Pattern & Logical Reasoning
Module 04 of 06 · Full Solution Edition
Arithmetic Sequences
Core Concept: The n-th term formula
An arithmetic sequence is a list of numbers where each consecutive pair differs by the same constant d (the common difference):
The n-th term is given by:
Example. For 3, 8, 13, 18, 23, … with a = 3, d = 5: the 46-th term is a₄₆ = 3 + 45 · 5 = 228.
Two common questions:
-
Find a specific term: plug n into
aₙ = a + (n−1)d. -
Find the position of a value: solve
aₙ = valuefor n.
In a sequence of numbers, the first term is 3. Each new term is obtained by adding 5 to the previous term. The first four terms are 3, 8, 13, 18. What are the next three terms in the sequence?
Strategy
Keep adding the common difference d = 5 to the most recent term.
Steps
- 5th term:
18 + 5 = 23 - 6th term:
23 + 5 = 28 - 7th term:
28 + 5 = 33
Answer
C — 23, 28, 33
In an increasing list of consecutive integers, the 3rd and 4th numbers in the list add to 11. What is the 6th number in the list?
Strategy
Consecutive integers differ by 1. If the 3rd term is x, the 4th is x + 1; use the sum to find x, then count up.
Steps
- Let 3rd term =
x. Then 4th term =x + 1. -
x + (x + 1) = 11 ⇒ 2x = 10 ⇒ x = 5. So 3rd = 5, 4th = 6. - 5th = 7, 6th = 8.
Answer
D — 8
The positive integers are arranged in rows: Row 1 has just {1}, Row 2 has {2, 3}, Row 3 has {4, 5, 6}, Row 4 has {7, 8, 9, 10}, … Each new row contains one more integer than the previous. How many integers less than 2000 are in the column that contains the number 2000?
Strategy
Three steps: (1) find which row contains 2000, (2) find which column that row places 2000 in, (3) count how many earlier rows have an entry in that same column.
Key formula
Row k starts with the number 1 + (1 + 2 + … + (k−1)) = 1 + k(k−1)/2 and contains k integers.
Steps
-
Find row of 2000. Row k ends at
k(k+1)/2. For k = 62: ends at 1953. For k = 63: ends at 2016. So 2000 lives in Row 63, which starts at1 + 63·62/2 = 1954. -
Find column. Column index =
2000 − 1954 + 1 = 47. -
Earlier rows with column 47. Row k has column 47 only if
k ≥ 47. So eligible rows: 47, 48, …, 62 — that’s62 − 47 + 1 = 16rows. - (Row 63’s entry in column 47 IS 2000 itself, so it doesn’t count toward “less than 2000”.)
Pitfall
It’s tempting to say “47 rows have column 47” and then subtract — but the question asks for entries strictly less than 2000, so the row that contains 2000 must be excluded.
Answer
D — 16
The 26 letters of the alphabet are written in order, clockwise around a circle. The ciphertext of a message is created by replacing each letter of the message by the letter that is 4 letters clockwise from the original letter (a Caesar cipher). For example, ZAP has ciphertext DET. What is the ciphertext of the message WIN?
Strategy
Shift each letter forward by 4. When you pass Z, wrap around to A (modular arithmetic mod 26).
Steps
-
W → X → Y → Z → A(4 steps) → A -
I → J → K → L → M→ M -
N → O → P → Q → R→ R
Ciphertext: AMR.
Lesson
Anything cyclic on 26 letters obeys the same modular rule: position of a letter = ((original − 1 + shift) mod 26) + 1, with A = 1.
Answer
C — AMR
Cyclic Patterns
Core Concept: Use the remainder, not the quotient
A cyclic (periodic) pattern repeats every L terms. To find the n-th term:
- If
r ≠ 0→ the n-th term equals the r-th term of the cycle. - If
r = 0→ the n-th term equals the L-th (last) term of the cycle.
To count how often an item appears in the first N terms: let N = qL + r (quotient q, remainder r). Each full cycle contributes a fixed count; then add the appearances among the first r leftover terms.
