Patterns and page numbers · applications

Applications: Patterns and Page Numbers

11 problems

PSLE · GCSE Higher

01

Counting the Digits in the Page Numbers 1 to 165

heuristicGroup the pages by how many digits their numbers have

A storybook has 165 pages numbered consecutively from 1 to 165. How many individual printed digits were used in total to number all the pages of the book?

Pages10 to 99100 to 165165 pagesDigits9
Pages 1 to 9: nine pages, one digit each, 9 digits.
The first block is pages 1 to 9: nine pages, each numbered with one digit, 9 × 1 = 9 digits.
step 1 of 4

Split the book into three blocks by the length of the page number. Inside a block every page uses the same number of digits, so a block's digits are its pages times that number.

All steps
  1. The first block is pages 1 to 9: nine pages, each numbered with one digit, 9 × 1 = 9 digits.
  2. The second block is pages 10 to 99: the 99 pages up to page 99 minus the 9 already counted, 99 − 9 = 90 pages, each with two digits: 90 × 2 = 180 digits.
  3. The third block is pages 100 to 165: 165 − 99 = 66 pages, each with three digits: 66 × 3 = 198 digits.
  4. The three blocks use 9 + 180 + 198 = 387 digits. Check: if all 165 pages had three digits there would be 165 × 3 = 495; pages 1 to 9 use 2 × 9 = 18 fewer and pages 10 to 99 use 90 fewer, and 495 − 18 − 90 = 387.

answer387 digits

techniqueNumbers to Ten Million · Multiplying in Columns

examsPSLE · GCSE Higher

Common pitfalls

  • Counting pages 10 to 99 as 99 − 10 = 89. Taking one page number from the other counts the gaps between the pages; counting the pages themselves includes both ends: 99 − 10 + 1 = 90.
  • Multiplying all 165 pages by 3, which gives 495. Only the pages from 100 on have three digits; the 99 pages before them use fewer.
02

A Book Numbered with 492 Digits: How Many Pages?

heuristicSpend the digits on the one-digit pages, then the two-digit pages; what is left goes three at a time

A printer used 492 digits to number all the pages of a reference manual starting from page 1. How many numbered pages does the manual have in total?

Digits492 digitsPages?
Start with all 492 digits and spend them tier by tier.
Draw one bar for all 492 digits used to number the book.
step 1 of 5

Draw the 492 digits as one bar. Cut off the digits used by the one-digit pages and then by the two-digit pages; what remains numbers the three-digit pages, three digits each.

All steps
  1. Draw one bar for all 492 digits used to number the book.
  2. Pages 1 to 9 use one digit each, 9 digits. Cut them off: 492 − 9 = 483 digits are left.
  3. Pages 10 to 99 are 90 pages of two digits, 180 digits. Cut them off: 483 − 180 = 303 digits are left.
  4. Every page still to come has a three-digit number, so the 303 digits number 303 ÷ 3 = 101 pages.
  5. Those 101 pages come after the first 99, so the book has 99 + 101 = 200 pages, ending at page 200. Check: 9 + 180 + 101 × 3 = 492.

answer200 pages

techniqueNumbers to Ten Million · Multiplying and Dividing Undo Each Other

examsPSLE · GCSE Higher

Common pitfalls

  • Dividing all 492 digits by 3, which gives 164 pages. Pages 1 to 99 use one or two digits each, so their digits come off first.
  • Adding the 101 three-digit pages to 100. Pages 1 to 99 are 99 pages; the three-digit pages start at 100, so 101 of them end at page 200, not 201.
  • Taking off 90 digits for pages 10 to 99. There are 90 pages there, but each uses two digits: 180.
03

One Sheet Torn from a 40-Page Magazine

heuristicFind the total of every page; what is missing from it is the two consecutive pages on the torn sheet

A magazine has 40 pages, numbered from 1 to 40. One sheet of paper is torn out of it. The page numbers on the pages that are left add up to 749. What are the two page numbers on the torn-out sheet?

