How to Find the Unit Digit of a Number: Step-by-Step

The unit digit of any power depends on exactly two things: the unit digit of the base and the remainder when the exponent is divided by the cycle length of that digit.

This turns problems that look like they need a calculator into 10-second mental calculations. Unit digit questions appear in the quantitative aptitude rounds of nearly every campus placement test, from entry-level IT service firms to quant-intensive recruiters. The approach is identical regardless of how large the base or exponent becomes.

What Is the Unit Digit?

The unit digit is the rightmost digit of any integer, also called the ones digit or the ones place. It is the remainder when you divide the number by 10.

Examples:

• In 3,451, the unit digit is 1.
• In 2,700, it is 0.
• In 89, it is 9.
• In 10,006, it is 6.

The key insight for aptitude problems is that only the unit digits interact during multiplication and exponentiation. The unit digit of 684 times 759 is exactly the same as the unit digit of 4 times 9. The full values never need to be computed.

This matters because exam questions routinely ask for the unit digit of expressions like 2^100 or 987 times 413 times 726. Computing the actual result would take minutes and a calculator. Using the cyclicity method, both are answerable under 20 seconds.

Unit Digit of a Product

For any product of integers, the rule is straightforward:

• Step 1: Extract the unit digit from each factor.
• Step 2: Multiply those unit digits together.
• Step 3: Take the unit digit of the result.

Worked example: Find the unit digit of 684 times 759 times 413 times 676.

• Unit digits: 4, 9, 3, 6.
• 4 times 9 = 36. Unit digit = 6.
• 6 times 3 = 18. Unit digit = 8.
• 8 times 6 = 48. Unit digit = 8.

The same principle extends to sums. For a sum, add the unit digits of all terms and take the unit digit of the total. If the unit digits of a two-term sum are 7 and 4, their sum is 11, and the unit digit is 1.

This is also why you can evaluate chains of operations piece by piece: simplify each factor to its unit digit, operate, reduce again. The unit digit of the running total is all that ever matters.

Unit Digit of a Power: The Cyclicity Table

When a number is raised to successive powers, its unit digit repeats in a cycle. That cycle depends only on the unit digit of the base, not on the magnitude of the number.

The complete table for all ten unit digits:

Unit digit of base	Cycle of unit digits in powers	Cycle length
0	0	1
1	1	1
2	2, 4, 8, 6	4
3	3, 9, 7, 1	4
4	4, 6	2
5	5	1
6	6	1
7	7, 9, 3, 1	4
8	8, 4, 2, 6	4
9	9, 1	2

Digits 0, 1, 5, and 6: cycle length 1

These four unit digits stay constant regardless of the exponent:

• Any power of 0 ends in 0 (for exponent ≥ 1).
• Any power of 1 ends in 1.
• Any power of 5 ends in 5. Verify: 5^1 = 5, 5^2 = 25, 5^3 = 125. Unit digits: 5, 5, 5.
• Any power of 6 ends in 6. Verify: 6^1 = 6, 6^2 = 36, 6^3 = 216. Unit digits: 6, 6, 6.

No cycle lookup is needed for these four. If you see a base ending in 1, 5, or 6, the answer to any unit digit power question is the digit itself.

Digits 4 and 9: cycle length 2

• 4’s cycle: 4, 6. Odd exponent gives unit digit 4; even exponent gives unit digit 6.
• 9’s cycle: 9, 1. Odd exponent gives unit digit 9; even exponent gives unit digit 1.

Verify for 4: 4^1 = 4 (unit digit 4), 4^2 = 16 (unit digit 6), 4^3 = 64 (unit digit 4), 4^4 = 256 (unit digit 6). The pattern alternates cleanly.

Verify for 9: 9^1 = 9, 9^2 = 81 (unit digit 1), 9^3 = 729 (unit digit 9). Alternates between 9 and 1.

