When you multiply two numbers, the unit digit of the answer depends only on the unit digits of the numbers you multiplied. That single fact makes these questions easy. To find the last digit of 243¹⁷, you never need 243 itself; you only need 3¹⁷, and in fact only the last digit of that.

Keep multiplying a digit by itself and the unit digits soon start repeating. For 2 the sequence of unit digits is 2, 4, 8, 6, 2, 4, 8, 6 and so on. The repeating block is called the cycle, and its length is the cyclicity. Once you know the cycle, any power is a matter of finding the right position in it.

Unit digitCycle of unit digitsLength
0, 1, 5, 6the digit itself every time1
44, 62
99, 12
22, 4, 8, 64
33, 9, 7, 14
77, 9, 3, 14
88, 4, 2, 64

Two memory aids help. The cycles for 2 and 8 contain the same four digits, as do the cycles for 3 and 7. And 4 and 9 alternate: an odd power of 4 ends in 4 and an even power ends in 6, while an odd power of 9 ends in 9 and an even power ends in 1.

  1. Keep only the unit digit of the base.
  2. Divide the exponent by 4 and note the remainder.
  3. A remainder of 1, 2 or 3 picks the first, second or third digit of the cycle. A remainder of 0 picks the fourth.

Worked example 1: 2⁴³

43 ÷ 4 leaves remainder 3. The third digit in the cycle 2, 4, 8, 6 is 8.

Worked example 2: 3¹⁰²

102 ÷ 4 leaves remainder 2. The second digit in 3, 9, 7, 1 is 9.

Worked example 3: 7⁹⁶

96 is a multiple of 4, so the remainder is 0 and we take the fourth digit of 7, 9, 3, 1: 1.

Worked example 4: 8⁵⁷

57 ÷ 4 leaves remainder 1, so the answer is the first digit of 8, 4, 2, 6: 8.

For bases ending in 4 or 9, just check whether the exponent is odd or even. 4³⁵ has an odd exponent, so its unit digit is 4.

Likewise 9⁴⁸ has an even exponent, so its unit digit is 1.

For a product of powers, find each unit digit separately and multiply them, keeping only the last digit. For a sum, add the unit digits. For a difference, subtract them, borrowing 10 if the first digit is smaller, provided the first number is the larger one.

Worked example 5: a product

243¹⁷ × 158²² reduces to 3¹⁷ × 8²². 17 ÷ 4 leaves 1, giving 3; 22 ÷ 4 leaves 2, giving 4. Then 3 × 4 = 12, so the unit digit is 2.

Worked example 6: a sum

2²⁰ + 3²⁰: 20 is a multiple of 4, so the unit digits are 6 and 1. Their sum is 7.

Questions sometimes give a huge exponent written as an expression, such as 7^(4k + 3). You do not need the value of k: the remainder on division by 4 is 3, so the unit digit is the third in the cycle of 7, which is 3.

The same thinking settles questions about factorials. From 5! onwards every factorial ends in 0, so in a sum such as 1! + 2! + … + 50! only the first four terms affect the unit digit: 1 + 2 + 6 + 24 = 33, giving a unit digit of 3.

  • ✘ Taking the first digit of the cycle when the remainder is 0. Remainder 0 means the fourth (last) digit.
  • ✘ Working with the whole base. Only the unit digit of the base matters.
  • ✘ Adding exponents when multiplying different bases. Find each unit digit separately, then multiply.
  • ✘ Forgetting to borrow in a difference. 2 − 7 becomes 12 − 7 = 5.

Cyclicity also drives remainder questions on large powers, and it pairs with divisibility rules in number-system sets. The core formulas are listed in the Quant cheat code.

Q1. What is the unit digit of 3⁶⁵? easy

✅ Answer: (d) 3

💡 Explanation: The cycle of 3 is 3, 9, 7, 1. 65 ÷ 4 leaves remainder 1, so the unit digit is the first in the cycle, 3.

Q2. What is the unit digit of 22⁴⁷? easy

✅ Answer: (a) 8

💡 Explanation: Only the unit digit 2 matters. Its cycle is 2, 4, 8, 6. 47 ÷ 4 leaves remainder 3, so the unit digit is the third in the cycle, 8.

Q3. What is the unit digit of 57⁸⁴? medium

✅ Answer: (b) 1

💡 Explanation: The cycle of 7 is 7, 9, 3, 1. 84 is a multiple of 4, so take the last term of the cycle, 1.

Q4. What is the unit digit of 39⁴¹ × 44³⁰? medium

✅ Answer: (c) 4

💡 Explanation: 9 to an odd power ends in 9, and 4 to an even power ends in 6. Multiplying the unit digits, 9 × 6 = 54, so the answer is 4.

Q5. What is the unit digit of 6⁹⁹ + 5⁹⁹? easy

✅ Answer: (a) 1

💡 Explanation: Every power of 6 ends in 6 and every power of 5 ends in 5. 6 + 5 = 11, so the unit digit of the sum is 1.

Q6. What is the unit digit of 18²³ − 13²³? hard

✅ Answer: (d) 5

💡 Explanation: 23 ÷ 4 leaves remainder 3. The third term in the cycle of 8 is 2 and in the cycle of 3 it is 7. Since 18²³ is the larger number, borrow: 12 − 7 = 5.

Why do I divide the exponent by 4 and not by the length of each cycle?

Every digit's cycle has length 1, 2 or 4, and each of those divides 4. Using 4 for every digit therefore always lands on the right place in the cycle, so you only need to remember one rule.

What if the remainder is zero?

A remainder of zero means the exponent is a whole number of complete cycles, so take the last digit of the cycle, not the first. For 2 that is 6, for 3 it is 1, for 7 it is 1 and for 8 it is 6.

All problems and solutions on this page are original. No outside facts are relied on; every numeric answer is recomputed by software before publishing.

How this page was prepared. Drafted with AI assistance (Claude (Anthropic AI model)) from the sources listed above. 15 calculations on this page were recomputed by software and matched. Automated checks also covered structure, internal links and duplicate text. Software checks cannot confirm facts. Reviewed by praveenkumar on 2026-10-20 against the sources listed above.

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