Unit Digit and Cyclicity

"What is the units digit of 7^53?" is a classic 30-second placement question (TCS, Infosys, Wipro). The units digit of any power repeats in a short cycle, so you only need the exponent's position in that cycle.

Cyclicity of the units digit

Base ends in	Cycle of units digits	Length
0, 1, 5, 6	same digit	1
4	4, 6	2
9	9, 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

Method: for a length-4 cycle, take the exponent mod 4. If the remainder is 1,2,3 pick that position; if 0, pick the 4th (last) digit.

Example: 7^53. 53 mod 4 = 1 → first digit of (7,9,3,1) = 7. ✓
Example: 2^34. 34 mod 4 = 2 → second of (2,4,8,6) = 4. ✓

Exam Tricks & Tips

• 🎯 Only the units digit of the base matters — 173^53 behaves like 3^53.
• 🎯 For 4 and 9 (length-2 cycles): odd power keeps the base's digit, even power flips it (4→6, 9→1).
• 🎯 Exponent ÷ 4 remainder 0 → use the LAST digit of the 4-cycle, not the first.
• 🎯 Units digit of a product = product of the units digits (then take units).
• 🎯 0,1,5,6 raised to any power keep the same units digit.
• ❌ Don't take the exponent mod 10 or mod 5 — the cycle length is 4 (or 2), so reduce mod 4.

Expected exam pattern: units digit of a single power, of a product of powers, or of a sum like 2^n + 3^n. 1 question, ~30 seconds.

Quick recap: Keep only the base's units digit; use its cycle (length 1, 2 or 4); reduce the exponent mod the cycle length; remainder 0 means the last digit of the cycle.

Q1. Units digit of 7^53?
A1. 53 mod 4 = 1 → 7.

Q2. Cycle of units digits for base ending in 2?
A2. 2, 4, 8, 6 (length 4).

Q3. Units digit of 2^34?
A3. 34 mod 4 = 2 → 4.

Q4. Which digits keep the same units for every power?
A4. 0, 1, 5 and 6.

Q5. Units digit of 4^odd vs 4^even?
A5. Odd power → 4, even power → 6.

Q6. Units digit of 9^even?
A6. 1 (9^odd is 9).

Q7. Exponent mod 4 gives 0 — which cycle position?
A7. The 4th (last) digit of the cycle.

Q8. Units digit of 173^53?
A8. Same as 3^53; 53 mod 4 = 1 → 3.

Q9. Units digit of 13^12 × 17^9?
A9. 3^12 → (12 mod 4 = 0) 1; 7^9 → (9 mod 4 = 1) 7; 1×7 = 7.

Q10. Cycle for base ending in 8?
A10. 8, 4, 2, 6 (length 4).

Q11. Why reduce the exponent mod 4, not mod 10?
A11. The units-digit pattern repeats every 4 powers (for 2,3,7,8).

Q12. Units digit of 5^100 + 6^100?
A12. 5 + 6 = 11 → units digit 1.

"What is the last digit of 2^57 × 3^41?" appears in almost every company test because it is a 15-second question — if you know cyclicity, and a 3-minute disaster if you try to compute the power.

What this topic tests: finding the unit (last) digit of large powers and products using the fact that unit digits repeat in short cycles.

The method

Beginner — the unit digit only depends on the unit digit

The last digit of a product depends only on the last digits of the factors. 137 × 48 ends in 7 × 8 = 56 → 6. So for powers, only the base's last digit matters.

Intermediate — the cycle table

Unit digits of powers repeat with period at most 4:

Last digit	Cycle	Length
0,1,5,6	itself	1
4	4,6	2
9	9,1	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

Advanced — the fast approach

1. Look at the base's last digit. If it is 0,1,5,6, the answer is that digit — done.
2. Otherwise take the exponent mod the cycle length (4, or 2 for 4 and 9). If the result is 0, use the last entry of the cycle.
   Example: 2^57. Cycle of 2 has length 4; 57 mod 4 = 1 → first entry → 2.

Worked example

Unit digit of 2^57 × 3^41.

• 2^57: 57 mod 4 = 1 → cycle 2,4,8,6 → 1st = 2.
• 3^41: 41 mod 4 = 1 → cycle 3,9,7,1 → 1st = 3.
• Product ends in 2 × 3 = 6.

Unit digit of 7^84. 84 mod 4 = 0 → take last of 7,9,3,1 → 1.

Where it appears

TCS NQT, Cognizant, Wipro, Capgemini quant; and as a shortcut to eliminate MCQ options in bigger problems (if the answer must end in 4, drop every option that doesn't).

Speed tricks and shortcuts

• 0,1,5,6 are "fixed points" — any power ends in the same digit. Spot these instantly.
• For 4 and 9, just ask odd or even exponent: 4^odd→4, 4^even→6; 9^odd→9, 9^even→1.
• Mnemonic for the four-cycles: "2-4-8-6" and "3-9-7-1" — the others reverse these.
• Remainder 0 means "end of the cycle", not the first term.

Key formulas

• Unit digit of powers cycles with period ≤ 4: 2→(2,4,8,6), 3→(3,9,7,1), 7→(7,9,3,1), 8→(8,4,2,6).
• 0,1,5,6 always keep the same unit digit.
• 4→(4,6) period 2; 9→(9,1) period 2.
• For aⁿ: take n mod 4 (use 4 if remainder 0) to pick the cycle position.
• Unit digit of a product = unit digit of the product of unit digits.
• Unit digit of n! is 0 for n ≥ 5.

Usage: reduce the exponent mod 4 and read the unit digit off the cycle.

Worked Example

Problem: Find the units digit of 7^95 × 3^58.

Solution: Units digits of powers repeat in cycles of length 4.

For 7: the cycle is 7, 9, 3, 1 (for exponents 1, 2, 3, 4, then it repeats).

• 95 mod 4 = 3 → take the 3rd term → units digit of 7^95 is 3.

For 3: the cycle is 3, 9, 7, 1.

• 58 mod 4 = 2 → take the 2nd term → units digit of 3^58 is 9.

Multiply just the units digits: 3 × 9 = 27, whose units digit is 7.

(Note: when the exponent is a multiple of 4, remainder 0 means you take the 4th/last term of the cycle.)

Answer: 7.