Digits 2, 3, 7, and 8: cycle length 4

These four digits have the longest cycle and form the majority of exam questions. Each repeats after 4 powers:

• 2: 2, 4, 8, 6. Check: 2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16 (unit digit 6), 2^5 = 32 (unit digit 2, cycle restarts).
• 3: 3, 9, 7, 1. Check: 3^1 = 3, 3^2 = 9, 3^3 = 27 (unit digit 7), 3^4 = 81 (unit digit 1), 3^5 = 243 (unit digit 3, restarts).
• 7: 7, 9, 3, 1. Check: 7^1 = 7, 7^2 = 49 (unit digit 9), 7^3 = 343 (unit digit 3), 7^4 = 2401 (unit digit 1), 7^5 = 16807 (unit digit 7, restarts).
• 8: 8, 4, 2, 6. Check: 8^1 = 8, 8^2 = 64 (unit digit 4), 8^3 = 512 (unit digit 2), 8^4 = 4096 (unit digit 6), 8^5 = 32768 (unit digit 8, restarts).

Memorise these four cycles. Most exam questions come from this group.

The Mod Method and the Off-by-One Trap

For cycle-4 digits (2, 3, 7, 8), the standard method is:

• Step 1: Identify the unit digit of the base.
• Step 2: Look up the 4-position cycle for that digit.
• Step 3: Divide the exponent by 4 and note the remainder.
• Step 4: Map the remainder to the cycle position using this table:

Remainder after dividing exponent by 4	Use cycle position
1	Position 1 (first element)
2	Position 2
3	Position 3
0	Position 4 (last element)

For cycle-2 digits (4, 9), apply the same logic dividing by 2: remainder 1 gives position 1, remainder 0 gives position 2 (the last element).

The most common exam error: remainder 0 is not position 1

When the exponent is exactly divisible by 4, the remainder is 0. A common error is to treat 0 as “back to the start,” selecting position 1 in the cycle. That is wrong.

Remainder 0 means the full 4-step cycle ran through completely. You land on position 4, the last element.

Direct verification for digit 2: 2^4 = 16, unit digit 6. The cycle [2, 4, 8, 6] places 6 at position 4. If you used position 1 by mistake, you would answer 2, which is incorrect.

Direct verification for digit 3: 3^4 = 81, unit digit 1. The cycle [3, 9, 7, 1] places 1 at position 4. Correct answer: 1.

Direct verification for digit 7: 7^4 = 2401, unit digit 1. The cycle [7, 9, 3, 1] places 1 at position 4. Correct answer: 1.

The quantitative sections in D.E. Shaw’s placement process regularly include problems designed to catch this exact mistake, where the exponent is a multiple of 4 and the correct answer is the last cycle element, not the first.

This method is a direct application of modular arithmetic: dividing the exponent by the cycle length and using the remainder to select a position. The same operation drives much of number theory and computing.

Three Worked Examples

Work through these step by step to lock in the method before the exam.

Example 1: Find the unit digit of (3547)^153 times (251)^72.

• Unit digit of 3547 = 7. Cycle for 7: [7, 9, 3, 1], length 4.
• 153 divided by 4: quotient 38, remainder 1. Position 1 = 7.
• Unit digit of 251 = 1. Any power of 1 stays 1. Unit digit = 1.
• Product unit digit: 7 times 1 = 7.

Example 2: Find the unit digit of (128)^37.

• Unit digit of 128 = 8. Cycle for 8: [8, 4, 2, 6], length 4.
• 37 divided by 4: quotient 9, remainder 1. Position 1 = 8.
• Unit digit of (128)^37 = 8.

Example 3: Find the unit digit of (43)^44 + (53)^53.

• Both 43 and 53 have unit digit 3. Cycle for 3: [3, 9, 7, 1], length 4.
• (43)^44: 44 divided by 4 = 11, remainder 0. Position 4 (not position 1) = 1.
• (53)^53: 53 divided by 4 = 13, remainder 1. Position 1 = 3.
• Sum: 1 + 3 = 4. Unit digit = 4.

Example 3 combines both a remainder-0 case and a sum operation in one expression, which is the standard format for harder placement questions. The Sopra Steria aptitude test pattern includes combined exponent-plus-sum problems of this type in its quantitative section.

The mod-4 pattern from this article is a specific case of modular arithmetic, the same mathematical structure that handles positional encoding in transformer models and hash functions in cryptographic signing. If the path from “mod 4 gives the cycle position” to “mod N handles token position in a language model” is one worth taking, TinkerLLM starts that at ₹299 with hands-on experiments on actual models, starting from the same number reasoning covered here.