Last Digit
The last (unit) digit of a power depends only on the unit digit of the base and the exponent, never on the other digits — so 2467^85 has the same unit digit as 7^85. Every digit cycles: 0,1,5,6 are "self" digits (their powers always end in the same digit, cycle length 1); 4 and 9 alternate with cycle length 2; and 2,3,7,8 have cycle length 4. The CAT method is simple. Take the exponent modulo 4 (the longest cycle). If the remainder is 1,2,3 pick that term of the cycle; if the remainder is 0, use the 4th (last) term — a remainder of 0 does NOT mean the digit is the base. For products and sums, find the unit digit of each piece and combine: the unit digit of a product is the unit digit of the product of unit digits; for a sum, add the unit digits and take the last digit of that total.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
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Formula Reference Sheet
Cyclicity & last digit
| Cycle length 1 (digits 0,1,5,6) | last digit of d^n = d for all n ≥ 1 |
|---|---|
| Cycle length 2 (digits 4,9) | 4: 4,6,4,6… ; 9: 9,1,9,1… |
| Cycle length 4 (digits 2,3,7,8) | 2:2,4,8,6 | 3:3,9,7,1 | 7:7,9,3,1 | 8:8,4,2,6 |
| Reduce exponent (cycle 4) | use n mod 4; if remainder 0, use the 4th term |
| Last digit of a product | multiply only the unit digits, then take its unit digit |
Last two digits power-tools
| Base ends in 1 | (…a1)^n → last two digits = (a×n mod 10) then 1 |
|---|---|
| Base ends in 76 | 76^n always ends in 76 |
| Even base via 24/76 | 2^10 = …24 ; 24^(even) ends 76, 24^(odd) ends 24 |
| Base ends in 25 | 25^n ends in 25 for all n ≥ 1 (odd-power 25 / even patterns) |
| Binomial near 100 | (100±a)^n ≡ (±a)^n + 100·n·(±a)^(n−1) (mod 100) |