Patterns and page numbers · applications

Applications: Patterns and Page Numbers

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

PSLE · GCSE Higher

01

Page Number Digit Counting (Pages to Digits)

heuristicTiered Place-Value Grouping

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.
Draw Tier 1: Block of numbers [1 to 9] containing 9 pages. Each page has 1 digit: 9 × 1 = 9.
step 1 of 4

Construct a 3-tier segmented bar representing the place-value blocks and calculate the digit contribution for each tier.

  1. Draw Tier 1: Block of numbers [1 to 9] containing 9 pages. Each page has 1 digit: 9 × 1 = 9.
  2. Draw Tier 2: Block of numbers [10 to 99] containing 90 pages. Each page has 2 digits: 90 × 2 = 180.
  3. Draw Tier 3: Block of numbers [100 to 165] containing 165 − 99 = 66 pages. Each page has 3 digits: 66 × 3 = 198.
  4. Combine all three tier blocks: 9 + 180 + 198 = 387 digits.

answer387 digits

techniqueNumbers to Ten Million · Multiplying in Columns

examsPSLE · GCSE Higher

Common pitfalls

  • Calculating the count of 2-digit pages as 99 − 10 = 89 instead of 99 − 10 + 1 = 90 (fence-post error).
  • Multiplying 165 directly by 3 (165 × 3 = 495), assuming every page has 3 digits.
  • Counting pages 100 to 165 as 165 − 100 = 65 pages instead of 66.
02

Reverse Digit Counting (Digits to Pages)

heuristicDeductive Range Allocation / Working Backwards

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 digits
Start with all 492 digits and spend them tier by tier.
Start with total digit supply bar: 492.
step 1 of 5

Deplete a total bar of 492 digits step-by-step by carving out full lower-tier blocks until reaching the target page count.

  1. Start with total digit supply bar: 492.
  2. Carve out Tier 1 (pages 1 to 9): Remove 9 digits ⟹ 492 − 9 = 483 left.
  3. Carve out Tier 2 (pages 10 to 99): Remove 180 digits ⟹ 483 − 180 = 303 left.
  4. The remaining 303 digits belong entirely to 3-digit page blocks: 303 ÷ 3 = 101 blocks.
  5. Add the 101 pages to the 99 completed pages: 99 + 101 = 200 pages.

answer200 pages

techniqueNumbers to Ten Million · Multiplying and Dividing Undo Each Other

examsPSLE · GCSE Higher

Common pitfalls

  • Dividing 492 directly by 3 (492 ÷ 3 = 164), assuming all pages are 3-digit numbers.
  • Adding 101 to 100 instead of 99, getting 201 pages.
  • Subtracting 90 digits instead of 180 digits for the 2-digit range.
03

Torn-Out Sheet / Missing Page Numbers

heuristicConsecutive Parity Pairing and Triangular Sum Invariance

A student accidentally tears out 1 single sheet of paper from a magazine that originally contained 40 consecutive numbered pages (pages 1 to 40). The sum of all the page numbers on the remaining pages is 749. What are the two page numbers printed 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.
Sum of 40 pages using pairing: 20 pairs of (1 + 40 = 41) ⟹ 20 × 41 = 820.
step 1 of 5

Pair the pages from both ends to calculate the complete sum, find the missing gap, and split the gap using a two-unit comparison model with a difference of 1.

  1. Sum of 40 pages using pairing: 20 pairs of (1 + 40 = 41) ⟹ 20 × 41 = 820.
  2. Missing page sum = 820 − 749 = 71.
  3. Draw two comparison bars for the two sides of the single sheet: Front page [u], Back page [u][1].
  4. Value of 2u = 71 − 1 = 70 ⟹ u = 35.
  5. Front page = 35, Back page = 35 + 1 = 36.

answerPages 35 and 36

techniqueThe Sum of an Arithmetic Series · Odd and Even Numbers

examsH2

Common pitfalls

  • Assuming a torn sheet can carry an even page on the front and an odd page on the back (e.g., pages 34 and 35), violating standard book pagination where recto is odd and verso is even.
  • Using 40 × 40 = 1600 instead of the triangular summation formula n(n+1)2.
  • Dividing 71 by 2 and rounding arbitrarily without checking consecutive integers.
04

Linear Matchstick / Border Tile Pattern

heuristicCommon Difference Generalization ($T_n = dn + c$)

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.
Deconstruct Figure 1: 1 initial vertical stick + 1 group of 3 sticks = 1 + 1(3) = 4.
step 1 of 6

Isolate the first base stick (fixed anchor) and observe that every additional square adds a 'C-shaped' group of 3 matchsticks.