The sequence of five symbols ○ ◀ ⊗ △ ★ repeats to form the pattern: ○ ◀ ⊗ △ ★ ○ ◀ ⊗ △ ★ … If the pattern is continued, the 23rd symbol in the pattern is:
Steps
- Cycle length
L = 5. Compute23 mod 5 = 3. - So the 23rd symbol = the 3rd symbol in the cycle = ⊗.
Answer
C — ⊗
A list of five numbers repeats to form the pattern 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, … What is the 221st number in the pattern?
Steps
- Cycle length
L = 5.221 = 44 × 5 + 1, so221 mod 5 = 1. - The 221st number = the 1st number in the cycle = 5.
Answer
A — 5
The numbers 2, 0, 2, 5 are repeated to form the pattern 2, 0, 2, 5, 2, 0, 2, 5, … If a total of 50 numbers are written, how many times will the number 5 appear?
Strategy
Count appearances per cycle, multiply by complete cycles, then add appearances in the leftover.
Steps
- Cycle = (2, 0, 2, 5), length
L = 4. The 5 sits at position 4 of the cycle (exactly one 5 per cycle). -
50 ÷ 4 = 12remainder2. So 12 complete cycles → 12 fives. - Leftover 2 numbers = positions 1, 2 of the cycle = 2, 0. No more 5s.
- Total: 12 fives.
Pitfall
Don’t double-count when the leftover happens to include the position of the target symbol — check by tracing the cycle positions, not by guessing.
Answer
C — 12
Complex Cyclic Problems
Core Concept: When the “cycle” isn’t constant
Some patterns aren’t simple period-L cycles. Two common upgrades:
-
Variable-length blocks. If the n-th block has
f(n)symbols, find the cumulative lengthS(k) = f(1) + f(2) + … + f(k), locate the block containing the target position, then drill into that block. - Recurrence with eventual period. A sequence defined by a rule (each new term depends on previous ones) often “settles” into a repeating tail. Generate terms until the pattern repeats, then exploit it.
The digits 1 to 9 are written in order, with digit n written n times. This forms the block of digits 1 22 333 4444 … 999999999. The block is written 100 times. What is the 1953rd digit written?
Strategy
Find the block length first, then reduce 1953 mod (block length) to locate the position within a single block.
Steps
- Length of one block:
1 + 2 + 3 + … + 9 = 45digits. -
1953 ÷ 45 = 43remainder18. So the 1953rd digit is the 18th digit of the 44th block. - Cumulative digit positions within a block:
- “1” → position 1
- “22” → positions 2–3
- “333” → positions 4–6
- “4444” → positions 7–10
- “55555” → positions 11–15
- “666666” → positions 16–21
- Position 18 falls in the “6” group → digit is 6.
Pitfall
Don’t confuse “the 44th block” with “the 43rd block + remainder zero”. When remainder = 0, you’d be at the END of the 43rd block (position 45 of that block). Here remainder = 18 ≠ 0, so we ARE inside the 44th block.
Answer
C — 6
A Gareth sequence is a sequence of numbers in which each number after the second is the non-negative difference between the two previous numbers. For example, starting 15, 12, the next terms are 15 − 12 = 3, 12 − 3 = 9, 9 − 3 = 6, giving 15, 12, 3, 9, 6, … If a Gareth sequence begins 10, 8, what is the sum of the first 30 numbers?
Strategy
Generate terms until you spot a repeating tail, then count cycles inside the first 30 terms.
Steps
- Generate:
10, 8, |10−8|=2, |8−2|=6, |2−6|=4, |6−4|=2, |4−2|=2, |2−2|=0, |2−0|=2, |0−2|=2, |2−2|=0, … - From term 6 onward the sequence is
2, 2, 0, 2, 2, 0, 2, 2, 0, …— a period-3 cycle summing to4per cycle. - Sum of first 5 terms (the “preamble” before the cycle settles):
10 + 8 + 2 + 6 + 4 = 30. - Remaining terms: indices 6 to 30 = 25 terms of pattern (2, 2, 0). That’s
8full cycles (24 terms) + 1 extra term. - Cycles contribute
8 × 4 = 32. The extra term is the 1st of the cycle =2. - Total:
30 + 32 + 2 = 64.
