using the HCF of 12, 8 and 6, which is 2 cm. A cube with 2 cm edges is smaller than one block. The blocks are put together to build the cube, so its edge is a multiple of each side of a block, not a factor.
07

LCM: The Least Spent on Equal Numbers of Buns and Sausages

heuristicList the Multiples of Each Pack Size / The First Number on Both Lists

At a supermarket, hot dog buns are sold in packs of 6 for $4, and sausages are sold in packs of 8 for $5. Melissa wants to buy exactly the same number of buns and sausages, spending as little as she can. (a) How many buns and how many sausages does she buy? (b) What is the least amount of money she can spend?

Buns66666packs of 6
Buns come in packs of 6: she can buy 6, 12, 18, 24 or 30 buns.
Buns come in packs of 6, so she can buy 6, 12, 18, 24 or 30 buns, and so on. These are the multiples of 6.
step 1 of 5

List how many buns she can buy and how many sausages she can buy, in whole packs. The first number on both lists is the LCM of 6 and 8, the smallest equal number. Then work out the packs and the cost.

All steps
  1. Buns come in packs of 6, so she can buy 6, 12, 18, 24 or 30 buns, and so on. These are the multiples of 6.
  2. Sausages come in packs of 8, so she can buy 8, 16, 24 or 32 sausages, and so on. These are the multiples of 8.
  3. (a) The first number on both lists is 24, the LCM of 6 and 8. She buys 24 buns and 24 sausages.
  4. 24 buns is 24 ÷ 6 = 4 packs, which cost 4 × $4 = $16. 24 sausages is 24 ÷ 8 = 3 packs, which cost 3 × $5 = $15.
  5. (b) The least she can spend is $16 + $15 = $31. The next number on both lists, 48, would need twice as many packs and cost twice as much.

answer(a) 24 buns and 24 sausages; (b) $31

techniqueThe Lowest Common Multiple · Common Multiples

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Finding the LCM of the prices, 4 and 5, and answering $20. The buns and the sausages must match in number, so the LCM comes from the pack sizes, 6 and 8, not from the prices.
  • Multiplying 24 by the price of a pack, 24 × $4 = $96. $4 buys a pack of 6 buns, not one bun, so first find the number of packs: 24 ÷ 6 = 4.
08

LCM Plus a Remainder: Eggs That Leave 2 Over in Cartons of 6, 8 or 9

heuristicTake the Eggs Left Over Away / The Rest Is a Common Multiple

A farmer has a basket of eggs. When he packs them into cartons of 6, there are 2 eggs left over. The same happens when he packs them into cartons of 8, and when he packs them into cartons of 9. He has more than 2 eggs. (a) If the 2 eggs left over are taken out, what is the smallest number of eggs there could be in the rest? (b) What is the smallest possible number of eggs in the basket?

Eggsfills cartons of 6, 8 or 9 exactly2take 2 out
Take the 2 eggs left over out. The rest fill cartons of 6, of 8 or of 9 exactly.
Take the 2 eggs left over out of the basket. The eggs that remain fill cartons of 6, of 8 or of 9 exactly, so their number is a common multiple of 6, 8 and 9.
step 1 of 4

Take the 2 eggs left over out of the basket. The rest fill cartons of 6, 8 or 9 exactly, so the rest is a common multiple of 6, 8 and 9. Find the smallest one, then put the 2 eggs back.

All steps
  1. Take the 2 eggs left over out of the basket. The eggs that remain fill cartons of 6, of 8 or of 9 exactly, so their number is a common multiple of 6, 8 and 9.
  2. Count in 9s, the largest carton: 9, 18, 27, 36, 45, 54 and 63 are not multiples of 8, but 72 = 8 × 9 is. 72 is also a multiple of 6: 72 = 6 × 12.
  3. (a) The smallest number of eggs in the rest is 72, the LCM of 6, 8 and 9.
  4. (b) Put the 2 eggs back: 72 + 2 = 74. The smallest possible number of eggs is 74. Check: 74 = 6 × 12 + 2 = 8 × 9 + 2 = 9 × 8 + 2.

