Base Systems, Cyclicity & Unit Digits

This topic bundles three fast-scoring CAT tools: unit-digit cyclicity (last digit of large powers), base conversion, and related digit logic. Together they yield ~1–2 quick QA questions and speed up remainder problems.

Unit-digit cyclicity

The last digit of powers repeats with period ≤ 4:

Base ends in	Cycle of unit digits	Period
2	2, 4, 8, 6	4
3	3, 9, 7, 1	4
7	7, 9, 3, 1	4
8	8, 4, 2, 6	4
4	4, 6	2
9	9, 1	2
0,1,5,6	same digit	1

To find the unit digit of a^n: take n mod 4 (use 4 if remainder is 0) and read the cycle.

Base systems

A number in base b: digits × powers of b. E.g. (101)₂ = 1·4 + 0·2 + 1·1 = 5. To convert decimal→base b, divide repeatedly by b and read remainders bottom-up.

Method

1. Unit digit: reduce the exponent mod 4 (mod cycle length), read the corresponding cycle digit.
2. Base→decimal: multiply each digit by b^position and sum.
3. Decimal→base: successive division by b; remainders (bottom to top) are the digits.

Exam Tricks & Tips

• 🎯 Exponent mod 4 gives the unit digit. 7^100: 100 mod 4 = 0 → last digit of the cycle (3,9,7,1) = 1. (Check: 7⁴ = 2401 ends in 1.)
• 🎯 A remainder of 0 means the LAST cycle digit, not the first — the commonest slip.
• 🎯 0,1,5,6 are "self" digits: any power ends in the same digit — instant.
• 🎯 Base b digits are 0…b−1. A digit ≥ b signals a wrong base or an error.
• 🎯 Number of trailing zeros in n! = ⌊n/5⌋+⌊n/25⌋+⌊n/125⌋+… (count the 5s). 100! has 20+4 = 24 zeros.
• ❌ Common mistake: taking exponent mod 4 but then reading the cycle's first digit for a 0 remainder — 0 maps to the fourth (last) digit.

Expected exam pattern

1–2 QA questions on last digit of a large power/product, base conversions or base-arithmetic, and trailing zeros in factorials. MCQ + TITA. Cyclicity also shortcuts remainder questions.

Quick recap

Unit digit of a^n: reduce n mod 4 (remainder 0 → last cycle digit) and read the base's 4-cycle. Convert bases by positional powers (b→10) or repeated division (10→b). Trailing zeros in n! = sum of ⌊n/5^k⌋.

Cover the answer, recall, then check. 12 cards on cyclicity and base systems.

Q1. What is the max period of unit-digit cyclicity?
A1. 4 — the last digit of powers repeats every 4 (or fewer).

Q2. Unit-digit cycle for a base ending in 2?
A2. 2, 4, 8, 6 (period 4).

Q3. Unit-digit cycle for a base ending in 3?
A3. 3, 9, 7, 1 (period 4).

Q4. Unit digit of 7^100?
A4. 100 mod 4 = 0 → last digit of (7,9,3,1) = 1.

Q5. Which final digits give period 1 (self)?
A5. 0, 1, 5, 6 — any power ends in the same digit.

Q6. When the exponent mod 4 = 0, which cycle digit?
A6. The last (4th) digit of the cycle, not the first.

Q7. Convert (101)₂ to decimal.
A7. 1·4 + 0·2 + 1·1 = 5.

Q8. How do you convert decimal to base b?
A8. Divide repeatedly by b; read the remainders from bottom to top.

Q9. Allowed digits in base b?
A9. 0 to b−1 (a digit ≥ b signals an error).

Q10. Trailing zeros in n! — formula?
A10. ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + … (count the factors of 5).

Q11. How many trailing zeros in 100!?
A11. 20 + 4 = 24.

Q12. The classic cyclicity mistake?
A12. Reading the first cycle digit for a 0 remainder — 0 maps to the last digit.

This topic bundles three CAT favourites that all exploit patterns and place-value: non-decimal base systems (binary, base-7, etc.), the cyclicity of unit digits (the last digit of huge powers repeats in short cycles), and finding the units digit of monstrous expressions. None require heavy computation — they require spotting the repeating cycle and reading the right place value.

What this tests

Positional and cyclic reasoning — converting between number bases, and using the periodicity of last digits to evaluate the units digit of large powers and products.

