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Cyclicity of Digits (Last Digit Problems)

Introduction

Last-digit (cyclicity) problems involve finding the last digit of large powers or repeated multiplications without performing the full calculation. These problems are common in aptitude tests because the last digit follows a repeating pattern or cycle. Once you know the cycle, you can quickly find the last digit for huge exponents.

Pattern: Cyclicity of Digits (Last Digit Problems)

Pattern

The last digit of powers repeats in a fixed cycle. Reduce the base to its last digit, find the cycle, and use the remainder of the exponent divided by the cycle length to find the last digit.

  • Step 1: Reduce base to last digit: Only the last digit of the base affects the result. Replace the base with (base mod 10).
  • Step 2: Remember these standard cycles:
    • 0 → [0] (length 1)
    • 1 → [1] (length 1)
    • 2 → [2, 4, 8, 6] (length 4)
    • 3 → [3, 9, 7, 1] (length 4)
    • 4 → [4, 6] (length 2)
    • 5 → [5] (length 1)
    • 6 → [6] (length 1)
    • 7 → [7, 9, 3, 1] (length 4)
    • 8 → [8, 4, 2, 6] (length 4)
    • 9 → [9, 1] (length 2)
  • Step 3: Reduce the exponent: If the cycle length = L, find remainder r = (exponent mod L). If r = 0, use r = L. The r-th number in the cycle gives the last digit.
  • Step 4: Handle special cases: For base ending in 0, 1, 5, or 6 → last digit always remains the same.

Step-by-Step Example

Question

Find the last digit of 72024.

Solution

  1. Step 1: Identify the cycle:

    The last digit of powers of 7 repeats as [7, 9, 3, 1]. Hence, the cycle length L = 4.

  2. Step 2: Reduce the exponent:

    2024 mod 4 = 0 → when remainder is 0, use r = 4.

  3. Step 3: Find the r-th term in the cycle:

    The 4th element in [7, 9, 3, 1] is 1.

  4. Final Answer:

    Last digit of 72024 is 1.

  5. Quick Check:

    Since every 4th power of 7 ends with 1, 72024 (multiple of 4) → last digit 1 ✅

Quick Variations

1. Find the last digit of (am × bn): compute both separately and multiply the last digits.

2. For expressions like a(bc), reduce the top exponent mod the cycle length.

3. If the base ends with 0, 1, 5, or 6 → no need to calculate further, the last digit stays the same.

Trick to Always Use

  • Step 1: Focus only on the last digit of the base.
  • Step 2: Memorize the standard cycles (especially for 2, 3, 7, 8).
  • Step 3: Reduce the exponent mod cycle length and pick that term.
  • Step 4: If remainder is 0, take the last number in the cycle.

Summary

Summary

In Cyclicity of Digits problems:

  • Only the last digit matters.
  • Each digit (0-9) follows a fixed repetition cycle.
  • Reduce the exponent modulo the cycle length to locate the correct last digit.
  • This method works for huge exponents instantly without actual multiplication.

Practice

(1/5)
1. Find the last digit of 2^9.
easy
A. 2
B. 4
C. 8
D. 6

Solution

  1. Step 1: Use only the last digit of the base:

    The last digit of 2^k depends on the cycle for 2: [2, 4, 8, 6].

  2. Step 2: Find cycle position:

    Cycle length L = 4. Compute 9 mod 4 = 1, so use the 1st element of the cycle.

  3. Step 3: Read the last digit:

    Cycle[1] = 2 → last digit = 2.

  4. Final Answer:

    Last digit = 2 → Option A.

  5. Quick Check:

    2^9 = 512 → last digit 2 ✅
Hint: For base 2, use cycle [2,4,8,6] and take exponent mod 4.
Common Mistakes: Forgetting to use exponent mod cycle length or indexing the cycle wrongly.
2. Find the last digit of 3^7.
easy
A. 3
B. 7
C. 9
D. 1

Solution

  1. Step 1: Reduce base to its last digit:

    Use last-digit cycle for 3: [3, 9, 7, 1].

  2. Step 2: Reduce exponent modulo cycle length:

    Cycle length L = 4. 7 mod 4 = 3, so use the 3rd element of the cycle.

  3. Step 3: Read the last digit:

    Cycle[3] = 7 → last digit = 7.

  4. Final Answer:

    Last digit = 7 → Option B.

  5. Quick Check:

    3^7 = 2187 → last digit 7 ✅
Hint: Remember 3's cycle [3,9,7,1]; take exponent mod 4 to choose position.
Common Mistakes: Using the wrong cycle index when exponent modulo result is 0 or off-by-one indexing.
3. Find the last digit of 7^25.
easy
A. 7
B. 9
C. 3
D. 1

Solution

  1. Step 1: Use last-digit cycle for 7:

    7 → [7, 9, 3, 1], cycle length L = 4.

  2. Step 2: Reduce exponent modulo cycle length:

    25 mod 4 = 1, so use the 1st element of the cycle.

  3. Step 3: Read the last digit:

    Cycle[1] = 7 → last digit = 7.

  4. Final Answer:

    Last digit = 7 → Option A.

  5. Quick Check:

    Powers repeat every 4; 7^25 has same last digit as 7^1 → 7 ✅
Hint: For base 7, cycle [7,9,3,1]; if exponent ≡ 1 (mod 4) answer is 7.
Common Mistakes: Not reducing exponent modulo 4 or misreading cycle order.
4. Find the last digit of 8^16.
medium
A. 2
B. 4
C. 6
D. 8

Solution

  1. Step 1: Last-digit cycle for 8:

    8 → [8, 4, 2, 6], cycle length L = 4.

  2. Step 2: Reduce exponent modulo cycle length:

    16 mod 4 = 0. When remainder = 0, use the 4th element of the cycle.

  3. Step 3: Read the 4th element:

    Cycle[4] = 6 → last digit = 6.

  4. Final Answer:

    Last digit = 6 → Option C.

  5. Quick Check:

    Because 16 is a multiple of 4, 8^16 ends with the same digit as 8^4, which is 6 ✅
Hint: If exponent mod cycle = 0, pick last cycle element; 8's cycle last element is 6.
Common Mistakes: Using remainder 0 as index 0 instead of mapping to the last cycle element.
5. Find the last digit of (2^5 × 3^7).
medium
A. 2
B. 6
C. 8
D. 4

Solution

  1. Step 1: Find last digit of each factor separately:

    For 2^5: cycle [2,4,8,6], 5 mod 4 = 1 → last digit = 2. For 3^7: cycle [3,9,7,1], 7 mod 4 = 3 → last digit = 7.

  2. Step 2: Multiply the last digits and take final last digit:

    2 × 7 = 14 → last digit of product = 4.

  3. Final Answer:

    Last digit = 4 → Option D.

  4. Quick Check:

    Compute small powers if needed: 2^5 ends with 2, 3^7 ends with 7, 2×7 gives 14 → last digit 4 ✅
Hint: Compute last digits separately, multiply them, then take last digit of the result.
Common Mistakes: Multiplying full large numbers instead of using last-digit cycles for each factor.

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