20 pairs20 × 41 = 820
Pair the pages from both ends: 1 + 40, 2 + 39, ... twenty pairs of 41, so all 40 pages sum to 820.
Pair the first page with the last, the second with the second to last, and so on: 1 + 40 = 41, 2 + 39 = 41, and there are 40 ÷ 2 = 20 pairs. All 40 pages add up to 20 × 41 = 820.
step 1 of 5

Pair the pages from both ends to add them all. Then draw the two pages of the torn sheet as two bars, the back one longer by 1.

All steps
  1. Pair the first page with the last, the second with the second to last, and so on: 1 + 40 = 41, 2 + 39 = 41, and there are 40 ÷ 2 = 20 pairs. All 40 pages add up to 20 × 41 = 820.
  2. The pages that are left add up to 749, so the torn sheet carried 820 − 749 = 71.
  3. A sheet has two consecutive page numbers. Draw the front page as one unit, u, and the back page as one unit and 1 more.
  4. Together they make 2u + 1 = 71, so 2u = 70 and u = 35.
  5. The torn sheet is pages 35 and 36. Check: 820 − 35 − 36 = 749.

answerPages 35 and 36

techniqueThe Sum of an Arithmetic Series · Odd and Even Numbers

examsH2

Common pitfalls

  • Halving 71 and getting 35.5. The two pages are whole numbers 1 apart, so take the 1 away first: 71 − 1 = 70, and the front page is 70 ÷ 2 = 35.
  • Adding the 40 pages as 40 × 40 = 1600 or 40 × 41 = 1640. Pairing the pages gives 20 pairs of 41, which is half of 40 × 41: 820.
  • Taking only one page away. A sheet of paper has two sides, so tearing out one sheet removes two page numbers.
04

A Row of Matchstick Squares: Figure 45, and the Figure with 241 Sticks

heuristicCount the first stick on its own; every square then adds the same three sticks

The pattern below shows a sequence of squares formed by matchsticks: Figure 1 uses 4 matchsticks to form 1 square. Figure 2 uses 7 matchsticks to form 2 connected squares. Figure 3 uses 10 matchsticks to form 3 connected squares. (a) How many matchsticks are needed to form Figure 45? (b) Which Figure number is made of 241 matchsticks?

1 first stickFigure 1: 1 + 1 × 3 = 4 sticks
Figure 1: one first stick, then a C of three: 1 + 1 × 3 = 4.
Figure 1 is the first upright stick and one C of three: 1 + 1 × 3 = 4 sticks.
step 1 of 6

Start every figure with the single upright stick at its left end. Each square then needs three more sticks in the shape of a C: its top, its right side and its bottom.

All steps
  1. Figure 1 is the first upright stick and one C of three: 1 + 1 × 3 = 4 sticks.
  2. Figure 2 is the same first stick and two Cs: 1 + 2 × 3 = 7 sticks. The second square uses the first square's right side as its left side, so it needs only three new sticks.
  3. Figure 3 is the first stick and three Cs: 1 + 3 × 3 = 10 sticks. So Figure n is the first stick and n Cs.
  4. (a) Figure 45 is the first stick and 45 Cs: 1 + 45 × 3 = 136 matchsticks.
  5. (b) Take the first stick away from 241: 241 − 1 = 240 sticks are in Cs.
  6. Each C is 3 sticks, so there are 240 ÷ 3 = 80 Cs, one for each square: Figure 80. Check: 1 + 80 × 3 = 241.

answer(a) 136 matchsticks; (b) Figure 80

techniqueThe nth Term · Term-to-Term Rules

examsPSLE · O-Level · GCSE Higher · H2

Common pitfalls

  • Multiplying 45 by 4 because one square takes 4 sticks, which gives 180. Neighbouring squares share a side, so after the first square each one needs only 3 new sticks.
  • Dividing 241 by 3 at once, which does not divide exactly. The first stick belongs to no C; take it away first: 240 ÷ 3 = 80.
05

Square Arrays of Tiles: Figure 25, and the Figure with 1089 Tiles

heuristicFollow the side of each square, which is one more than the figure number

A pattern of square tile arrays is shown: Figure 1 has a total of 4 tiles (2 × 2). Figure 2 has 9 tiles (3 × 3). Figure 3 has 16 tiles (4 × 4). (a) How many tiles are in Figure 25? (b) Which figure number consists of exactly 1089 tiles?