The method

Beginner — base conversion both ways

A number in base b uses place values …b³, b², b¹, b⁰. To decimal: multiply each digit by its place value and sum — (243)₇ = 2·49 + 4·7 + 3 = 98 + 28 + 3 = 129. From decimal: repeatedly divide by b, collecting remainders bottom-up — 129 ÷ 7 = 18 r3, 18 ÷ 7 = 2 r4, 2 ÷ 7 = 0 r2 → read up: (243)₇. In base b, the largest single digit is b−1 (so base-7 uses only digits 0–6).

Intermediate — unit-digit cyclicity

The last digit of a^n cycles with a small period:

• Digits 0, 1, 5, 6 → always end in themselves (cycle length 1).
• Digits 4, 9 → cycle length 2 (4: 4,6,4,6…; 9: 9,1,9,1…).
• Digits 2, 3, 7, 8 → cycle length 4.
  For length-4 digits, find n mod 4: e.g., unit digit of 7^n depends on n mod 4 → 7¹=7, 7²=9, 7³=3, 7⁴=1, then repeats. If n mod 4 = 0, use the 4th term (the last in the cycle).

Advanced — the FAST CAT approach: reduce exponents mod the cycle length

• To get the units digit of a^b: take the units digit of a, find its cycle length (1, 2, or 4), reduce b mod (cycle length) (treating remainder 0 as the full cycle position), and read the corresponding last digit.
• For products/sums, the units digit = units digit of the product/sum of the parts' units digits.
• For last two digits, cycles have length dividing 20 (mod 100) — the powers of the unit digit mod 100 repeat; or use binomial expansion for numbers ending in 1.
• Base + remainder link: the last digit of N in base b is simply N mod b — a neat bridge between bases and remainders.

Worked example

Question: What is the units digit of 3^{47} + 7^{53}?
For 3^47: units-digit cycle of 3 is {3, 9, 7, 1}, length 4. 47 mod 4 = 3 → 3rd term = 7.
For 7^53: cycle of 7 is {7, 9, 3, 1}, length 4. 53 mod 4 = 1 → 1st term = 7.
Sum of units digits = 7 + 7 = 14 → units digit = 4. No power was ever computed; reducing 47 and 53 mod 4 and reading the cycle gave both last digits instantly. Answer: 4.

CAT relevance

Unit-digit and base-system questions appear in CAT QA and are pure-speed items — a full mark in under 30 seconds once you know the cycles. Cyclicity also underlies remainder problems (last digit is mod 10). Base systems occasionally appear in DILR-style number puzzles too, making this a cross-utility topic.

Speed tricks & shortcuts

• Cycle lengths: {0,1,5,6}→1; {4,9}→2; {2,3,7,8}→4. Reduce the exponent mod the cycle length.
• Remainder 0 → use the last term of the cycle, not the first.
• Units digit of a product = product of the units digits (then take its units digit).
• Last digit in base b = N mod b.
• Mnemonic — "Cycle length 4 for 2,3,7,8; halve the rest."

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: (a) Find the unit digit of 7^123. (b) Convert the binary number 1011 (base 2) to decimal, and convert decimal 45 to base 2.

Solution:
(a) Unit digits of powers repeat in a short CYCLE. For 7:
7^1 = 7, 7^2 = 49 (→9), 7^3 = 343 (→3), 7^4 = 2401 (→1), then 7^5 ends in 7 again. The unit-digit cycle of 7 is (7, 9, 3, 1) with length 4.
To find 7^123's unit digit, find where 123 lands in the cycle: 123 mod 4 = 3 (since 120 is divisible by 4, remainder 3). A remainder of 3 corresponds to the 3rd term of the cycle → 3.
(If the remainder were 0, you take the LAST term of the cycle, i.e. the 4th, which is 1.)
So 7^123 ends in 3.

(b) Binary 1011 to decimal: multiply each bit by its place value (powers of 2, right to left: 2^0, 2^1, 2^2, 2^3):
1×2^3 + 0×2^2 + 1×2^1 + 1×2^0 = 8 + 0 + 2 + 1 = 11. So 1011(base 2) = 11(base 10).

Decimal 45 to binary: repeatedly divide by 2 and read remainders bottom-up:
45 ÷ 2 = 22 r1
22 ÷ 2 = 11 r0
11 ÷ 2 = 5 r1
5 ÷ 2 = 2 r1
2 ÷ 2 = 1 r0
1 ÷ 2 = 0 r1
Reading remainders from last to first: 101101. Check: 32 + 8 + 4 + 1 = 45 ✓.

Answer: (a) 7^123 ends in 3 (cycle 7,9,3,1 of length 4; 123 mod 4 = 3 → third term). (b) 1011(2) = 11(10); 45(10) = 101101(2).