Unit 3

Repeating Pattern Sequence

About this unit

Identify the repeating cycle or growth rule in number, shape, and symbol sequences — then use it to find any term, no matter how far along the sequence. This unit also covers growing patterns where you need to find a formula for the nth figure.

What types of questions will you face?

1Find the element at a large position (e.g. 999th, 1004th) in a repeating symbol or number cycle
2Calculate how many elements (cubes, circles, matchsticks) will be in figure N of a growing pattern
3Identify the rule operating in an arithmetic or geometric number sequence
4Work out the position of a specific element within a pattern diagram
5Decode rule-based sequences where the rule toggles or transforms a state

Skills you will build

Finding the cycle length of a repeating sequence
Using division with remainder (modular arithmetic) to locate the nth term
Identifying and extending linear and quadratic growing patterns
Recognising Fibonacci, triangular, and other classic sequences
Translating a visual pattern into a numerical formula

By the end of this unit, you will be able to

Find the element at any position in a repeating sequence in seconds using remainders
Generate a formula for how many elements will be in the nth figure of a growing pattern
Recognise common sequence types and apply the right rule immediately
Decode complex toggle-based or rule-based state sequences

Difficulty profile

Ranges from Very Easy (simple repeating cycles) to Difficult (complex growing patterns with combined rules). The Fibonacci sequence and toggle-based puzzles are the hardest questions in this unit.

Exam tip: Repeating Pattern Sequence

For repeating sequences: count the cycle length (e.g. 8 symbols), divide the position by that length, and the remainder tells you which symbol it is. If remainder = 0, it's the last in the cycle.

Sample Questions

Lesson 1 of 2Repeating Pattern SequenceIntroductory

Welcome to one of the most satisfying question types in OC Thinking Skills — repeating sequences. Once you spot the cycle, finding any term in the sequence takes just a few seconds of arithmetic.

This exact format is a consistent presence on OC tests, and the technique is always the same whether the sequence is made of numbers, letters, or shapes. Many students gain a quick mark here by applying the method correctly.

The examiner wants to confirm that you understand a repeating sequence wraps around and that you can use division with remainder to locate any term without laboriously counting every single element up to that position.

A sequence that clearly repeats (e.g. 5, 6, 7, 8, 9, 5, 6, 7, 8, 9…) is shown, and you are asked to name the element at a large position — typically in the hundreds or thousands — where counting one by one would take too long.

Best approach: Count how many elements are in one full cycle. Divide the target position by the cycle length. The remainder tells you which position in the cycle your answer is at. Special case: if the remainder is 0, the answer is the LAST element of the cycle (not the first).

Question

Study this pattern. What is the 51st number? 5 6 7 8 9 5 6 7 8 9 5 6 \dots

A $5$
B $6$
C $7$
D $9$

Decided on your answer? Check how you went below.

Correct Answer

5

Step-by-Step Explanation

Step 1: Identify the repeating cycle. The sequence is: 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 5, 6, \dots The same group 5, 6, 7, 8, 9 repeats over and over. The cycle has 5 numbers . Step 2: Divide 51 by the cycle length. 51 \div 5 = 10 remainder 1 (Check: 10 \times 5 = 50, and 50 + 1 = 51 ✓) Step 3: Use the remainder to find the position in the cycle. A remainder of 1 means the 51st number sits at the 1st position of the cycle. Position in cycle: 1 2 3 4 5 Number: 5 6 7 8 9 The 1st number in the cycle is 5 . The 51st number in the pattern is 5.

Lesson 2 of 2Repeating Pattern SequenceIntermediate

Growing pattern questions take sequences to an entirely different level. Instead of repeating, each figure gets larger according to a mathematical rule — and you're asked about a figure so far along the sequence that you could never draw it. The key is finding the formula rather than counting term by term.

Figure growth questions appear regularly in OC TS at medium difficulty and are among the most elegant question types in the paper. Students who spot the algebraic relationship solve them in under thirty seconds; students who try to extend the pattern step-by-step run out of time at figure 367.

The examiner is checking whether you can recognise an invariant property — a quantity that remains constant across all figures regardless of how large n grows. The examiner deliberately uses a large figure number (like 367) to force students to find the general rule rather than compute directly.

A sequence of four or five figures is shown visually. Each figure contains multiple element types (squares and circles, triangles and lines, etc.) growing at different rates. You are asked about the relationship between the element counts at a very large figure number.

Best approach: Build a small table: count each element type for figures 1, 2, 3, 4. Compute the quantity the question asks for (here, squares minus circles) for each figure. If the result is the same every time, you have found an invariant — that value is your answer, no matter what figure number is asked about. Verify by writing the formula for each element type.

Question

Study the figures below. Each figure is made of a grid of squares and a stacked pyramid of circles.

How many more squares than circles will there be in Figure 367 ?

A $367$
B $1$
C $734$
D $2$

Decided on your answer? Check how you went below.

Correct Answer

1

Step-by-Step Explanation

Step 1: Count squares and circles in each figure. Figure Squares Circles Squares - Circles 1211243136514871 The difference is always exactly 1 — an invariant. Step 2: Find the formulas (to confirm). Each figure n contains: Squares: 2 rows \times n columns = 2n squares Circles: n on the bottom row + (n - 1) on the next = 2n - 1 circles Step 3: Prove the difference is always 1. Squares - Circles = 2n - (2n - 1) = 1 The n cancels completely. The difference is 1 for every figure, no matter how large. Step 4: Apply to Figure 367. Figure 367 has 2 \times 367 = 734 squares and 2 \times 367 - 1 = 733 circles. Difference = 734 - 733 = 1 Why the other options are wrong: Option The mistake A — 367 Multiplied the figure number itself — ignores the structure entirely C — 734 Gives the number of squares in Figure 367, not the difference D — 2 Off by one — perhaps mistakenly counted circles as 2n - 2 rather than 2n - 1 The invariant insight: When a question asks about Figure 367, always check whether the requested quantity is the same for every figure first. If it is, you never need to compute the actual counts.

Up next: Practice Set 1

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Option	The mistake
A — 367	Multiplied the figure number itself — ignores the structure entirely
C — 734	Gives the number of squares in Figure 367, not the difference
D — 2	Off by one — perhaps mistakenly counted circles as 2n − 2 rather than 2n − 1