Lesson
For any recurrence on bounded integers, the sequence MUST eventually repeat (pigeonhole on consecutive pairs). The trick is generating enough terms to see the repeat — usually 10–20 is plenty.
Answer
E — 64
Calendar Problems
Core Concept: The calendar is just modular arithmetic mod 7
- Days of the week repeat every 7 days. To advance D days from a known weekday, compute
D mod 7and shift. - From day
xof a month to dayy: there arey − xdays between them (NOTy − x + 1). - Month lengths: Jan 31 · Feb 28/29 · Mar 31 · Apr 30 · May 31 · Jun 30 · Jul 31 · Aug 31 · Sep 30 · Oct 31 · Nov 30 · Dec 31.
- The k-th occurrence of a given weekday in a month falls on day
(first occurrence) + 7(k − 1).
Dalia’s birthday is on a Wednesday and Bruce’s birthday is 60 days after Dalia’s. On what day of the week is Bruce’s birthday?
Steps
-
60 mod 7 = 4(since60 = 8 × 7 + 4). - Wednesday + 4 days → Thursday → Friday → Saturday → Sunday.
Answer
E — Sunday
The month of June has 30 days. If in a certain year June 1 is on a Tuesday, on which day of the week is June 30?
Pitfall
The gap from June 1 to June 30 is 29 days, not 30. (Day 1 IS the starting day; you only advance 29 to reach day 30.)
Steps
- Days to advance:
30 − 1 = 29. -
29 mod 7 = 1. - Tuesday + 1 day = Wednesday.
Answer
C — Wednesday
A public holiday is always celebrated on the third Wednesday of a certain month. In that month, the holiday cannot occur on which of the following days?
Strategy
Find the range of dates the 3rd Wednesday can land on. Anything outside that range is impossible.
Steps
- The 1st Wednesday of a month falls on some date between day 1 and day 7 inclusive.
- The 3rd Wednesday = (1st Wednesday) + 14 days. So it falls between day 15 and day 21 inclusive.
- Check the options: 16 ✓, 18 ✓, 19 ✓, 21 ✓, but 22 is outside the range.
Answer
B — 22nd
Logical Reasoning
Core Concept: The arrow method for ordering problems
When a problem gives several “more / fewer / faster / slower” comparisons, draw an arrow from “fewer / slower” on the left to “more / faster” on the right. Place each person on the arrow as constraints accumulate.
Worked example. Suppose: (1) Frank eats more than Julia, (2) Matt eats fewer than Julia, (3) Chloe eats more than Frank.
- From (1): place Julia, then Frank to its right.
- From (2): place Matt to Julia’s left.
- From (3): place Chloe to Frank’s right.
Final order: Matt < Julia < Frank < Chloe.
For seating / arrangement problems: use case analysis. Fix one element, list the cases for each remaining constraint, eliminate contradictions.
Five students ran a race. Ryan was faster than Henry and Faiz. Henry was slower than Faiz. Toma was faster than Ryan but slower than Omar. Which student finished fourth?
Strategy
Combine all clues into a single order, slowest to fastest.
Steps
- Ryan > Henry, Ryan > Faiz, Faiz > Henry → so far:
Henry < Faiz < Ryan. - Omar > Toma > Ryan → combined:
Henry < Faiz < Ryan < Toma < Omar. - From fastest to slowest: Omar (1st), Toma (2nd), Ryan (3rd), Faiz (4th), Henry (5th).
Answer
A — Faiz
Kathy owns more cats than Alice and more dogs than Bruce. Alice owns more dogs than Kathy and fewer cats than Bruce. Which of the statements MUST be true?
Strategy
Treat cats and dogs separately. Build the arrow for each.
Steps
- Cats: Kathy > Alice, and Bruce > Alice. So Alice has the fewest cats. But we don’t know whether Kathy > Bruce or Bruce > Kathy. → A, B, C are NOT guaranteed.
-
Dogs: Kathy > Bruce, and Alice > Kathy. Combined:
Bruce < Kathy < Alice. So Alice has the MOST dogs and Bruce has the fewest. → D is true; E is false.
Pitfall
Problems with two parallel quantities (cats AND dogs) trick you into mixing them. Keep them on separate arrows; the answer comes from whichever arrow gives a complete order.
