EQN-01-12

Remainders & Divisibility Patterns

Work with remainders directly and use cyclic patterns to answer "what is left over" questions.

~5 min read•Difficulty 2/5

What a remainder is

dividend = divisor \times quotient + remainder, 0 \leq remainder < divisor

17 = 5 \times 3 + 2 \to 17 \div 5 leaves remainder 2

The remainder is always smaller than the divisor.

Remainders cycle

When you divide consecutive numbers, the remainders repeat in a fixed cycle of length equal to the divisor.

Dividing by 4 : remainders go 0, 1, 2, 3, 0, 1, 2, 3, \dots

So the remainder of $N \div d$ depends only on where N sits in the cycle — useful for "day of the week" and pattern questions.

Divisibility tests (quick recall)

2 : last digit even 4 : last two digits divisible by 4

3 : digit sum \div 35 : ends in 0 or 5

6 : divisible by 2 AND 39 : digit sum \div 910 : ends in 0

Last-digit and units cycles

Powers repeat their last digit in cycles (e.g. powers of 2: 2,4,8,6,2,4,…). Find the cycle length, then use the remainder of the exponent to locate the units digit.

Worked examples

Example 1 (difficulty 2/5)

What is the remainder when 100 is divided by 7?

◎ Predict the answer first:

Example 2 (difficulty 2/5)

Today is Wednesday. What day will it be after 45 days?

◎ Predict the answer first:

Tips

Remainder $=$ number $- ($ largest multiple of the divisor below it).
For cyclic "after N steps" problems, use the remainder of $N \div$ cycle length.
A remainder is always less than its divisor — a quick sanity check.

Common mistakes

Giving a remainder equal to or larger than the divisor

Why: Subtraction stops one multiple too early.

Fix: Keep subtracting until what is left is below the divisor.

Counting all N days instead of N mod 7 for day-of-week questions

Why: The full count seems necessary.

Fix: Reduce N by multiples of the cycle length first.

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