Figure 1: 2 × 2 = 4 tilesside = figure number + 1
Figure 1 is 2 × 2, Figure 2 is 3 × 3, Figure 3 is 4 × 4.
Figure 1 is a 2 × 2 square, Figure 2 is 3 × 3, and Figure 3 is 4 × 4.
step 1 of 5

Look at the side of each square, not at its number of tiles. The side grows by one tile from each figure to the next.

All steps
  1. Figure 1 is a 2 × 2 square, Figure 2 is 3 × 3, and Figure 3 is 4 × 4.
  2. Each side is one tile longer than the figure number: Figure n is a square of side n + 1.
  3. (a) Figure 25 has a side of 25 + 1 = 26 tiles, so it has 26 × 26 = 676 tiles.
  4. (b) A square of 1089 tiles has a side that multiplied by itself makes 1089. The number ends in 9, so the side ends in 3 or 7, and 33 × 33 = 1089: the side is 33.
  5. The side is one more than the figure number, so this is Figure 33 − 1 = 32.

answer(a) 676 tiles; (b) Figure 32

techniqueSequences Worth Knowing · Quadratic Sequences

examsPSLE · GCSE Higher

Common pitfalls

  • Squaring the figure number: 25 × 25 = 625. Figure 25's side is 26 tiles, not 25.
  • Answering Figure 33. 33 is the side of the square; the figure number is one less, 32.
06

A Staircase of Dots: Figure 20, and the Figure with 120 Dots

heuristicTwo copies of a staircase fit together into a rectangle one dot longer than it is wide

Dots are arranged in a triangular staircase formation: Figure 1 has 1 dot. Figure 2 has 3 dots (1 + 2). Figure 3 has 6 dots (1 + 2 + 3). Figure 4 has 10 dots (1 + 2 + 3 + 4). (a) How many dots are there in Figure 20? (b) If a figure has 120 dots, what is its figure number?

Figure 1: 1 rowstwo copies make 1 × 2 = 2 dotshalf: 1 dots
Figure 1 with its mirror copy: a 1 × 2 rectangle, 2 dots, so the figure is 1.
Figure 1 and its copy make a rectangle 1 by 2: 2 dots, so Figure 1 has 2 ÷ 2 = 1 dot.
step 1 of 6

Turn a copy of the staircase upside down and fit it against the original. The two make a rectangle one dot longer than it is wide, so each staircase is half of the rectangle.

All steps
  1. Figure 1 and its copy make a rectangle 1 by 2: 2 dots, so Figure 1 has 2 ÷ 2 = 1 dot.
  2. Figure 2 and its copy make a rectangle 2 by 3: 6 dots, so Figure 2 has 6 ÷ 2 = 3 dots.
  3. Figure 3 and its copy make a rectangle 3 by 4: 12 dots, so Figure 3 has 12 ÷ 2 = 6 dots. Figure n and its copy always make a rectangle n by n + 1.
  4. (a) Figure 20 and its copy make 20 × 21 = 420 dots, so Figure 20 has 420 ÷ 2 = 210 dots.
  5. (b) A figure of 120 dots and its copy make a rectangle of 120 × 2 = 240 dots.
  6. The rectangle's sides are two consecutive numbers that multiply to 240: 15 × 16 = 240. The figure number is the shorter side, 15.

