Unit Digit and Cyclicity of Powers
Unit Digit and Cyclicity of Powers
The unit digit of any product or power depends only on the unit digits of the bases. Combine that with the fact that unit digits of powers repeat in a cycle of at most 4, and a question like 2137⁷⁵⁴ becomes a two-line calculation.
1. Cyclicity Table (memorise this)
| Unit digit of base | Cycle of unit digits | Cycle length |
|---|---|---|
| 0 | 0 | 1 |
| 1 | 1 | 1 |
| 5 | 5 | 1 |
| 6 | 6 | 1 |
| 4 | 4, 6 (odd power 4, even power 6) | 2 |
| 9 | 9, 1 (odd power 9, even power 1) | 2 |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
2. The Method in One Line
Take the unit digit of the base. Divide the power by 4 and keep the remainder r. If r = 0, use power 4. The answer is the unit digit of (unit digit)ʳ.
Only the last two digits of the power are needed to find the remainder on division by 4 (because 100 is a multiple of 4).
Worked check: 7⁹⁵ → 95 ÷ 4 leaves 3 → 7³ = 343 → unit digit 3.
3. Rules for Expressions
| Expression | Rule |
|---|---|
| Product of powers | Find each unit digit, multiply them, keep the last digit |
| Sum of powers | Add the unit digits, keep the last digit |
| Difference of powers | Subtract; if negative, add 10 (the first number is larger, so borrow) |
| Any product containing a 5 and an even number | Unit digit 0 |
| Product of odd numbers containing a 5 | Unit digit 5 |
| n! for n ≥ 5 | Unit digit 0 |
| 1! + 2! + ... + n! for n ≥ 4 | Unit digit 3 (1 + 2 + 6 + 24 = 33) |
| Power tower a^(b^c) | Reduce b^c on division by 4 first |
✗ 487⁷⁵ − 233³⁹: 3 − 7 = −4, unit digit 4 | ✓ Borrow 10: 13 − 7 = 6, unit digit 6
✗ Power remainder 0 means use power 0 | ✓ Remainder 0 means use the 4th term of the cycle (2⁴ ends in 6, not 1)
हिंदी नोट: इकाई अंक केवल आधार के इकाई अंक और घात को 4 से भाग देने पर बचे शेषफल पर निर्भर करता है। शेषफल 0 आए तो घात 4 मानिए।
Exam Pointer: Unit digit was asked in the 2016 cycle: a product of two powers (12 April 2016), two consecutive powers of the same base added (28 April 2016) and a product of three powers in the January 2017 CBT-2. We did not find a verified 2020-21 or 2025-26 shift for it, so treat it as an occasional 20-second mark; patterns without a verified shift are tagged syllabus-based.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Unit digit of a single power
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the unit digit of 7⁹⁵.
Cycle of 7 is 7, 9, 3, 1. 95 ÷ 4 leaves 3, so the unit digit equals that of 7³ = 343, which is 3.
Answer: 3
EXAMATLAS LEVEL
Q. Find the unit digit of 2137⁷⁵⁴.
Only the base's unit digit 7 matters. 754 ÷ 4: look at the last two digits 54, which leave 2. So the answer is the unit digit of 7² = 49, which is 9.
Answer: 9
Pattern 2: Unit digit of a product of powers
[PYQ: NTPC CBT-1 12-Apr-2016 Shift-2 | NTPC CBT-2 18-Jan-2017 Shift-3]
EXAM LEVEL
Q. Find the unit digit of 13²⁷ × 24¹⁸ × 17¹³.
3²⁷: 27 leaves 3, so 3³ ends in 7. 4¹⁸: even power of 4 ends in 6. 7¹³: 13 leaves 1, so it ends in 7. Product of unit digits 7 × 6 × 7 = 294, unit digit 4.
Answer: 4
EXAMATLAS LEVEL
Q. Find the unit digit of 4⁶¹ × 9⁶² × 2⁶³ × 7⁶⁴.
4⁶¹ is an odd power of 4, ends in 4. 9⁶² is an even power of 9, ends in 1. 2⁶³: 63 leaves 3, 2³ = 8. 7⁶⁴: 64 leaves 0, so use 7⁴, which ends in 1. Product 4 × 1 × 8 × 1 = 32, unit digit 2.
Answer: 2
Pattern 3: Unit digit of a sum or difference of powers
[PYQ: NTPC CBT-1 28-Apr-2016 Shift-2]
EXAM LEVEL
Q. Find the unit digit of 52⁴³ + 39²².
2⁴³: 43 leaves 3, 2³ = 8. 9²² is an even power of 9, ends in 1. 8 + 1 = 9.
Answer: 9
EXAMATLAS LEVEL
Q. Find the unit digit of 487⁷⁵ − 233³⁹.
7⁷⁵: 75 leaves 3, unit digit of 7³ is 3. 3³⁹: 39 leaves 3, unit digit of 3³ is 7. The first number is larger, so subtract with a borrow: 13 − 7 = 6.
Answer: 6
Pattern 4: Factorial and self-power series
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the unit digit of 1! + 2! + 3! + ... + 50!.
From 5! onwards every factorial ends in 0. 1 + 2 + 6 + 24 = 33, so the unit digit is 3.
Answer: 3
EXAMATLAS LEVEL
Q. Find the unit digit of 1¹ + 2² + 3³ + 4⁴ + ... + 10¹⁰.
Unit digits term by term: 1, 4, 7 (27), 6 (256), 5, 6, 3 (7⁷: 7 leaves 3, 7³ ends in 3), 6 (8⁸: 8 leaves 0, use 8⁴ ending 6), 9 (odd power of 9), 0. Sum = 1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 + 9 + 0 = 47, unit digit 7.
Answer: 7
Pattern 5: Variable or stacked exponents
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the unit digit of 234¹⁰⁰ + 234¹⁰¹.
Even power of 4 ends in 6, odd power ends in 4. 6 + 4 = 10, unit digit 0.
Answer: 0
EXAMATLAS LEVEL
Q. Find the unit digit of 7^(7⁷).
First find 7⁷ on division by 4. 7 leaves 3 (that is, −1), and an odd power of −1 is −1, which is 3. So the exponent behaves like 3, and the unit digit equals that of 7³ = 343, which is 3.
Answer: 3
CBT-2 LEVEL
Pattern 6: Last two digits of a power
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the last two digits of 31²⁵.
For a base ending in 1, the last digit is 1 and the tens digit = (tens digit of base × unit digit of power), last digit only. 3 × 5 = 15, so tens digit 5. Last two digits 51.
Answer: 51
EXAMATLAS LEVEL
Q. Find the last two digits of 49²⁰ + 24¹⁵.
49² = 2,401 ends in 01, so any even power of 49 ends in 01. For 24: odd powers end in 24 and even powers end in 76. 24¹⁵ ends in 24. Sum 01 + 24 = 25.
Answer: 25
60-Second Revision
- 0, 1, 5, 6 never change; 4 and 9 alternate; 2, 3, 7, 8 cycle in 4.
- Divide the power by 4 (use its last two digits); remainder 0 means take the 4th cycle term.
- Products: multiply unit digits; sums: add; differences: borrow 10 if negative.
- n! ends in 0 for n ≥ 5; 1! + ... + n! ends in 3 for n ≥ 4.
- Stacked power: reduce the exponent on division by 4 first.
- CBT-2: (…1)ⁿ tens digit = tens × n's unit digit; 49 even power ends 01; 24 odd 24, even 76.