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Special Properties & Number Puzzles

Introduction

Special-number properties and puzzles are a favourite in aptitude and olympiad-style questions. These problems test recognition of well-known classes (perfect, amicable, narcissistic/Armstrong, Kaprekar, automorphic, Harshad, triangular/square numbers, repunits, etc.) and ability to apply simple checks or constructions to decide membership or produce the required result.

Learning the definitions, short tests, and a few constructive procedures (like Kaprekar's routine) helps solve many "trick" and pattern problems quickly.

Pattern: Special Properties & Number Puzzles

Pattern

Each special property has a short defining test - apply the test step-by-step and use small computations (digit sums, factor sums, powering digits, rearrangements) to verify the property.

  • Armstrong / Narcissistic number (n-digit): Sum of each digit^n equals the number.
    Test: If number has n digits, compute Σ (digitn) and compare to the number.
    Example formula: for 153 (3-digit): 1³ + 5³ + 3³ = 1 + 125 + 27 = 153.
  • Perfect number: Sum of proper divisors equals the number.
    Test: Compute sum of divisors d where d < n and d|n; check equality.
    Example: 28: 1 + 2 + 4 + 7 + 14 = 28.
  • Amicable pair (a, b): sum of proper divisors of a = b and sum for b = a.
    Test: Compute σ(a)-a and σ(b)-b and compare.
  • Kaprekar’s routine (4-digit example → 6174): For a 4-digit number with at least two different digits: sort digits desc (D) and asc (A), compute D - A, repeat; many numbers reach 6174 in ≤ 7 iterations.
    Procedure: Repeat until fixed point or loop.
  • Automorphic number: Square ends with the number itself.
    Test: Compute n² and check last k digits (k = number of digits of n).
  • Harshad / Niven number: Number divisible by sum of its digits.
    Test: Compute digit sum s; check n mod s = 0.
  • Triangular / Square / Pentagonal tests: Use inverse formulas: triangular n ↔ solve k(k+1)/2 = n and check integral k; square ↔ sqrt(n) integer, etc.
  • Repunits / Repeating-digit numbers: Recognize forms (111..1) and use divisibility or formula R_k = (10^k - 1)/9 for algebraic checks.
  • Digital-root tricks: Useful for quick divisibility checks (mod 9), repeated-sum tests, and spotting impossibilities.

Step-by-Step Example

Question

Starting from 3524, apply Kaprekar’s 4-digit routine until you reach the fixed point; show steps and final result.

Solution

  1. Step 1: Ensure 4 digits:

    3524 is already 4 digits (if fewer digits pad with leading zeros).
  2. Step 2: Form descending and ascending numbers:

    Descending D = 5432 (digits sorted high→low).
    Ascending A = 2345 (digits sorted low→high).
  3. Step 3: Compute D - A:

    5432 - 2345 = 3087.
  4. Step 4: Repeat with 3087:

    D = 8730, A = 0378 (i.e. 378), 8730 - 378 = 8352.
  5. Step 5: Repeat with 8352:

    D = 8532, A = 2358 → 8532 - 2358 = 6174.
  6. Step 6: One more iteration shows fixed point:

    From 6174: D = 7641, A = 1467 → 7641 - 1467 = 6174 (fixed).

    Final Answer:

    Kaprekar routine from 3524 reaches 6174 in 3 steps and then remains fixed.
  7. Quick Check:

    Verify each subtraction and leading-zero handling (e.g., treat 0378 as 378 but keep it 4-digit during ordering). The routine ended at 6174 which is the known Kaprekar constant for 4-digit numbers with at least two distinct digits. ✅

Quick Variations

1. Kaprekar’s routine for 3-digit numbers often ends at 495 (3-digit Kaprekar constant) with similar procedure.

2. To test automorphic property for n (k digits), compute n² mod 10^k and compare with n.

3. For Armstrong numbers, n-digit power changes with n: 4-digit Armstrong example 9474: 9⁴+4⁴+7⁴+4⁴ = 9474.

4. For perfect/amicable checks, compute proper-divisor sums using prime factorization to speed up divisors sum calculation.

Trick to Always Use

  • Step 1 → Read the definition first (precise test), then reduce the problem to a small computation (digit-sum, powering digits, divisor-sum, modular check).
  • Step 2 → Use modular arithmetic (mod 9, mod 10^k) for quick impossibility checks before heavy computation.
  • Step 3 → For repeated routines (Kaprekar, iterations), keep numbers in fixed-width form (pad leading zeros) and track until fixed point or a short cycle appears.
  • Step 4 → For divisor sums, use prime factorization: if n = Π p_i^{a_i} then sum-of-divisors σ(n) = Π (p_i^{a_i+1} - 1)/(p_i - 1); proper-sum = σ(n) - n.

