Pattern Period
For the period-4 digits (2, 3, 7, 8) you locate the answer by dividing the exponent by 4 and reading the remainder: remainder 1, 2, 3 points to the 1st, 2nd, 3rd term of the cycle, and remainder 0 (exponent divisible by 4) points to the 4th, last term. Example: 7^203 — the cycle is (7,9,3,1) and 203 mod 4 = 3, so take the 3rd term ⇒ 3. Only the last two digits of the exponent matter for "mod 4", since 100 is divisible by 4, which keeps the arithmetic tiny. The very same idea — a quantity that repeats with a fixed period — drives remainder problems: a^n mod m cycles too. For instance 3^n mod 7 runs 3, 2, 6, 4, 5, 1 and then repeats every 6, so any 3^n remainder reduces to n mod 6. Spotting the period turns a monstrous power into one short division.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
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Formula Reference Sheet
Unit-digit cycle lengths
| Period 1 (digit never changes) | 0, 1, 5, 6 → unit digit stays the same for every power |
|---|---|
| Period 2 | 4 → 4, 6, 4, 6… and 9 → 9, 1, 9, 1… |
| Period 4 | 2, 3, 7, 8 each cycle through 4 distinct unit digits |
| Cycle of 2 | 2¹=2, 2²=4, 2³=8, 2⁴=6, then repeat (2,4,8,6) |
| Cycle of 3 | 3¹=3, 3²=9, 3³=7, 3⁴=1, then repeat (3,9,7,1) |
Picking the unit digit by power mod 4
| Reduce the exponent | For period-4 digits, use n mod 4 to choose the term |
|---|---|
| Remainder 1, 2, 3 | unit digit = 1st, 2nd, 3rd term of the cycle |
| Remainder 0 (n divisible by 4) | unit digit = 4th (last) term of the cycle |
| Last digit of a product/sum | depends only on the last digits of the parts |
| Remainder cyclicity | a^n mod m also repeats with a fixed period (the order of a) |