answer(a) 72 eggs; (b) 74 eggs

techniqueThe Lowest Common Multiple · Remainders

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Taking the 2 away from the LCM, 72 − 2 = 70. In cartons of 6, 70 = 6 × 11 + 4 leaves 4 over, not 2. The eggs are 2 more than a full-carton count, so the 2 is added: 72 + 2 = 74.
  • Adding 2 to each carton size and finding the LCM of 8, 10 and 11. The 2 is a number of eggs left over, not a change in the size of a carton, so it is added to the total, not to 6, 8 and 9.
09

LCM Less a Shortage: Badges That Leave the Last Bag 3 Short

heuristicAdd the Missing Badges / The Total Is a Common Multiple

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. (a) What is the smallest number of badges that would fill bags of 5, bags of 6 or bags of 8 with no bag short? (b) What is the smallest possible number of badges the teacher has?

Badgesbadges33 short
Each time, the last bag needs 3 more badges to be full.
Each time, the last bag needs 3 more badges to be full. So with 3 more badges, every bag would be full, whether it holds 5, 6 or 8.
step 1 of 5

The last bag is 3 short every time, so 3 more badges would fill every bag exactly. The badges plus 3 make a common multiple of 5, 6 and 8. Find the smallest one, then take the 3 missing badges away.

All steps
  1. Each time, the last bag needs 3 more badges to be full. So with 3 more badges, every bag would be full, whether it holds 5, 6 or 8.
  2. That means the number of badges plus 3 is a common multiple of 5, 6 and 8.
  3. Count in 8s. A multiple of 8 is even, so it is also a multiple of 5 only if it ends in 0: 40, 80, 120, and so on. 40 ÷ 6 and 80 ÷ 6 leave remainders, but 120 = 6 × 20.
  4. (a) The smallest number that fills every bag size exactly is 120, the LCM of 5, 6 and 8.
  5. (b) The teacher has 3 fewer: 120 − 3 = 117 badges. Check: in bags of 5, 6 and 8 the last bag holds 2, 3 and 5 badges, since 117 = 5 × 23 + 2 = 6 × 19 + 3 = 8 × 14 + 5, and each is 3 short of full.

answer(a) 120 badges; (b) 117 badges

techniqueThe Lowest Common Multiple · Remainders

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Adding the 3 and answering 120 + 3 = 123. In bags of 5, 123 = 5 × 24 + 3, so the last bag holds 3 badges and is 2 short, not 3 short. The badges are 3 fewer than a full count, so the 3 is taken away.
  • Adding the badges that are in the last bag instead of the badges missing from it. In bags of 5 the last bag holds 2 badges, and 117 + 2 = 119 is not a multiple of 5. It is the 3 missing badges that complete the bag: 117 + 3 = 120.
10

Common Multiples: Roses That Leave 3 Over in Fives and 4 Over in Sevens

heuristicList Both Conditions and Find the First Match / Then Step On by the LCM

A florist has fewer than 100 roses. When she puts them into bunches of 5, she has 3 roses left over. When she puts them into bunches of 7, she has 4 roses left over. (a) What is the smallest number of roses she could have? (b) What are all the possible numbers of roses she could have?

roses, 0 to 100sevens + 4411182532394653606774818895
Leaving 4 over in bunches of 7: start at 4 and count on in 7s. 4, 11, 18, 25, 32, and so on.
List the numbers that leave 4 over in bunches of 7. Start at 4 and count on in 7s: 4, 11, 18, 25, 32, and so on.
step 1 of 6

List the numbers that fit each condition and find the first number on both lists. 35, the LCM of 5 and 7, is a whole number of fives and a whole number of sevens, so adding 35 roses leaves both remainders the same. Step on from the first match by 35 until the numbers pass 100.