answer(a) 210 dots; (b) Figure 15

techniqueSequences Worth Knowing · The Sum of an Arithmetic Series

examsPSLE · GCSE Higher

Common pitfalls

  • Adding the same step every time. The counts go up by 2, then 3, then 4, so a rule such as 1 + 19 × 2 for Figure 20 does not hold.
  • In (b), looking for two consecutive numbers that multiply to 120. The figure is half of the rectangle, so the rectangle's sides multiply to 2 × 120 = 240.
07

A Repeating String of Beads: the 147th Bead, and the Blue Beads in the First 200

heuristicCut the string into repeating blocks; the remainder says where a bead falls in its block

A string of decorative beads is arranged in a repeating pattern: 2 Red, 3 Blue, 1 Yellow, 2 Green, followed by 2 Red, 3 Blue, 1 Yellow, 2 Green, and so on. (a) What is the colour of the 147th bead? (b) How many Blue beads are there in the first 200 beads?

1 block of 8RRBBBYGG
One block: 2 red, 3 blue, 1 yellow, 2 green. Eight beads, three of them blue.
One block is red, red, blue, blue, blue, yellow, green, green: 8 beads, 3 of them blue.
step 1 of 5

Draw one block of 8 beads. Whole blocks are counted by multiplying; only the beads after the last whole block are counted one at a time.

All steps
  1. One block is red, red, blue, blue, blue, yellow, green, green: 8 beads, 3 of them blue.
  2. (a) 147 ÷ 8 = 18 remainder 3: the first 18 × 8 = 144 beads are 18 whole blocks, and 3 beads are left over.
  3. Count those 3 beads into the next block: red, red, blue. The 147th bead is blue.
  4. (b) 200 ÷ 8 = 25 with nothing left over: the first 200 beads are exactly 25 whole blocks.
  5. Each block has 3 blue beads, so there are 25 × 3 = 75 blue beads.

answer(a) Blue; (b) 75 Blue beads

techniqueRemainders · Modular Arithmetic on a Clock Face

examsPSLE

Common pitfalls

  • Reading a remainder of 0 as the first bead of a block. A remainder of 0 means the bead ends a whole block: it is the 8th bead, green.
  • Starting the blue beads at the 2nd place. Write the block out in order; the blue beads are its 3rd, 4th and 5th beads.
08

Two Sequences Taking Turns: the 49th and 50th Numbers

heuristicGroup the numbers in twos; each group is the one before it with 4 added to both numbers

Consider the number sequence: 3, 20, 7, 24, 11, 28, 15, 32, … (a) What is the 49th number in the sequence? (b) What is the 50th number in the sequence?

Group 1320positions 1, 2
Read the sequence in pairs. Group 1 is 3 and 20.
Group 1 holds the 1st and 2nd numbers: 3 and 20.
step 1 of 6

Group the numbers in twos: (3, 20), (7, 24), (11, 28), and so on. Every group is the group before it with 4 added to both numbers.

All steps
  1. Group 1 holds the 1st and 2nd numbers: 3 and 20.
  2. Group 2 holds the 3rd and 4th numbers: 7 and 24, each 4 more than in Group 1.
  3. Group 3 holds the 5th and 6th numbers: 11 and 28, 4 more again.
  4. Group g comes g − 1 steps of 4 after Group 1: its first number is 3 + (g − 1) × 4 and its second is 20 + (g − 1) × 4.
  5. (a) The first 48 numbers fill 24 groups, so the 49th number is the first number of Group 25: 3 + 24 × 4 = 99.
  6. (b) The 50th number is the second number of Group 25: 20 + 24 × 4 = 116.

answer(a) 99; (b) 116

techniqueThe nth Term · Term-to-Term Rules

examsPSLE · O-Level · GCSE Higher · H2

Common pitfalls

  • Following the steps between neighbours, +17 then −13, all the way to the 49th number. It works, but it takes 48 steps; each list goes up by 4 and reaches its 25th number in one.
  • Counting 49 steps of 4. The 49th number is only the 25th number in the odd-place list, 24 steps after its first.
09

