Solved Examples

Example 1: Find the last two digits of 21^25. - Base ends in 1: tens digit = 2, n = 25. - Tens digit of answer = (2 × 25) mod 10 = 50 mod 10 = 0 - Last two digits = 01

Example 2: Find the last two digits of 7^298. - 7⁴ = 2401 → last two digits “01” (this is now an “ends in 1” number with tens digit 0) - 298 = 4×74 + 2, so 7^298 = (7⁴)^74 × 7² - (…01)^74: tens digit = (0 × 74) mod 10 = 0 → stays “01” - 7² = 49 - Last two digits = 01 × 49 → 49

Example 3: Find the last two digits of 43^43. - 43² = 1849 → last two digits 49 - 43⁴ = 49² = 2401 → last two digits 01 (tens digit 0) - 43 = 4×10 + 3, so 43^43 = (43⁴)^10 × 43³ - (…01)^10: tens digit = (0×10) mod 10 = 0 → stays “01” - 43³ = 43² × 43 = 1849 × 43 = 79,507 → last two digits 07 - Last two digits = 01 × 07 → 07

Example 4: Find the last two digits of 29^29. - 29² = 841 → last two digits 41 (an “ends in 1” number, tens digit 4) - 29 = 2×14 + 1, so 29^29 = (29²)^14 × 29¹ - (…41)^14: tens digit = (4 × 14) mod 10 = 56 mod 10 = 6 → gives “61” - Multiply 61 × 29 = 1769 → last two digits 69 Answer: Last two digits = 69

Example 5: Find the last two digits of 71^95. - Base ends in 1: tens digit = 7, n = 95. - Tens digit of answer = (7 × 95) mod 10 = 665 mod 10 = 5 Answer: Last two digits = 51

Example 6 (Edge case — base ends in 7, large exponent): Find the last two digits of 27^35. - 27² = 729 → last two digits 29. - 27⁴ = 29² = 841 → last two digits 41 (an “ends in 1” number, tens digit 4). - 35 = 4×8 + 3, so 27^35 = (27⁴)^8 × 27³. - (…41)^8: tens digit = (4 × 8) mod 10 = 32 mod 10 = 2 → gives “21” - 27³ = 27² × 27 = 729 × 27 = 19,683 → last two digits 83 - Multiply the two last-two-digit results: 21 × 83 = 1,743 → last two digits 43 Answer: Last two digits = 43 (Verified independently via repeated squaring mod 100: 27²≡29, 27⁴≡41, 27⁸≡81, 27¹⁶≡61, 27³²≡21; 27³⁵ = 27³²×27²×27¹ ≡ 21×29×27 ≡ 609×27 ≡ 9×27 ≡ 243 ≡ 43 (mod 100) — matches.)