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Mathematics Dictionary
Dr. K. G. Shih
Perfect Numbers

Properties



  • Q01 | - Perfect number
  • Q02 | - Prove that 496 is a perfect number
  • Q03 | - prove that 8128 is a perfecr number
  • Q04 | - References

    • Q01. Perfect numbers

    Definition and examples
    • Definition : The sum of factors of a number = number itself
    • Examples
      • [Example] 1, 2 ,3 are factor of 6 and 1 + 2 + 3 = 6
      • [Example] 6 is also a factor of 6 but not used in sum
      • [Example] Prove that 28 is a perfect number
    • Perfecr numbers
      • 1st perfect number is 6
      • 2nd perfect number is 28
      • 3rd peferct number is 496
      • 4th perfect number is 8128
      • 5th perfect number is 33550336
      • 6th perfect number is 8589869056
      • 7th perfect number is 137438691328
      • 8th perfect number is 2305843008139952128

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    Q02. Prove that 496 is a perfect number

    Method 1 : Sum of factors
    • Factors of 496 are
    • 1, 2, 4, 8, 16, 31, 62, 124, 248, 496
    • Sum the factors without 496
    • Sum = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496
    • Hence 496 is a perfect number
    • This is the 3rd perfect number
    Method 2 : Use properties
    • Property 1 : P1 = Sum[(2*n -1)^3]
      • If n = 4, P1 = 1^3 + 3^3 + 5^3 + 7^3
      • P1 = 1 + 27 + 125 + 343
      • P1 = 496
    • P2 = (2^(n - 1))*(2^n - 1)
      • If n = 5, P2 = (2^4)*(31) = 496
    • Since P1 = P2, Hence 496 is the 3rd perfect number

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    Q03. Prove that 8128 is 4th perfecr number

    Find the 4th perfect number using properties of perfect number
    • P1 = 1^3 + 3^3 + 5^3 + 7^3 + .....
      • First 4 terms P1 = 1 + 27 + 125 + 343 = 496
      • First 5 terms P1 = 496 + 9^3 = 1225
      • First 6 terms P1 = 1225 + 11^3 = 2556
      • First 7 terms P1 = 2556 + 13^3 = 4753
      • First 8 terms P1 = 4753 + 15^3 = 8128
    • P2 = (2^(n - 1))*(2^n - 1)
      • If n = 5, P2 = (2^4)*(31) = 496
      • If n = 6, P2 = (2^5)*(61) = 1952
      • If n = 7, P2 = (2^6)*(127) = 8128
    • Since P1 = P2, Hence 8128 is the 4th perfect number
    Prove that 8128 is perfect number
    • Factors of 8128 :
      • 1 .... 8128
      • 2 .... 4064
      • 4 .... 2032
      • 8 .... 1016
      • 16 .... 508
      • 32 .... 254
      • 64 .... 127
    • Sum = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 127 + 254 + 508 + 1016 + 2032 + 4064
    • Sum = 8128

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    Q4. References

    Reference :
    • P234-p235 Computer Mathematics by Dr shih
    • Mathematics Dictionary Chapter 1
    • Keyword : Perfect number
    • Subject : Algebra section 3

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