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Highest Power of a Prime in n!

Introduction

Number Theory और combinatorics में कई problems पूछती हैं: n! को divide करने वाला किसी prime का सबसे बड़ा power कितना है? यह trailing zeros निकालने, divisibility समझने और factorial-based expressions को simplify करने में useful होता है।

ऐसे problems को solve करने के लिए इस्तेमाल होता है Legendre’s Formula, जो बताता है कि n! को किसी prime से कितनी बार divide किया जा सकता है।

Pattern: Highest Power of a Prime in n!

Pattern

Legendre’s Formula prime p का exponent n! में इस तरह देती है:

vₚ(n!) = ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + …

Sum तब तक चलता है जब तक pᵏ > n न हो जाए। Result बताता है कि p का highest power जो n! को exactly divide करता है, कितना है। यानी p^vₚ(n!) n! को divide करता है, लेकिन p^(vₚ(n!)+1) divide नहीं करेगा।

  • Use Cases: Trailing zeros निकालना (p = 5 लें), divisibility check करना, factorial ratios simplify करना।
  • Intuition: p के हर multiple में एक p होता है, p² के multiples extra p देते हैं, और इसी तरह आगे।

Step-by-Step Example

Question

100! को divide करने वाला 2 का सबसे बड़ा power निकालें।

Solution

  1. Step 1: Legendre’s Formula apply करें:

    v₂(100!) = ⌊100/2⌋ + ⌊100/4⌋ + ⌊100/8⌋ + ⌊100/16⌋ + ⌊100/32⌋ + ⌊100/64⌋ + …

  2. Step 2: हर term calculate करें:

    ⌊100/2⌋ = 50
    ⌊100/4⌋ = 25
    ⌊100/8⌋ = 12
    ⌊100/16⌋ = 6
    ⌊100/32⌋ = 3
    ⌊100/64⌋ = 1
    ⌊100/128⌋ = 0 (यहां रुकें)

  3. Step 3: सबको add करें:

    v₂(100!) = 50 + 25 + 12 + 6 + 3 + 1 = 97

  4. Final Answer:

    100! को divide करने वाला 2 का highest power है 2⁹⁷ (exponent e = 97)।

  5. Quick Check:

    Trailing zeros के लिए हम 2 और 5 की powers compare करते हैं। v₅(100!) = ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24. क्योंकि 5 की तुलना में 2 ज्यादा होते हैं, trailing zeros = 24 ✅

Quick Variations

1. Highest power of 5 in n!: Trailing zeros के लिए उपयोग → v₅(n!) = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …

2. Highest power of p in (n choose r): Formula: vₚ(C(n, r)) = vₚ(n!) - vₚ(r!) - vₚ((n-r)!).

3. जब pᵏ > n हो जाए, terms add करना बंद कर दें क्योंकि आगे contributions नहीं मिलेंगे।

Trick to Always Use

  • Step 1: Legendre’s Formula apply करें - n को p, p², p³… से divide करते जाएँ और integer parts को add करें।
  • Step 2: Trailing zeros निकालने के लिए हमेशा p = 5 लें क्योंकि 2s ज्यादा होते हैं।
  • Step 3: Combination problems के लिए factorial exponents का difference लें।
  • Step 4: Division result 0 होते ही रुक जाएँ - आगे कोई contribution नहीं मिलेगा।

Summary

Summary

  • Legendre’s Formula से prime p का exponent n! में repeated division से निकाला जाता है - p, p², p³…
  • जब division result 0 हो जाए, आगे terms जोड़ने की जरूरत नहीं।
  • निकला हुआ exponent vₚ(n!) बताता है कि p का highest power exactly n! को divide करता है।
  • Trailing zeros (p = 5), factorial divisibility और combination exponent problems में यह method बहुत useful है।

याद रखने वाला example:
v₂(100!) = 50 + 25 + 12 + 6 + 3 + 1 = 97 → highest power = 2⁹⁷.

