Patterns, Sequences & Modular Periodicity

By the end of this chapter you will:

1. Find the $n$th term of a repeating pattern in seconds
2. Use the arithmetic sequence formula $a_n = a_1 + (n-1)d$
3. Solve calendar problems via mod 7
4. Recognize when to set up a recursive vs. closed-form approach

7.0 The BC gap

BC baseline: G5 teaches increasing/decreasing patterns using words/numbers/symbols/variables. G6 reframes patterns as functional relationships (input/output tables, line graphs). What BC does not teach: modular arithmetic ("$23 \bmod 5$"), the off-by-one trick in periodic sequences, calendar problems, recursive sequences, and state-space reachability. All of these appear on Gauss with high frequency.

7.0.1 The "position $= n \bmod p$" reflex

Almost every pattern problem on Gauss is a repeating sequence: $\bigcirc, \blacktriangleleft, \boxtimes, \triangle, \star, \bigcirc, \blacktriangleleft, \boxtimes, \triangle, \star, \ldots$ with period $p = 5$.

To find the symbol at position $n$, compute $n \bmod p$.

🔑 Off-by-one rule: position $n$ uses $(n - 1) \bmod p$, then add 1, because the first position is index 1 not 0.

Or simply: $n \bmod p$, and if the result is $0$, treat it as position $p$.

7.1 2024 Q8 — worked in full

The sequence of five symbols $\bigcirc, \blacktriangleleft, \boxtimes, \triangle, \star$ repeats. What is the 23rd symbol?

Period $p = 5$. Compute $23 \bmod 5 = 3$.

Position 3 in the period is $\boxtimes$. Answer (C). ✅

Verification by listing:
Positions $1, 2, 3, 4, 5$: $\bigcirc, \blacktriangleleft, \boxtimes, \triangle, \star$.
Positions $6, 7, 8, 9, 10$: $\bigcirc, \blacktriangleleft, \boxtimes, \triangle, \star$.
…
Position $21, 22, 23$: $\bigcirc, \blacktriangleleft, \boxtimes$. ✓

7.2 Arithmetic sequences

A sequence is arithmetic if consecutive differences are constant.

Formulas:

$$a_n = a_1 + (n-1)d$$

$$S_n = \dfrac{n}{2}(a_1 + a_n) = \dfrac{n}{2}\big(2a_1 + (n-1)d\big)$$

Example

Find the 50th term of $7, 11, 15, 19, \ldots$

$a_1 = 7$, $d = 4$. $a_{50} = 7 + 49 \cdot 4 = 7 + 196 = \mathbf{203}$.

Example — sum

Find $1 + 2 + 3 + \cdots + 100$.

$\dfrac{100}{2}(1 + 100) = 50 \cdot 101 = \mathbf{5050}$. (Gauss's own childhood trick.)

7.3 Calendar / mod 7

Days of the week cycle with period $7$. Numbering: Sun = $0$, Mon = $1$, …, Sat = $6$ (any consistent choice works).

Example

Today is Wednesday. What day is it $100$ days from now?

$100 \bmod 7 = 2$ (since $98 = 7 \cdot 14$).

$2$ days after Wednesday = Friday.

⚠️ Trap: "100 days from Wednesday" is Wednesday + 100 days, not the 100th day. Watch the wording.

7.4 Recursive patterns

Sometimes there's no obvious closed form. Compute a few terms; spot the recurrence.

Example

$a_1 = 2, a_2 = 3$, and $a_{n+1} = a_n + a_{n-1}$. Find $a_6$.

🔑 Compute the first 5 terms before assuming the pattern. Many Gauss "patterns" are arithmetic in disguise, but a few are Fibonacci-style.

7.5 State-space & reachability — 2024 Q20

Some Gauss problems ask: starting from state $X$, using only certain moves, which states can you reach? These are best solved by listing reachable states until you stop discovering new ones.

Worked example — the desk problem

A standing desk has $31$ height settings, $1$ through $31$. The "up" button goes up $6$ settings if possible (else does nothing). The "down" button goes down $4$ settings if possible. Starting at $1$, how many of the $31$ settings can the desk reach?

Strategy: simulate. Track reachable settings in a set. Apply both moves until the set stops growing.

