Unit digit and remainder questions are among the easiest to score on — once we know the patterns. The entire concept boils down to this: unit digits repeat in cycles, and remainders follow predictable rules. Learn the patterns, and these become free marks.
Unit Digit Cyclicity
Every digit, when raised to successive powers, repeats its unit digit in a cycle. This is called cyclicity.
Key observations:
- Digits 0, 1, 5, 6 → always give the same unit digit (period = 1)
- Digits 4, 9 → alternate between two values (period = 2)
- Digits 2, 3, 7, 8 → cycle through four values (period = 4)
How to Find the Unit Digit of a^n
- Look at the unit digit of a (only the unit digit matters)
- Find the cycle length for that digit
- Divide n by the cycle length and find the remainder
- Use the remainder to pick the right digit from the cycle
If the remainder is 0, we use the LAST digit in the cycle (not the first).
Step-by-Step Method
Example: Find the unit digit of 7^253.
- Unit digit of base = 7
- Cycle of 7: 7, 9, 3, 1 (period = 4)
- Divide power by cycle: 253 ÷ 4 = 63 remainder 1
- Remainder 1 → 1st digit in cycle → 7
Example: Find the unit digit of 3^100.
- Unit digit of base = 3
- Cycle of 3: 3, 9, 7, 1 (period = 4)
- 100 ÷ 4 = 25 remainder 0
- Remainder 0 → last digit in cycle → 1
Example: Find the unit digit of 2^347 + 3^128.
Handle each separately:
- 2^347: Cycle of 2 is (2,4,8,6), period 4. 347 ÷ 4 = remainder 3. Third in cycle → 8
- 3^128: Cycle of 3 is (3,9,7,1), period 4. 128 ÷ 4 = remainder 0. Last in cycle → 1
- Sum’s unit digit: 8 + 1 = 9
Remainder Shortcuts
Remainder when dividing by 2, 5, 10
- ÷ 2: Just look at the last digit
- ÷ 5: Just look at the last digit
- ÷ 10: Just look at the last digit
Remainder when dividing by 3 or 9
- ÷ 3: Sum of all digits, find remainder of that sum ÷ 3
- ÷ 9: Sum of all digits, find remainder of that sum ÷ 9
Example: Remainder when 5847 is divided by 9.
- Sum of digits: 5 + 8 + 4 + 7 = 24
- 24 ÷ 9 = 2 remainder 6
Remainder when dividing by 11
Use the alternating sum trick (same as divisibility by 11):
- Take digits from right, alternately add and subtract
Example: Remainder when 86345 is divided by 11.
- From right: 5 - 4 + 3 - 6 + 8 = 6
- Remainder = 6
If the alternating sum is negative, keep adding 11 until it’s positive.
Remainder of Powers — The Cycle Method
To find remainder of a^n ÷ d:
- Find the cycle of remainders that a^1, a^2, a^3… give when divided by d
- Use the cycle to find the answer
Example: Remainder when 2^100 is divided by 7.
- 2¹ ÷ 7 → remainder 2
- 2² ÷ 7 → remainder 4
- 2³ ÷ 7 → remainder 1
- Cycle: (2, 4, 1) with period 3
- 100 ÷ 3 = remainder 1
- So remainder = first in cycle = 2
Fermat’s Little Theorem (Useful Shortcut)
In simple language, Fermat’s theorem tells us: if we raise any number to the power (prime - 1) and divide by that prime, the remainder is always 1.
Example: Remainder when 3^40 is divided by 41.
41 is prime, and 3 is not divisible by 41. By Fermat’s: 3^(41-1) = 3^40 ≡ 1 (mod 41). Remainder = 1.
Worked Examples
Example 1: Find the unit digit of 17^256 × 13^124.
- 17^256: Unit digit of 17 is 7. Cycle of 7: (7,9,3,1), period 4. 256 ÷ 4 = 0 remainder. Last in cycle → 1
- 13^124: Unit digit of 13 is 3. Cycle of 3: (3,9,7,1), period 4. 124 ÷ 4 = 0 remainder. Last in cycle → 1
- Product: 1 × 1 = 1
Example 2: Find the remainder when 37^73 is divided by 9.
Instead of working with 37, find 37 mod 9 first: 37 ÷ 9 = 4 remainder 1. So 37 ≡ 1 (mod 9). Therefore 37^73 ≡ 1^73 ≡ 1 (mod 9). Remainder = 1.
This is a huge shortcut — always reduce the base first!
Example 3: What is the unit digit of 4^4^4?
First, we need to figure out the power: 4^4 = 256. So we need unit digit of 4^256. Cycle of 4: (4, 6), period 2. 256 is even → unit digit = 6
Example 4: Find the remainder when 2^256 is divided by 17.
17 is prime. By Fermat’s: 2^16 ≡ 1 (mod 17). 256 = 16 × 16, so 2^256 = (2^16)^16 ≡ 1^16 = 1 (mod 17). Remainder = 1
Example 5: Find the unit digit of 1! + 2! + 3! + … + 100!
- 1! = 1 (unit digit 1)
- 2! = 2 (unit digit 2)
- 3! = 6 (unit digit 6)
- 4! = 24 (unit digit 4)
- 5! = 120 (unit digit 0)
- 6! = 720 (unit digit 0)
- Every factorial from 5! onwards ends in 0 (because they contain 2×5=10 as a factor)
So unit digit = unit digit of (1 + 2 + 6 + 4 + 0 + 0 + …) = unit digit of 13 = 3
Common Exam Patterns
- “Unit digit of a^n” → Find cycle, divide power by period
- “Unit digit of sum/product” → Find unit digit of each term separately, then add/multiply
- “Remainder when large power ÷ prime” → Use Fermat’s Little Theorem
- “Remainder when N ÷ d” → Reduce base first (base mod d), then find cycle of remainders
- “Unit digit of factorial sum” → Only first 4 terms matter (5! onwards ends in 0)
Practice Problems
Q1: Find the unit digit of 8^2048 + 9^2048.
Q2: Find the remainder when 7^99 is divided by 5.
Q3: What is the unit digit of 2^2 × 3^3 × 4^4 × 5^5 × 6^6?
Answers:
A1: 8^2048: Cycle of 8 is (8,4,2,6), period 4. 2048 ÷ 4 = 0 remainder → last in cycle → 6. 9^2048: Cycle of 9 is (9,1), period 2. 2048 ÷ 2 = 0 remainder → last in cycle → 1. Unit digit of sum: 6 + 1 = 7
A2: 7 mod 5 = 2. So we need 2^99 mod 5. Cycle of 2 mod 5: 2, 4, 3, 1 (period 4). 99 ÷ 4 = remainder 3. Third in cycle = 3.
A3: 2² = 4, 3³ = 27 (unit 7), 4⁴ = 256 (unit 6), 5⁵ ends in 5, 6⁶ ends in 6. Product of unit digits: 4 × 7 = 28(unit 8) → 8 × 6 = 48(unit 8) → 8 × 5 = 40(unit 0) → 0 × 6 = 0