Practice

(1/5)
1. Find the highest power of 2 that divides 50!.
easy
A. 47
B. 46
C. 48
D. 49

Solution

  1. Step 1: Apply Legendre’s formula:

    v₂(50!) = ⌊50/2⌋ + ⌊50/4⌋ + ⌊50/8⌋ + ⌊50/16⌋ + ⌊50/32⌋ + …
  2. Step 2: Compute each term:

    ⌊50/2⌋ = 25, ⌊50/4⌋ = 12, ⌊50/8⌋ = 6, ⌊50/16⌋ = 3, ⌊50/32⌋ = 1, next term ⌊50/64⌋ = 0 (stop).
  3. Step 3: Sum:

    v₂(50!) = 25 + 12 + 6 + 3 + 1 = 47.
  4. Final Answer:

    Highest power exponent = 47 → 2⁴⁷ divides 50! (Option A).
  5. Quick Check:

    All multiples of 2 contribute one 2, multiples of 4 add an extra, multiples of 8 another, etc. Summing these gives 47 ✅
Hint: Sum ⌊n/2⌋ + ⌊n/4⌋ + ⌊n/8⌋ … until terms are 0.
Common Mistakes: Forgetting to include higher powers like 16 and 32 which still contribute.
2. Find the highest power of 5 that divides 100!.
easy
A. 24
B. 23
C. 25
D. 22

Solution

  1. Step 1: Use Legendre’s formula for p = 5:

    v₅(100!) = ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ + …
  2. Step 2: Compute terms:

    ⌊100/5⌋ = 20, ⌊100/25⌋ = 4, ⌊100/125⌋ = 0 (stop).
  3. Step 3: Sum:

    v₅(100!) = 20 + 4 = 24.
  4. Final Answer:

    Highest power exponent = 24 → 5²⁴ divides 100! (Option A).
  5. Quick Check:

    This equals the number of trailing zeros of 100! because 2s are more abundant than 5s → 24 zeros ✅
Hint: Divide by 5, 25, 125, … and sum integer quotients.
Common Mistakes: Ignoring the contribution of 25, 125, etc., which add extra factors of 5.
3. Find the highest power of 3 that divides 80!.
easy
A. 35
B. 36
C. 37
D. 38

Solution

  1. Step 1: Apply Legendre’s formula for p = 3:

    v₃(80!) = ⌊80/3⌋ + ⌊80/9⌋ + ⌊80/27⌋ + ⌊80/81⌋ + …
  2. Step 2: Compute terms:

    ⌊80/3⌋ = 26, ⌊80/9⌋ = 8, ⌊80/27⌋ = 2, ⌊80/81⌋ = 0 (stop).
  3. Step 3: Sum:

    v₃(80!) = 26 + 8 + 2 = 36.
  4. Final Answer:

    Highest power exponent = 36 → 3³⁶ divides 80! (Option B).
  5. Quick Check:

    Counting multiples of 3, 9, and 27 captures every factor 3 in 80!; total 36 ✅
Hint: Divide by 3, 9, 27, … until quotient 0 and sum results.
Common Mistakes: Stopping after ⌊n/3⌋ and forgetting contributions from 9 and 27.
4. Find the highest power of 7 that divides 200!.
medium
A. 31
B. 33
C. 32
D. 34

Solution

  1. Step 1: Legendre’s formula for p = 7:

    v₇(200!) = ⌊200/7⌋ + ⌊200/49⌋ + ⌊200/343⌋ + …
  2. Step 2: Compute terms:

    ⌊200/7⌋ = 28, ⌊200/49⌋ = 4, ⌊200/343⌋ = 0 (stop).
  3. Step 3: Sum:

    v₇(200!) = 28 + 4 = 32.
  4. Final Answer:

    Highest power exponent = 32 → 7³² divides 200! (Option C).
  5. Quick Check:

    Terms beyond 49 (e.g., 343) exceed 200 so they contribute 0; total 32 ✅
Hint: Divide by 7, 49, 343, … until terms are zero and sum them.
Common Mistakes: Including p^k larger than n or forgetting the 49-term.
5. Find the highest power of 11 that divides 500!.
medium
A. 48
B. 50
C. 47
D. 49

Solution

  1. Step 1: Apply Legendre’s formula for p = 11:

    v₁₁(500!) = ⌊500/11⌋ + ⌊500/121⌋ + ⌊500/1331⌋ + …
  2. Step 2: Compute terms:

    ⌊500/11⌋ = 45, ⌊500/121⌋ = 4, ⌊500/1331⌋ = 0 (stop).
  3. Step 3: Sum:

    v₁₁(500!) = 45 + 4 = 49.
  4. Final Answer:

    Highest power exponent = 49 → 11⁴⁹ divides 500! (Option D).
  5. Quick Check:

    Only 11 and 121 contribute (11³ = 1331 > 500), giving total 49 ✅
Hint: Add ⌊n/11⌋ + ⌊n/121⌋ + … until zero.
Common Mistakes: Forgetting the 121 term which adds extra factors of 11.

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