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Cyclicity of Digits (Last Digit Problems)

Introduction

Last-digit (cyclicity) problems में बड़े powers या repeated multiplications की last digit बिना पूरा calculation किए निकाली जाती है। Aptitude exams में ये common हैं क्योंकि last digit हमेशा एक repeating pattern या cycle follow करती है। Cycle पता होने पर बड़े से बड़े exponent की last digit तुरंत मिल जाती है।

Pattern: Cyclicity of Digits (Last Digit Problems)

Pattern

Powers की last digit एक fixed cycle में repeat होती है। Base की last digit लें, cycle पहचानें, और exponent को cycle length से mod करके last digit तय करें।

  • Step 1: Base की last digit लें: सिर्फ last digit ही matter करती है। Base को (base mod 10) से replace कर दें।
  • Step 2: Standard cycles याद रखें:
    • 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: Exponent reduce करें: अगर cycle length = L है, तो r = exponent mod L लें। अगर r = 0, तो r = L मानें। Cycle के r-th term से last digit मिल जाता है।
  • Step 4: Special cases: Base की last digit 0, 1, 5 या 6 हो → last digit हमेशा वही रहती है।

Step-by-Step Example

Question

72024 की last digit निकालें।

Solution

  1. Step 1: Cycle पहचानें

    7 के powers की last digits → [7, 9, 3, 1] इसलिए cycle length L = 4.

  2. Step 2: Exponent reduce करें

    2024 mod 4 = 0 → r = 4 लें (क्योंकि remainder 0 होने पर cycle का आखिरी term लेते हैं)।

  3. Step 3: r-th term चुनें

    Cycle का 4th element = 1

  4. Final Answer:

    72024 की last digit = 1

  5. Quick Check:

    हर 4th power of 7 की last digit 1 होती है → 2024 multiple of 4 → last digit 1. ✅

Quick Variations

1. (am × bn) की last digit → दोनों की last digit अलग निकालें और multiply करें।

2. a(bc) जैसे expressions में top exponent को cycle length से mod करें।

3. Base ending with 0, 1, 5, 6 → कोई calculation ज़रूरी नहीं, last digit हमेशा same रहती है।

Trick to Always Use

  • Step 1: सिर्फ base की last digit पर ध्यान दें।
  • Step 2: Standard cycles याद रखें (खासकर 2, 3, 7, 8 के)।
  • Step 3: Exponent को cycle length से mod करें और r-th term चुनें।
  • Step 4: यदि remainder 0 हो, cycle का आखिरी term लें।

Summary

Summary

Cyclicity of Digits problems में:

  • सिर्फ base की last digit matter करती है।
  • हर digit (0-9) की powers एक fixed cycle follow करती हैं।
  • Exponent को cycle length से mod करके correct term मिल जाता है।
  • Huge exponents होने पर भी last digit instantly मिल जाती है।

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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