All steps
  1. List the numbers that leave 4 over in bunches of 7. Start at 4 and count on in 7s: 4, 11, 18, 25, 32, and so on.
  2. List the numbers that leave 3 over in bunches of 5. Start at 3 and count on in 5s: 3, 8, 13, 18, 23, and so on.
  3. (a) The first number on both lists is 18, so the smallest number of roses is 18.
  4. 35 is the LCM of 5 and 7: it makes 7 bunches of 5 and 5 bunches of 7. So adding 35 roses adds whole bunches of both sizes, and the roses left over stay at 3 and 4.
  5. Step on from 18 by 35: 18 + 35 = 53 and 53 + 35 = 88. The next one, 88 + 35 = 123, is more than 100.
  6. (b) The possible numbers of roses are 18, 53 and 88. Check: 88 = 5 × 17 + 3 = 7 × 12 + 4.

answer(a) 18 roses; (b) 18, 53 or 88 roses

techniqueCommon Multiples · Modular Arithmetic on a Clock Face

examsPSLE

Common pitfalls

  • Giving only 18. The question asks for all the possible numbers, and adding 35 roses, which is 7 bunches of 5 and 5 bunches of 7, keeps both remainders the same, so 53 and 88 fit as well.
  • Stepping on by 5 + 7 = 12 and getting 30. 30 = 5 × 6 leaves nothing over in bunches of 5. Only a common multiple of 5 and 7 adds whole bunches of both sizes, and the smallest is 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 4hundreds46no
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.

All steps
  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.

All steps
  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?

123468121624324896
The factor pairs of 96: 1 × 96, 2 × 48, 3 × 32, 4 × 24, 6 × 16 and 8 × 12. 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.

All steps
  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?

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960
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.

All steps
  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.
15

Divisibility: A Code 5d8k with Two Missing Digits, Found from Four Conditions

heuristicDivisibility Tests / Apply the Conditions One at a Time

A 4-digit security code is written as 5d8k, where d and k are single digits from 0 to 9. The code meets four conditions. Condition 1: it is divisible by 6. Condition 2: it is not divisible by 4. Condition 3: the 2-digit number 5d is a prime number. Condition 4: it is divisible by 9. (a) State the divisibility rule for 6. (b) Using conditions 1 and 2 only, find all the possible values of k. (c) Find the code.

5d8krule for 6even, and digit sum a multiple of 3
(a) 6 = 2 × 3, so a number divisible by 6 is even and has a digit sum that is a multiple of 3.
(a) A number is divisible by 6 when it is divisible by both 2 and 3, because 6 = 2 × 3. So it must be even, and the sum of its digits must be a multiple of 3.
step 1 of 6

Apply the conditions one at a time, each with its divisibility test. Conditions 1 and 2 are about the last digits, so they narrow down k first. The digit sum for condition 4 then fixes d for each k, and condition 3 decides between the codes that are left.

All steps
  1. (a) A number is divisible by 6 when it is divisible by both 2 and 3, because 6 = 2 × 3. So it must be even, and the sum of its digits must be a multiple of 3.
  2. Condition 1 needs the code to be even, so k is 0, 2, 4, 6 or 8. For condition 2, only the last two digits decide, because 100 is a multiple of 4. Of 80, 82, 84, 86 and 88, the numbers 80, 84 and 88 are multiples of 4, so k cannot be 0, 4 or 8.
  3. (b) k = 2 or k = 6. The digit-sum part of condition 1 depends on d as well, so on its own it rules out no value of k. Check: 82 = 4 × 20 + 2 and 86 = 4 × 21 + 2, so neither is a multiple of 4.
  4. Condition 4: the digit sum 5 + d + 8 + k = 13 + d + k must be a multiple of 9. It lies between 15 and 28, so it is 18 or 27. If k = 2, then 15 + d = 18 and d = 3, giving 5382. If k = 6, then 19 + d = 27 and d = 8, giving 5886.
  5. Both codes are even with a digit sum that is a multiple of 3, so both still meet condition 1. Condition 3 decides: 53 is not divisible by 2, 3, 5 or 7, and 8 × 8 = 64 is more than 53, so 53 is prime. But 58 = 2 × 29 is not prime.
  6. (c) The code is 5382. Check: 5382 = 6 × 897 = 9 × 598, and 82 is not a multiple of 4.