Answer
D — Alice owns the most dogs
Jack, Kelly, Lan, Mihai, and Nate are sitting in the 5 chairs around a circular table. Lan and Mihai are sitting beside each other. Jack and Kelly are NOT sitting beside each other. The 2 people who are seated on either side of Nate are:
Strategy
Treat Lan + Mihai as a fixed “block” of two adjacent seats. Three seats remain for Jack, Kelly, Nate. Then test which arrangement satisfies “Jack and Kelly not adjacent”.
Steps
- Label the 5 seats around the circle 1–2–3–4–5–1. Put the Lan-Mihai block in seats 1–2. The remaining seats 3, 4, 5 form an arc: 3 is adjacent to 2 (Mihai), 5 is adjacent to 1 (Lan), and 3-4-5 are adjacent in sequence.
- Place Jack, Kelly, Nate in seats 3, 4, 5. The adjacencies among them are:
3↔4and4↔5. (Seats 3 and 5 are NOT adjacent — seat 4 sits between them.) - Constraint: Jack and Kelly are NOT beside each other. So Jack and Kelly cannot occupy any adjacent pair → they must take seats 3 and 5 (the two non-adjacent seats). Nate goes in seat 4 (the middle).
- Nate’s neighbours are seats 3 and 5 = Jack and Kelly.
Lesson
For circular seating, glue adjacent constraints into a block, then count cases for the rest. The “non-adjacent” pair always sits across the gap from each other.
Answer
B — Jack and Kelly
Homework
108, 97, 86, 75, 64, … Which number comes next in the sequence?
Steps
Common difference: 97 − 108 = −11. Verify: 86 − 97 = −11, 75 − 86 = −11, 64 − 75 = −11. Next term: 64 − 11 = 53.
Answer
A — 53
What is the next number in this sequence? 1, 2, 3, 6, 11, 20, 37, ___ ?
Strategy
Test whether each term is the sum of the previous few.
Steps
- Check “sum of previous three”:
1+2+3 = 6 ✓,2+3+6 = 11 ✓,3+6+11 = 20 ✓,6+11+20 = 37 ✓. - Next term:
11 + 20 + 37 = 68.
Answer
D — 68
What is the next number in the sequence 3, 5, 9, 17, 33, …?
Steps
- Differences:
5−3 = 2,9−5 = 4,17−9 = 8,33−17 = 16. They double each time (powers of 2). - Next difference:
32. Next term:33 + 32 = 65.
Lesson
Equivalently, aₙ = 2·aₙ₋₁ − 1: 2·3 − 1 = 5, 2·5 − 1 = 9, 2·9 − 1 = 17, 2·17 − 1 = 33, 2·33 − 1 = 65. Same answer, different lens.
Answer
D — 65
Which of the following is the sum of three consecutive integers?
Concept
Three consecutive integers n − 1, n, n + 1 sum to 3n — always a multiple of 3.
Steps
Test each option for divisibility by 3: 17 ✗, 11 ✗, 25 ✗, 21 = 3 × 7 ✓ (so 21 = 6 + 7 + 8), 8 ✗.
Answer
D — 21
The integers 1 to 32 are spaced evenly and in order around the outside of a circle. Straight lines that pass through the centre of the circle join these numbers in pairs. Which number is paired with 12?
Strategy
A line through the centre splits the circle into two halves. With 32 numbers placed evenly, each number is paired with the one directly opposite — exactly 16 positions away.
Steps
Partner of 12: 12 + 16 = 28.
Pitfall
If the partner number went past 32 we’d wrap around: e.g. partner of 20 would be 20 + 16 = 36 ≡ 4 (mod 32).
Answer
A — 28
The pattern below is formed by groups of pictures. Group k contains k smiley faces (☺) followed by ♡, ✿, △: ☺ ♡ ✿ △ ☺☺ ♡ ✿ △ ☺☺☺ ♡ ✿ △ … What is the 80th picture in the pattern?
Strategy
Length of group k is k + 3. Find the cumulative count S(k) until it first reaches 80, locate which group, then drill in.