  1. Deconstruct Figure 1: 1 initial vertical stick + 1 group of 3 sticks = 1 + 1(3) = 4.
  2. Deconstruct Figure 2: 1 initial vertical stick + 2 groups of 3 sticks = 1 + 2(3) = 7.
  3. Deconstruct Figure 3: 1 initial vertical stick + 3 groups of 3 sticks = 1 + 3(3) = 10.
  4. (a) Figure 45 has 1 initial stick + 45 groups of 3 sticks: 1 + (45 × 3) = 1 + 135 = 136 sticks.
  5. (b) For 241 sticks, remove the 1 initial stick: 241 − 1 = 240.
  6. Divide by group size of 3: 240 ÷ 3 = 80 groups ⟹ Figure 80.

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

techniqueThe nth Term · Term-to-Term Rules

examsPSLE · O-Level · GCSE Higher · H2

Common pitfalls

  • Multiplying figure number directly by 4 (45 × 4 = 180), ignoring shared internal matchsticks.
  • Dividing 241 by 3 directly without subtracting the starting anchor stick of 1.
05

Quadratic / Square Array Pattern

heuristicSquare Number Mapping ($n^2$)

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.
Notice the grid dimensions: Fig 1 is 2 × 2, Fig 2 is 3 × 3, Fig 3 is 4 × 4.
step 1 of 5

Observe the dimensional expansion of the square grid by tracking row and column counts relative to the figure number.

  1. Notice the grid dimensions: Fig 1 is 2 × 2, Fig 2 is 3 × 3, Fig 3 is 4 × 4.
  2. Pattern rule: The side length of the square is always (Figure Number + 1).
  3. (a) Figure 25 side length = 25 + 1 = 26. Total tiles = 26 × 26 = 676.
  4. (b) To find the figure with 1089 tiles, find its side length: √1089 = 33.
  5. Deduct 1 from the side length: 33 − 1 = 32 ⟹ Figure 32.

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

techniqueSequences Worth Knowing · Quadratic Sequences

examsPSLE · GCSE Higher

Common pitfalls

  • Squaring the figure number directly: 252 = 625 instead of (25+1)2 = 676.
  • Confusing side length with figure number (stating Figure 33 instead of Figure 32).
06

Triangular Number Pattern (Staircase / Handshakes)

heuristicTriangular Summation Form $\left[\frac{n(n+1)}{2}\right]$

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.
Join two identical Figure 1 triangles: forms a 1 × 2 rectangle of 2 dots ⟹ 2 ÷ 2 = 1.
step 1 of 6

Double the triangular formation to form a rectangular grid of dimensions n × (n + 1), then take half the area.

  1. Join two identical Figure 1 triangles: forms a 1 × 2 rectangle of 2 dots ⟹ 2 ÷ 2 = 1.
  2. Join two identical Figure 2 triangles: forms a 2 × 3 rectangle of 6 dots ⟹ 6 ÷ 2 = 3.
  3. Join two identical Figure 3 triangles: forms a 3 × 4 rectangle of 12 dots ⟹ 12 ÷ 2 = 6.
  4. (a) For Figure 20: Double the dots to form a rectangle of 20 × 21 = 420. Divide by 2: 420 ÷ 2 = 210 dots.
  5. (b) For 120 dots: Double the dots to find the rectangle: 120 × 2 = 240.
  6. Find consecutive numbers that multiply to 240: 15 × 16 = 240. Figure number is the smaller number: 15.

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

techniqueSequences Worth Knowing · The Sum of an Arithmetic Series

examsPSLE · GCSE Higher

Common pitfalls

  • Treating the sequence as linear and attempting to find a constant difference (3 − 1 = 2, 6 − 3 = 3).
  • Forgetting to multiply by 2 before factoring for part (b) (e.g., trying to factor n(n+1) = 120 to get n ≈ 10.4).
07

Periodic / Cyclic Remainder Pattern

heuristicModulo Arithmetic / Cycle Size Division

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 color 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.
Draw 1 Unit Block = [R][R][B][B][B][Y][G][G] (8 beads total, containing 3 Blue).
step 1 of 5

Represent the sequence as standardized blocks of 8 beads and evaluate full blocks plus the partial residual block.

  1. Draw 1 Unit Block = [R][R][B][B][B][Y][G][G] (8 beads total, containing 3 Blue).
  2. (a) Find number of blocks in 147 beads: 147 ÷ 8 = 18 blocks with a remainder of 3 beads.
  3. Trace first 3 beads of the next block: [R], [R], [B] ⟹ 3rd bead is Blue.
  4. (b) Number of complete blocks in 200 beads: 200 ÷ 8 = 25 complete blocks.
  5. Each block contains 3 Blue beads: 25 × 3 = 75 Blue beads.