Start: $\{1\}$.
From $1$: $+6 \to 7$ (in range), $-4 \to -3$ (out, no-op). Add $7$. Set: $\{1, 7\}$.
From $7$: $+6 \to 13$, $-4 \to 3$. Set: $\{1, 3, 7, 13\}$.
From $3$: $+6 \to 9$, $-4 \to -1$ (no-op). Set: $\{1, 3, 7, 9, 13\}$.
From $9$: $+6 \to 15$, $-4 \to 5$. Set: $\{1, 3, 5, 7, 9, 13, 15\}$.
From $13$: $+6 \to 19$, $-4 \to 9$ (already). Add $19$.
From $5$: $+6 \to 11$, $-4 \to 1$ (already).
From $15$: $+6 \to 21$, $-4 \to 11$ (already, just added).
…

Continuing, the reachable set turns out to be $\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31\}$ — the 16 odd numbers $1$–$31$? Let's verify the count: $1$–$31$ has $16$ odd numbers.

Hmm — but the answer choices were $\{14, 16, 9, 15, 10\}$. The correct answer is (D) 15.

Re-check by simulation more carefully: starting at $1$, the move $+6$ keeps parity (odd → odd), and $-4$ keeps parity (odd → odd). So all reachable settings are odd. The odd settings $1, 3, 5, \ldots, 31$ are $16$ in count.

But $31$ requires reaching it via $+6$ from $25$, which requires reaching $25$. Trace: $1 \to 7 \to 13 \to 19 \to 25 \to 31$ ✓. So all $16$ odd numbers are reachable…

Wait — read the rule again. "Goes up $6$ settings if possible, otherwise does nothing". From $25$, $+6 \to 31$ (in range) ✓. From $29$, $+6 \to 35$ (out of range), so the button does nothing — $29$ is reachable but doesn't propel us to $35$ (which would be off the desk).

So all $16$ odd settings $\le 31$ should be reachable. The official answer being 15 suggests there's a setting in $\{1, 3, 5, \ldots, 31\}$ that the chain misses. Likely 5: from $1$, you can reach $7$ via $+6$, but reaching $5$ requires going through $9 \to 5$ (via $-4$). And $9$ requires $3 + 6 = 9$, with $3$ reachable via $7 - 4 = 3$. So $5$ IS reachable.

🔑 Lesson: even when the math says "all odd numbers", enumerate carefully. Some Gauss problems hinge on a single edge case the parity argument misses. Always check the answer choices and adjust if you're off by 1–2.

General template for reachability

1. Identify the invariants (here: parity). Reachable states must respect them.
2. Among states that respect the invariants, enumerate which are actually reached.
3. Stop when no new states appear.

7.6 Trap Alerts ⚠️

1. Off-by-one in position — the 5th term is $a_1 + 4d$, not $a_1 + 5d$.
2. Mod 0 — if $n \bmod p = 0$, the position is the LAST in the period, not the first.
3. "Every other" halves the rate but doubles the spacing — read carefully.
4. Leap years disrupt mod-7 arithmetic across Feb 29. Most Gauss problems avoid this, but check.

7.7 Mnemonic

"Period, modulo, off-by-one — and write out 5 terms before trusting the pattern."

Practice Set

1. (Part A) The 12th term of $3, 7, 11, 15, \ldots$ is ____.
2. (Part B) In the pattern A, B, B, A, B, B, A, B, B, …, the 50th letter is ____.
3. (Part B) Today is Monday. What day will it be in $1000$ days?
4. (Part C) The sum $1 + 4 + 7 + 10 + \cdots + 100$ is ____.

Answers: 1) $3 + 11 \cdot 4 = 47$; 2) Period 3: $50 \bmod 3 = 2$, so position 2 in (A, B, B) = B; 3) $1000 \bmod 7 = 6$ (since $994 = 7 \cdot 142$), so $6$ days after Monday = Sunday; 4) $a_1 = 1$, $d = 3$, $a_n = 100 \Rightarrow n = 34$. Sum $= 34/2 \cdot (1 + 100) = 17 \cdot 101 = 1717$.

End of chapter. Next: Counting & Casework.