A White Square with a Grey Border: Figure 30, and the Figure with 1600 Tiles

heuristicThe border is a strip of n tiles along each side of the white square, and the four corners

Square grids are made of white (unshaded) and grey (shaded) unit square tiles: Figure 1 has a 1 × 1 white tile enclosed by 8 grey border tiles (Total 3 × 3 = 9). Figure 2 has a 2 × 2 white tile core enclosed by 12 grey border tiles (Total 4 × 4 = 16). Figure 3 has a 3 × 3 white tile core enclosed by 16 grey border tiles (Total 5 × 5 = 25). (a) Find the number of grey tiles in Figure 30. (b) Find the number of white tiles in a figure that has a total of 1600 tiles.

Figure 3: 5 × 5 = 25 tileswhite 3 × 3 = 9grey: 4 sides of 3 + 4 corners = 16
The grey border is four sides of n tiles plus the four corners: 4n + 4. Slide n to see the corners stay four.
Each side of the white n × n square has a grey strip of n tiles along it, and one grey tile fills each corner: 4 × n + 4 grey tiles. Figure 1 has 4 + 4 = 8, as given.
step 1 of 5

Split the border into a strip along each side of the white square, n tiles long, and the four corner tiles left over.

All steps
  1. Each side of the white n × n square has a grey strip of n tiles along it, and one grey tile fills each corner: 4 × n + 4 grey tiles. Figure 1 has 4 + 4 = 8, as given.
  2. (a) Figure 30 has 4 × 30 + 4 = 124 grey tiles.
  3. (b) 1600 tiles make a square of side 40, because 40 × 40 = 1600.
  4. The border takes one tile from each end of every row, so the white core is 40 − 2 = 38 tiles wide.
  5. The white core has 38 × 38 = 1444 tiles. Check: the border is 4 × 38 + 4 = 156 tiles, and 1444 + 156 = 1600.

answer(a) 124 grey tiles; (b) 1444 white tiles

techniqueThe nth Term · Area and Perimeter

examsPSLE · O-Level · GCSE Higher · H2

Common pitfalls

  • Counting the border as 4n, four strips of n tiles. That leaves out the four corner tiles, which lie along no side of the white square.
  • Taking 1 from the side of 40 instead of 2. The border runs along both ends of every row, so the core is 40 − 2 = 38 tiles wide, not 39.
10

A 3-by-3 Window on a Calendar, with Dates Adding Up to 162

heuristicThe nine dates balance around the centre, so they add up to nine times the centre date

A 3 × 3 square window is placed over a monthly calendar page covering 9 dates. The sum of all 9 dates inside the window is 162. (a) What is the date in the centre of the window? (b) What are the dates in the four corners of this 3 × 3 window?

10111217?19242526sum = 9 × 18 = 162
Across a row the dates go up by 1, down a column by 7, so the nine dates sit symmetrically around the centre: every date above the centre is matched by one the same amount below.
Across a row the dates go up by 1, and down a column by 7. Every date in the window but the centre has a partner opposite it, as far above the centre date as it is below.
step 1 of 8

Every date above the centre is matched by one the same distance below it, and every date to its left by one to its right. So the centre date is the average of the nine dates.

All steps
  1. Across a row the dates go up by 1, and down a column by 7. Every date in the window but the centre has a partner opposite it, as far above the centre date as it is below.
  2. So the centre date is the average of the nine dates: 162 ÷ 9 = 18.
  3. (a) The centre date is 18. Check: the window holds 10 to 12, 17 to 19 and 24 to 26; the pairs 10 + 26, 11 + 25, 12 + 24 and 17 + 19 each make 36, and 4 × 36 + 18 = 162.
  4. (b) Each corner is one week and one day from the centre: 7 for the week and 1 for the day.
  5. The top-left corner is a week earlier and a day earlier: 18 − 7 − 1 = 10.
  6. The top-right corner is a week earlier and a day later: 18 − 7 + 1 = 12.
  7. The bottom-left corner is a week later and a day earlier: 18 + 7 − 1 = 24.
  8. The bottom-right corner is a week later and a day later: 18 + 7 + 1 = 26.