answer(a) The number must be even, and the sum of its digits must be a multiple of 3; (b) k = 2 or k = 6; (c) 5382

techniqueDivisibility Tests · The Sieve of Eratosthenes

examsPSLE

Common pitfalls

  • Testing condition 2 with the last digit alone and crossing out only k = 4 and k = 8. The test for 4 looks at the last two digits, and 80 = 4 × 20, so k = 0 makes the code divisible by 4 as well.
  • Stopping at the digit sum and giving 5886, or both codes. Conditions 1, 2 and 4 allow both 5382 and 5886; condition 3 is what separates them, because 58 = 2 × 29 is not prime.
16

LCM: Badges Between 50 and 100, Two Short of Full Packs, Then Packed in Sevens and Nines

heuristicCommon Multiple Less a Constant Shortage / Test Each Candidate Against the Last Condition

A teacher has between 50 and 100 reward badges in a box. When she puts them into packs of 6, 4 badges are left over. When she puts them into packs of 8, 6 badges are left over. When she sets them out in equal rows of 7, none are left over. (a) Explain why the number of badges plus 2 is a common multiple of 6 and 8. (b) How many badges does the teacher have? (c) She then packs all the badges into gift packets, some holding 7 badges and some holding 9, with none left over. What is the greatest number of 9-badge packets she can make?

badges, 0 to 1206s + 48s + 6
Packs of 6 leave 4 over and packs of 8 leave 6 over: each time, 2 more badges would fill another pack.
Packs of 6 leave 4 over, and 6 − 4 = 2 more badges would fill one more pack. Packs of 8 leave 6 over, and 8 − 6 = 2 more badges would fill one more pack as well.
step 1 of 6

Each remainder is 2 short of a full pack, so 2 more badges would make both packings exact. List the common multiples of 6 and 8, take 2 away from each, and keep the one between 50 and 100 that also fills rows of 7. For the packets, try the most 9-badge packets first.

All steps
  1. Packs of 6 leave 4 over, and 6 − 4 = 2 more badges would fill one more pack. Packs of 8 leave 6 over, and 8 − 6 = 2 more badges would fill one more pack as well.
  2. (a) With 2 more badges, both packings would come out exact. So the number of badges plus 2 is a multiple of 6 and a multiple of 8: a common multiple of 6 and 8.
  3. The lowest common multiple of 6 and 8 is 24, so the common multiples are 24, 48, 72, 96, 120 and so on. Taking 2 away from each gives 22, 46, 70, 94 and 118, and only 70 and 94 lie between 50 and 100.
  4. (b) Rows of 7: 70 = 7 × 10 exactly, but 94 = 7 × 13 + 3 leaves 3 over. The teacher has 70 badges. Check: 70 = 6 × 11 + 4 = 8 × 8 + 6.
  5. For the packets, start with as many 9-badge packets as fit: 70 ÷ 9 = 7 remainder 7. The 7 badges left over make exactly one packet of 7.
  6. (c) The greatest number of 9-badge packets is 7, with one packet of 7. Check: 7 × 9 + 1 × 7 = 63 + 7 = 70. Eight packets of 9 would need 72 badges, which is more than 70.

answer(a) 2 more badges would fill one more pack of 6 and one more pack of 8, so the number plus 2 is a common multiple of 6 and 8; (b) 70 badges; (c) 7 packets of 9

techniqueThe Lowest Common Multiple · Remainders · Common Multiples

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Adding the 2 instead of taking it away, and testing 72 + 2 = 74 or 96 + 2 = 98. The remainders are 2 short of a full pack, so the count is 2 less than a common multiple; 74 = 6 × 12 + 2 leaves 2 over, not 4.
  • Choosing 94 because it also fits the packs of 6 and 8 and lies between 50 and 100. 94 = 7 × 13 + 3 leaves 3 badges when they are set out in rows of 7, so it breaks the last condition.
17

Divisibility: Fun-Run Money Shared Among Charities, Before and After a Two-Dollar Gift

heuristicDivisibility Tests / Digit Sum for 3 and 9, Last Two Digits for 4

A school fun run raises $1854. The school wants to share all of the money equally among some charities, in whole dollars. (a) Without dividing, use divisibility tests to decide whether the money can be shared equally among 3 charities, among 4 charities and among 9 charities. (b) A parent then gives another $2, so the total is $1856. Which of the three ways of sharing work now?

