Cycle of Powers
Only the unit digit of the base decides the unit digit of any power, so 23^45 has the same last digit as 3^45. Sort the ten digits by how their unit digit behaves under multiplication. Digits 0, 1, 5 and 6 are fixed points — they reproduce themselves at every power (period 1), so 6^anything ends in 6. Digits 4 and 9 toggle between two values (period 2): 4 gives 4, 6, 4, 6… and 9 gives 9, 1, 9, 1…, so an even power of 9 ends in 1 and an odd power in 9. The four "hard" digits 2, 3, 7, 8 each run through a 4-long cycle: 2→(2,4,8,6), 3→(3,9,7,1), 7→(7,9,3,1), 8→(8,4,2,6). A neat CAT shortcut: 7 and 3 share the same set {1,3,7,9}, and 2 and 8 share {2,4,6,8}. Learn these five mini-cycles cold and most unit-digit questions become instant.
✅ 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) |