answer(a) 18; (b) 10, 12, 24 and 26

techniqueThe Mean · The nth Term

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Stepping 7 along a row and 1 down a column. The days of one week run along a row, so the next date to the right is 1 later and the date below is 7 later.
  • Looking for the corners straight from 162, for example 162 ÷ 4. Find the centre first; every corner is a fixed step from it.
11

A Border of Squares and Triangles Counted Three Ways

heuristicTrack What One More Square Adds to Each Count / Read the Perimeter Side by Side

A border is made of grey squares of side 2 cm and white equilateral triangles of side 2 cm. Figure 1 is one square with a triangle on its top edge and a triangle on its bottom edge. Each figure after it adds one square on the right, with a triangle on its top edge and a triangle on its bottom edge, and one more triangle fills each gap between neighbouring triangles: pointing down in the gap along the top, and pointing up in the gap along the bottom. Figure 1 has 1 square, 2 triangles and a perimeter of 12 cm. Figure 2 has 2 squares, 6 triangles and a perimeter of 16 cm. Figure 3 has 3 squares, 10 triangles and a perimeter of 20 cm. (a) Find the number of squares, the number of triangles and the perimeter of Figure 4 and of Figure 5. (b) Write an expression for the number of triangles in Figure n. (c) Which figure has a perimeter of 292 cm? (d) Each triangle has an area of 1.73 cm². Find the total area of the triangles in the figure that has 145 squares.

Figure 1: 1 square, 2 trianglesoutline: 6 sides of 2 cm = 12 cm
Figure 1: four slanting sides and two upright sides, 6 × 2 = 12 cm.
Figure 1: the outline is the four slanting sides of the two triangles and the two upright sides of the square, six sides of 2 cm: 6 × 2 = 12 cm.
step 1 of 6

Look at what one more square brings. It brings its own two triangles and two filling triangles, and it lengthens the outline along the top and along the bottom, while the two ends of the border stay as they were.

All steps
  1. Figure 1: the outline is the four slanting sides of the two triangles and the two upright sides of the square, six sides of 2 cm: 6 × 2 = 12 cm.
  2. Each new square brings 4 triangles: one on its top edge, one on its bottom edge and one filling each gap. The outline gains one 2 cm side along the top and one along the bottom, 4 cm in all, and the two ends stay the same.
  3. (a) Figure 4 has 4 squares, 10 + 4 = 14 triangles and a perimeter of 20 + 4 = 24 cm. Figure 5 has 5 squares, 14 + 4 = 18 triangles and a perimeter of 24 + 4 = 28 cm.
  4. (b) Figure n has Figure 1's 2 triangles and (n − 1) lots of 4 more: 2 + 4(n − 1) = 4n − 2.
  5. (c) The perimeter is 12 + 4(n − 1) = (4n + 8) cm. 4n + 8 = 292, so 4n = 284 and n = 71: Figure 71.
  6. (d) The figure with 145 squares is Figure 145: 4 × 145 − 2 = 578 triangles, and 578 × 1.73 = 999.94 cm².

answer(a) Figure 4: 4 squares, 14 triangles, 24 cm; Figure 5: 5 squares, 18 triangles, 28 cm; (b) 4n − 2; (c) Figure 71; (d) 999.94 cm²

techniqueThe nth Term · Continuing a Shape Pattern · Area and Perimeter

examsPSLE · O-Level · GCSE Higher · H2

Common pitfalls

  • Adding 8 cm for each new square, as if its two new triangles each put two slanting sides on the outline. The filling triangles cover the slanting sides where neighbours meet, so only their flat 2 cm sides join the outline: 4 cm more, not 8.
  • Writing the number of triangles as 4n. The list goes up by 4 but starts at 2, not 4, so each figure has 2 fewer than 4n: 4n − 2.