$1854 raisedby 3 and 91854yes
Test for 3 and for 9: the digit sum is 1 + 8 + 5 + 4 = 18, two nines, and 18 is a multiple of both 3 and 9.
Test for 3 and for 9: add the digits. 1 + 8 + 5 + 4 = 18, and 18 is a multiple of both 3 and 9. So $1854 can be shared equally among 3 charities and among 9 charities.
step 1 of 5

Test each number of charities with its divisibility test instead of dividing: the digit sum for 3 and for 9, and the last two digits for 4. Then test the new total the same way.

All steps
  1. Test for 3 and for 9: add the digits. 1 + 8 + 5 + 4 = 18, and 18 is a multiple of both 3 and 9. So $1854 can be shared equally among 3 charities and among 9 charities.
  2. Test for 4: only the last two digits decide, because 100 is a multiple of 4. 54 = 4 × 13 + 2, so 54 is not a multiple of 4.
  3. (a) The money can be shared equally among 3 or 9 charities, but not among 4. Check: 1854 ÷ 3 = 618 and 1854 ÷ 9 = 206.
  4. Test $1856 in the same way. Its last two digits are 56 = 4 × 14, a multiple of 4. Its digit sum is 1 + 8 + 5 + 6 = 20, which is not a multiple of 3 or of 9.
  5. (b) Now only the sharing among 4 charities works. Check: 1856 ÷ 4 = 464.

answer(a) Among 3 or among 9 charities, but not among 4; (b) only among 4 charities

techniqueDividing Exactly · Divisibility Tests

examsPSLE

Common pitfalls

  • Deciding that $1854 can be shared among 4 charities because 1854 is even. Being even shows only that it divides by 2. The test for 4 looks at the last two digits, and 54 is not a multiple of 4.
  • Thinking that sharing among 3 and 9 still works after the $2 gift. The digit sum changes from 18 to 20, and 20 is not a multiple of 3 or of 9, so neither sharing works any more.
18

Prime Factorization: Three Children's Ages from Their Product

heuristicFactor Tree / Three Ages Above 1 Share Out the Three Primes

Three children are each more than 1 year old. When their ages in years are multiplied together, the answer is 105. (a) Write 105 as a product of prime numbers. (b) How old is each child?

105521
105 ends in 5, so 5 is a factor: 105 = 5 × 21.
Start a factor tree. 105 ends in 5, so 5 is a factor: 105 = 5 × 21.
step 1 of 5

Split 105 in a factor tree until every branch ends in a prime. Then share the primes out among the three children, because an age above 1 must use at least one of them.

All steps
  1. Start a factor tree. 105 ends in 5, so 5 is a factor: 105 = 5 × 21.
  2. 5 is prime, so that branch stops. 21 is not prime: 21 = 3 × 7.
  3. (a) 3 and 7 are prime, so the tree is finished: 105 = 3 × 5 × 7.
  4. Each child is more than 1 year old, so each age is made by multiplying at least one of the primes. There are exactly three primes for the three children, so each age is one prime.
  5. (b) The children are 3, 5 and 7 years old. Check: 3 × 5 × 7 = 15 × 7 = 105.

answer(a) 105 = 3 × 5 × 7; (b) 3, 5 and 7 years old

techniquePrime Factorization · Factors

examsO-Level · GCSE Higher

Common pitfalls

  • Giving the ages as 1, 3 and 35, or 1, 5 and 21. These multiply to 105, but each child is more than 1 year old, so no age can be 1.
  • Stopping the factor tree at 105 = 5 × 21. 21 is not prime, because 21 = 3 × 7, so that branch must be split again.
19

Common Factors: Every Possible Number of Art Kits from Paintbrushes and Paint

heuristicList the Factors of Each Number / The Numbers in Both Lists Are the Possible Numbers of Kits

Ms Devi has 18 paintbrushes and 24 tubes of paint. She packs all of them into art kits, so that every kit has the same number of paintbrushes and the same number of tubes of paint, with none left over. (a) List every possible number of art kits she could make. (b) What is the greatest number of art kits she can make, and what is in each kit?

factors of 181236918
The number of kits must share out the 18 paintbrushes, so it is a factor of 18.
The number of kits must share the 18 paintbrushes out exactly, so it is a factor of 18: 1, 2, 3, 6, 9 and 18.
step 1 of 5

List the factors of 18 and the factors of 24. A number of kits must be in both lists, so the common factors are the possible numbers of kits, and the greatest of them is the most kits.

