Last Two Digits
The last two digits mean the number modulo 100. CAT favours four reliable cases. (1) Base ending in 1: for (…a1)^n the tens digit is (a × n) mod 10 and the unit digit is 1 — e.g. 31^17 ends in (3×17 = 51 → tens 1) and 1, giving 11. (2) Base ending in 76: 76 is "automorphic", so 76^n always ends in 76. (3) Even bases: anchor on 2^10 = 1024, so 2^10 ends in 24; note 24^odd ends in 24 and 24^even ends in 76, which lets you reduce any large power of an even number to a 24-or-76 form. (4) Base ending in 25: 25^n ends in 25 for every n ≥ 1, and similarly 5-ending numbers stabilise quickly. For bases near 100, use the binomial expansion: (100 ± a)^n keeps only the last two terms mod 100, i.e. (±a)^n + 100·n·(±a)^(n−1), and the second term vanishes mod 100, so you just need (±a)^n.
✅ 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) |