Steps
-
S(k) = (1+3) + (2+3) + … + (k+3) = k(k+1)/2 + 3k. - Cumulative:
S(9) = 72,S(10) = 85. So position 80 is in group 10 (positions 73–85). - Within group 10: position 73 is the 1st smile, 74 the 2nd, …, 82 the 10th smile. Then 83 = ♡, 84 = ✿, 85 = △.
- Position 80 is the 8th symbol of group 10 → still in the smiley block. ☺.
Answer
A — ☺
Some beads are arranged to form a necklace repeating the pattern: 4 purple, 3 red, 2 yellow, 4 purple, 3 red, 2 yellow, 4 purple, 3 red, 2 yellow, … If there are 100 beads in total, how many purple beads are there altogether?
Steps
- Cycle length:
4 + 3 + 2 = 9beads. Each cycle has 4 purples. -
100 ÷ 9 = 11remainder1. So 11 complete cycles contribute11 × 4 = 44purples. - Remainder 1: the first bead of the next cycle is purple (the cycle starts with 4 purples). So +1.
- Total purples:
44 + 1 = 45.
Answer
D — 45
In a certain month, there are 5 Tuesdays. The month does NOT start or end on a Tuesday. Which day of the week will the 5th day of the month fall on?
Strategy
List all possible “first Tuesday” dates, eliminate ones that break the constraints, then read off day 5.
Steps
- Let day
tbe the first Tuesday, with1 ≤ t ≤ 7. The 5 Tuesdays are:t, t+7, t+14, t+21, t+28. - For 5 Tuesdays to fit:
t + 28 ≤ 31→t ≤ 3. Sot ∈ {1, 2, 3}. - Month doesn’t START on Tuesday →
t ≠ 1. Sot ∈ {2, 3}. - Case
t = 3: 5th Tuesday is day 31, requiring a 31-day month — but then the month ENDS on Tuesday. Excluded. - Case
t = 2: 5 Tuesdays at days 2, 9, 16, 23, 30. In a 31-day month, day 31 = Wednesday (not Tuesday) ✓; in a 30-day month, day 30 = Tuesday (excluded). So month is 31 days, day 2 = Tuesday. - Day 2 = Tuesday → Day 5 = Tuesday + 3 = Friday.
Answer
D — Friday
When expressed as a repeating decimal, the fraction 1/7 is written as 0.142857 142857 … (the 6 digits 142857 repeat). The digit in the 3rd position to the right of the decimal point is 2. In which one of the following positions to the right of the decimal point will there also be a 2?
Strategy
The digit at position p of a period-6 decimal is determined by p mod 6. Find which positions have remainder 3 (since position 3 is the “2”).
Steps
- The 6 digits at positions 1, 2, 3, 4, 5, 6 are 1, 4, 2, 8, 5, 7. So digit = 2 happens when
position mod 6 = 3. - Check each option’s remainder mod 6:
- 119 mod 6 = 5 → digit 5
- 121 mod 6 = 1 → digit 1
- 123 mod 6 = 3 → digit 2 ✓
- 125 mod 6 = 5 → digit 5
- 126 mod 6 = 0 → digit 7 (position 6 of the cycle)
Answer
C — 123rd
A train consists of four carriages in four colours: red, green, white, and yellow. The green carriage is neither first nor last. The yellow carriage is not adjacent to the white or red carriages. The first carriage is white. What is the order of the carriages?
Strategy
Use the most restrictive constraints first. The yellow restriction is strongest: yellow can only sit next to green.
Steps
- Position 1 = white (given). Yellow’s only allowed neighbour is green.
- Yellow in position 2? Its neighbour would include position 1 = white → forbidden.
- Yellow in position 3? Its neighbours are positions 2 and 4 — both must be green. Impossible (only one green carriage).
- Yellow in position 4? Its only neighbour is position 3, which must be green. ✓ (Green at position 3 also satisfies “green not first or last”.)
- That leaves red for position 2. Final order: white, red, green, yellow.
Lesson
In constraint puzzles, attack the most restrictive variable first. “Yellow only adjacent to green” rules out 3 of 4 positions for yellow before you touch any other constraint.
Answer
E — white, red, green, yellow
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