All steps
  1. The number of kits must share the 18 paintbrushes out exactly, so it is a factor of 18: 1, 2, 3, 6, 9 and 18.
  2. It must also share the 24 tubes of paint out exactly, so it is a factor of 24: 1, 2, 3, 4, 6, 8, 12 and 24.
  3. (a) The numbers in both lists are the common factors of 18 and 24: 1, 2, 3 and 6. She could make 1, 2, 3 or 6 kits.
  4. The greatest common factor is 6. Six kits each get 18 ÷ 6 = 3 paintbrushes and 24 ÷ 6 = 4 tubes of paint.
  5. (b) She can make at most 6 kits, each with 3 paintbrushes and 4 tubes of paint. Check: 6 × 3 = 18 and 6 × 4 = 24.

answer(a) 1, 2, 3 or 6 kits; (b) 6 kits, each with 3 paintbrushes and 4 tubes of paint

techniqueCommon Factors · Factors · The Highest Common Factor

examsPSLE · GCSE Higher

Common pitfalls

  • Putting 9, 4 or 8 in the list. 9 is a factor of 18 but not of 24, and 4 and 8 are factors of 24 but not of 18. With 4 kits, for example, the 18 paintbrushes would leave 2 over.
  • Putting 6 paintbrushes in each kit because the answer is 6. The 6 is the number of kits; each kit gets 18 ÷ 6 = 3 paintbrushes.
20

Choosing HCF or LCM: Paper Strips Laid End to End, Then Cut into Equal Pieces

heuristicDecide from the Question / A Length Both Divide Needs the LCM, a Piece That Divides Both Needs the HCF

Ali has red paper strips that are each 12 cm long and green paper strips that are each 18 cm long. (a) He lays red strips end to end in one line and green strips end to end in a line beside it, both starting at the same place. What is the shortest length at which the two lines can end together? (b) He then takes one red strip and one green strip and cuts both into pieces that are all the same length, with nothing left over. What is the longest each piece can be, and how many pieces does he get?

Red12 cmGreen18 cm
Each line is made of whole strips, so its length is a multiple of 12 and a multiple of 18.
Part (a) asks for a length that 12 and 18 both divide into, because each line is made of whole strips. That is a common multiple, and the shortest is the LCM.
step 1 of 5

Ask each part which way round the dividing goes. In (a) both strip lengths divide the answer, so it is a common multiple. In (b) the answer divides both strip lengths, so it is a common factor.

All steps
  1. Part (a) asks for a length that 12 and 18 both divide into, because each line is made of whole strips. That is a common multiple, and the shortest is the LCM.
  2. Count in 18s and test each for 12: 18 is not a multiple of 12, but 36 = 3 × 12 is.
  3. (a) The two lines first end together at 36 cm, with 3 red strips and 2 green strips.
  4. Part (b) asks for a piece length that divides both 12 and 18. That is a common factor, and the longest is the HCF. The factors of 12 are 1, 2, 3, 4, 6 and 12, and the greatest of these that divides 18 is 6.
  5. (b) The longest piece is 6 cm. The red strip makes 12 ÷ 6 = 2 pieces and the green strip makes 18 ÷ 6 = 3, so he gets 5 pieces. Check: the HCF, 6, is less than both 12 and 18, and the LCM, 36, is more than both.

answer(a) 36 cm; (b) 6 cm, giving 5 pieces

techniqueChoosing Between HCF and LCM · The Highest Common Factor · The Lowest Common Multiple

examsGCSE Higher

Common pitfalls

  • Using the HCF, 6 cm, for part (a). A line of red strips can only be 12, 24, 36 and so on centimetres long, so it can never end at 6 cm.
  • Using the LCM, 36 cm, for part (b). A piece cut from a 12 cm strip cannot be longer than the strip; the piece length must divide both 12 and 18.