Summary

Summary

  • Memorise concise definitions and tests for special-number classes (Armstrong, perfect, automorphic, Harshad, Kaprekar, etc.) and apply the corresponding small computation (digit powers, divisor sums, modular checks, or iterative routines).
  • Use modular reductions (e.g., mod 9, mod 10ᵏ) and digit-based shortcuts to rule out impossibilities quickly before performing heavier calculations.
  • For iterative routines (Kaprekar, repeated digit-power checks), keep numbers in fixed width (pad leading zeros), iterate until a fixed point or short cycle, and document steps to avoid mistakes.
  • Always perform a quick verification step (e.g., recompute divisor sums for perfect/amicable checks, verify n² mod 10ᵏ for automorphic numbers) to confirm the result.

Example to remember:
Kaprekar’s routine starting from 3524 → 5432 - 2345 = 3087 → 8730 - 0378 = 8352 → 8532 - 2358 = 6174 (fixed point).

Practice

(1/5)
1. Which of the following numbers is an Armstrong number?
easy
A. 153
B. 154
C. 155
D. 156

Solution

  1. Step 1: Count digits:

    153 has 3 digits, so raise each digit to the power 3.
  2. Step 2: Compute:

    1³ + 5³ + 3³ = 1 + 125 + 27 = 153.
  3. Final Answer:

    153 equals the sum of its digit-cubes → 153 → Option A.
  4. Quick Check:

    Recompute 1 + 125 + 27 = 153 ✅
Hint: For an n-digit number, sum each digit^n and compare to the number.
Common Mistakes: Using wrong power (e.g., square instead of cube for 3-digit numbers).
2. Which of the following is a perfect number?
easy
A. 6
B. 8
C. 10
D. 12

Solution

  1. Step 1: Recall definition:

    A perfect number equals the sum of its proper divisors (divisors excluding the number itself).
  2. Step 2: Compute for 6:

    Proper divisors of 6 are 1, 2, 3 → sum = 1 + 2 + 3 = 6.
  3. Final Answer:

    Sum of proper divisors equals 6 → 6 → Option A.
  4. Quick Check:

    Other choices: 8 → 1+2+4=7 ≠ 8; 10 → 1+2+5=8 ≠ 10 ✅
Hint: Sum proper divisors and check equality with the number.
Common Mistakes: Including the number itself when summing divisors.
3. Which of the following numbers is automorphic (its square ends with the number itself)?
easy
A. 24
B. 76
C. 57
D. 98

Solution

  1. Step 1: Compute square and check ending digits:

    We need n² to end with n.
  2. Step 2: Check 76:

    76² = 5776 → last two digits = 76, so 76 is automorphic.
  3. Final Answer:

    76's square ends with 76 → 76 → Option B.
  4. Quick Check:

    Verify other options quickly: 24²=576 (ends 76), 57²=3249 (ends 49) → only 76 matches ✅
Hint: Check n² mod 10ᵏ (k = number of digits) and compare to n.
Common Mistakes: Checking only one digit for multi-digit numbers instead of k digits.
4. What is the Kaprekar constant for 4-digit numbers (the fixed point reached by the Kaprekar routine)?
medium
A. 495
B. 153
C. 6174
D. 9999

Solution

  1. Step 1: Recall Kaprekar routine:

    For 4-digit numbers (with at least two distinct digits) repeatedly form D (digits desc) and A (asc), compute D-A; iterate.
  2. Step 2: Known fixed point:

    For 4-digit numbers this routine converges to 6174 (Kaprekar constant) and then remains fixed: 7641-1467=6174.
  3. Final Answer:

    Kaprekar 4-digit constant is 6174 → Option C.
  4. Quick Check:

    Example: 3524 → 5432-2345=3087 → 8730-0378=8352 → 8532-2358=6174 → fixed point ✅
Hint: For 4-digit Kaprekar routine, the attractor (when it exists) is 6174.
Common Mistakes: Not padding with leading zeros for numbers with fewer than 4 digits during iterations.
5. Which of the following pairs is an amicable pair (sum of proper divisors of each equals the other)?
medium
A. (220, 285)
B. (1185, 1210)
C. (2620, 2925)
D. (220, 284)

Solution

  1. Step 1: Recall amicable definition:

    For pair (a,b): sum_proper_divisors(a)=b and sum_proper_divisors(b)=a.
  2. Step 2: Known classical pair:

    (220, 284) is the smallest known amicable pair: proper divisors of 220 sum to 284, and those of 284 sum to 220.
  3. Final Answer:

    (220, 284) is amicable → Option D.
  4. Quick Check:

    Proper divisors of 220: 1+2+4+5+10+11+20+22+44+55+110 = 284 ✅
Hint: Use known amicable pairs or compute proper-divisor sums via prime-factorization σ(n)-n.
Common Mistakes: Confusing amicable numbers with perfect numbers (where proper-divisor sum equals the same number).

Mock Test

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