Repeating Patterns | Moosa Academy

Repeating Patterns — Moosa Academy

Days of the week, hours on a clock, the last digit of a growing power — all of them run in a loop and start over. Once you know the length of the loop, reaching position $500$ takes one division instead of five hundred steps.

Theorem The remainder is the answer

Every repeating pattern has a fixed cycle length. To find what sits at any position:

Divide the position by the cycle length, keep the remainder, then count that many places.

"After" — count the remainder forward, starting from the next item.

"Before" — count the remainder backward, starting from the previous item.

"From" or "ends on" — count the remainder starting from the item itself.

If the remainder is $0$ , the position lands on the last item of the cycle — which is the same item you started from.

Example Days of the week — cycle 7

a) Today is Wednesday. What day is it in $80$ days?

80 \div 7 = 11

remainder

3

"After" — count

3

forward from the next day: Thursday, Friday, Saturday

Saturday

b) Today is Thursday. What day was it $45$ days ago?

45 \div 7 = 6

remainder

3

"Before" — count

3

backward: Wednesday, Tuesday, Monday

Monday

c) A Hijri year begins on a Tuesday. On what day does it end?

A Hijri year has

355

days

355 \div 7 = 50

remainder

5

"Ends on" — count

5

from the day itself: Tuesday, Wednesday, Thursday, Friday, Saturday

Saturday

A Hijri year is $355$ days and a Gregorian year $365$ . When a question does not say which calendar it means, treat it as Hijri.

Example Hours on a clock — cycle 12

It is $7$ o'clock now. What time will it be in $43$ hours?

43 \div 12 = 3

remainder

7

Move forward

7

hours:

7 + 7 = 14

14

is past

12

, so subtract:

14 - 12 = 2

$2$ o'clock

The clock face repeats every $12$ hours, and therefore also every $24$ , $48$ or $72$ . Any multiple of the cycle length can be discarded without changing the answer.

Example A repeating sequence

The numbers $1,\ 7,\ 2,\ 8,\ 9$ repeat in that order for ever. What is the $107$ th number?

The cycle length is

5

107 \div 5 = 21

remainder

2

Remainder

2

means the second item in the cycle

$7$

Twenty-one whole cycles pass by unnoticed. Only the leftover $2$ decides the answer.

Example The units digit of a power

The units digits of the powers of $3$ cycle as well:

3^{1} = 3

3^{2} = 9

3^{3} = 27

3^{4} = 81

3^{5} = 243

Units digits:

3,\ 9,\ 7,\ 1

— then it starts again, so the cycle length is

4

What is the units digit of $3^{22}$ ?

22 \div 4 = 5

remainder

2

Remainder

2

means the second digit in the cycle

The units digit is $9$

Note that the power itself is astronomically large, yet its final digit costs one division to find. That is the whole point of spotting the cycle.

Summary

Every repeating pattern has a fixed cycle length.
Divide the position by that length and keep only the remainder.
"After" counts forward from the next item; "before" counts backward; "from" counts from the item itself.
A remainder of $0$ lands on the last item of the cycle.
Days have cycle $7$ , clock hours cycle $12$ , and a Hijri year is $355$ days.
The units digits of a power repeat too — find the cycle, then use the remainder.