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

techniqueRemainders · Modular Arithmetic on a Clock Face

examsPSLE

Common pitfalls

  • Misinterpreting a remainder of 0 as the 0th bead instead of the very last bead of the cycle (Green).
  • Miscounting the cumulative positions within the cycle (e.g., assuming Blue begins at position 2 instead of position 3).
08

Interleaved / Alternating Two-Tier Sequence

heuristicParity Split (Odd/Even Index Branching)

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: (1st = 3, 2nd = 20).
step 1 of 6

Pair adjacent numbers (3, 20), (7, 24), (11, 28) … into 2-number compound groups to track rank easily.

  1. Group 1: (1st = 3, 2nd = 20).
  2. Group 2: (3rd = 7, 4th = 24).
  3. Group 3: (5th = 11, 6th = 28).
  4. Notice that in Group g, 1st number is 3 + (g − 1) × 4 and 2nd number is 20 + (g − 1) × 4.
  5. (a) 49th number is the 1st number of Group 25 (49 = 24 × 2 + 1): 3 + (24 × 4) = 3 + 96 = 99.
  6. (b) 50th number is the 2nd number of Group 25 (50 = 25 × 2): 20 + (24 × 4) = 20 + 96 = 116.

answer(a) 99; (b) 116

techniqueThe nth Term · Term-to-Term Rules

examsPSLE · O-Level · GCSE Higher · H2

Common pitfalls

  • Trying to find a single repeating rule between consecutive terms (+17, −13, +17, −13) and applying it 49 times.
  • Using 49 as the multiplier rather than half of 49 rounded up (25 terms).
09

Shaded vs Unshaded Grid / Compound Border Pattern

heuristicArea Subtraction / Nested Component Formulas

Square grids are made of white (unshaded) and gray (shaded) unit square tiles: Figure 1 has a 1 × 1 white tile enclosed by 8 gray border tiles (Total 3 × 3 = 9). Figure 2 has a 2 × 2 white tile core enclosed by 12 gray border tiles (Total 4 × 4 = 16). Figure 3 has a 3 × 3 white tile core enclosed by 16 gray border tiles (Total 5 × 5 = 25). (a) Find the number of gray 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 = 9gray: 4 sides of 3 + 4 corners = 16
The gray border is four sides of n tiles plus the four corners: 4n + 4. Slide n to see the corners stay four.
Deconstruct Gray Border: 4 sides of length n plus 4 fixed corners = 4 × n + 4.
step 1 of 5

Segment the outer perimeter border into 4 side strips of length n plus 4 single corner tiles.

  1. Deconstruct Gray Border: 4 sides of length n plus 4 fixed corners = 4 × n + 4.
  2. (a) For Figure 30: Gray tiles = (4 × 30) + 4 = 120 + 4 = 124 tiles.
  3. (b) When total tiles = 1600: Side of full square = √1600 = 40.
  4. White inner square side length = 40 − 2 (subtract borders) = 38.
  5. White tiles = 38 × 38 = 1444 tiles.

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

techniqueThe nth Term · Area and Perimeter

examsPSLE · O-Level · GCSE Higher · H2

Common pitfalls

  • Confusing total grid side (n + 2) with inner white square side (n), calculating 402 or (40−1)2.
  • Forgetting to add the 4 corner tiles when calculating the border: writing 4n instead of 4n + 4.
10

Consecutive Integer Window / Calendar Matrix

heuristicCentral Variable Balancing / Invariant Offset Summation

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 center 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 center: every date above the center is matched by one the same amount below.
Because calendar numbers have uniform horizontal differences (+1) and vertical differences (+7), the 9 numbers are balanced around the center box.
step 1 of 8

Apply average balancing: the average of any symmetrically distributed arithmetic grid is equal to its central number.

  1. Because calendar numbers have uniform horizontal differences (+1) and vertical differences (+7), the 9 numbers are balanced around the center box.
  2. Average of the 9 numbers = 162 ÷ 9 = 18.
  3. (a) Center box = 18.
  4. (b) Find corner dates from center date 18:
  5. Go up 1 week (−7) and left 1 day (−1): 18 − 8 = 10 (Top-left).
  6. Go up 1 week (−7) and right 1 day (+1): 18 − 6 = 12 (Top-right).
  7. Go down 1 week (+7) and left 1 day (−1): 18 + 6 = 24 (Bottom-left).
  8. Go down 1 week (+7) and right 1 day (+1): 18 + 8 = 26 (Bottom-right).

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

techniqueThe Mean · The nth Term

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Assuming a calendar row steps by 7 instead of 1, confusing rows with columns.
  • Dividing 162 by 4 to try to find the corners directly without finding the central anchor date.
Mr. Chalk Drill number sequences