Can you Solve Mathematical Problem from Harvard University Entrance Exam ?

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  • Опубліковано 21 лис 2024

КОМЕНТАРІ • 26

  • @KyyTyy
    @KyyTyy День тому +1

    42=32+8+2
    2¹+2³+2⁵
    (2^p)=2¹
    (2^q)=2³
    (2^r)=2⁵
    p=1, q=3,r=5

  • @oahuhawaii2141
    @oahuhawaii2141 18 годин тому +1

    There's no domain restriction on p, q, and r.
    2^p + 2^q + 2^r = 42 , p < q < r
    Let p = 0 and q = 1 .
    2⁰ + 2¹ + 2ʳ = 42
    2ʳ = 39
    r = log₂(39)
    One of an infinite number of solutions:
    p = 0 , q = 1 , r = log(39)/log(2) ≈ 5.2854022188622...
    Alternate solution:
    42 = 2Ah = 101010b = 2¹ + 2³ + 2⁵
    p = 1 , q = 3 , r = 5

    • @oahuhawaii2141
      @oahuhawaii2141 3 години тому

      I like this best:
      p = e
      q = π
      r = log₂(42-2^e-2^π) ≈ 4.73303627...

    • @oahuhawaii2141
      @oahuhawaii2141 2 години тому

      I like this solution best:
      p = e
      q = π
      r = log₂(42-2^e-2^π)
      = log(42--2^e-2^π)/log(2)
      ≈ 4.7330...

  • @kareolaussen819
    @kareolaussen819 День тому +1

    This is a question of finding the binary representation of 42. Take r>q>p and define r to be the largest number for which 2^r ≤ 42. Subtract and repeat the procedure for the remainder (here 10) until the remainder is zero. Here we find 42 = 32 + 8 + 2 = 2^5 + 2^3 + 2^1. I.e, r, q, p = 5, 3, 1.

    • @oahuhawaii2141
      @oahuhawaii2141 3 години тому

      No other permutation because of the relational condition.

  • @michaeledwards2251
    @michaeledwards2251 21 годину тому

    A Douglas Adams "Hitch Hikers" guide to the Galaxy reference. The computer gives the answer "42" to the question "life, the universe, and everything". Given computers today use binary, the question assumes binary is still used, and asks the question, How would the computer represent 42 ?

    • @oahuhawaii2141
      @oahuhawaii2141 3 години тому

      Some solid-state memory storage devices save data in di-bit form; each memory cell has 4 states. That's base 4, so 42 will be encoded as 222₄ . [My circuit used an MCU with internal EEPROM based on Fowler-Nordheim tunneling di-bit memory technology.]
      BTW, the story continues with the people designing an even more powerful computer to figure out what was the question that resulted in 42. The answer is "What is 6 times 9?" Note that: 6*9 = 54 = 42₁₃ . [I can see base 6 or 12 being used, but not base 13.]

    • @michaeledwards2251
      @michaeledwards2251 37 хвилин тому

      @@oahuhawaii2141
      The minimum number of states to represent numbers occurs with base 3 : too awkward to work with. Both base 2 and 4 are equally efficient.
      Base 13 covers the "Bakers Dozen" as a unit digit : binary 1101 : base 3 as 111 : base 13 makes sense as a triple digit base 3 number.

  • @에스피-z2g
    @에스피-z2g День тому +1

    Solution by insight
    2^p+2^q+2^r=42
    Among 2,4,8,16,32
    2+8+32=42
    The only possible combination.
    p=1, q=3,r=5

    • @superacademy247
      @superacademy247  День тому

      Thanks for sharing your insight 💯💕✅💪

    • @oahuhawaii2141
      @oahuhawaii2141 3 години тому

      Untrue. The requirement is p < q < r , but they can be real numbers. Here are 4 of an infinite number of valid solutions:
      p = 0 , q = 1 , r = log₂(39) = log(39)/log(2) ≈ 5.2854022188622...
      p = 2 , q = 4 , r = log₂(22) = log(22)/log(2) ≈ 4.4594316186372...
      p = 0 , r = 5 , q = log₂(9) = log(9)/log(2) ≈ 3.1699250014423...
      p = e , q = π , r = log₂(42-2^e-2^π) = log(42--2^e-2^π)/log(2) ≈ 4.7330362716713...

    • @oahuhawaii2141
      @oahuhawaii2141 3 години тому

      Untrue. The requirement is p < q < r , but they can be real numbers. Here are 4 of an infinite number of valid solutions:
      p = 0 , q = 1 , r = log₂(39) = log(39)/log(2) ≈ 5.2854022188622...
      p = 2 , q = 4 , r = log₂(22) = log(22)/log(2) ≈ 4.4594316186372...
      p = 0 , r = 5 , q = log₂(9) = log(9)/log(2) ≈ 3.1699250014423...
      p = e , q = π , r = log₂(42-2^e-2^π) = log(42--2^e-2^π)/log(2) ≈ 4.7330362716713...

    • @oahuhawaii2141
      @oahuhawaii2141 3 години тому

      Untrue. The requirement is p < q < r , but they can be real numbers. Here are 4 of an infinite number of valid solutions:
      p = 0 , q = 1 , r = log₂(39) = log(39)/log(2) ≈ 5.2854022188622...
      p = 2 , q = 4 , r = log₂(22) = log(22)/log(2) ≈ 4.4594316186372...
      p = 0 , r = 5 , q = log₂(9) = log(9)/log(2) ≈ 3.1699250014423...
      p = e , q = π , r = log₂(42-2^e-2^π) = log(42--2^e-2^π)/log(2) ≈ 4.7330362716713...

    • @oahuhawaii2141
      @oahuhawaii2141 3 години тому

      Untrue. The requirement is p < q < r , but they can be real numbers. Here are 4 of an infinite number of valid solutions:
      p = 0 , q = 1 , r = log₂(39) = log(39)/log(2) ≈ 5.2854022...
      p = 2 , q = 4 , r = log₂(22) = log(22)/log(2) ≈ 4.4594316...
      p = 0 , r = 5 , q = log₂(9) = log(9)/log(2) ≈ 3.1699250...
      p = e , q = π , r = log₂(42-2^e-2^π) = log(42--2^e-2^π)/log(2) ≈ 4.733036...

  • @key_board_x
    @key_board_x 15 годин тому +1

    2^(p) + 2^(q) + 2^(r) = 42
    2^(p) + 2^(q + p - p) + 2^(r + p - p) = 42
    2^(p) + 2^(p + q - p) + 2^(p + r - p) = 42
    2^(p) + [2^(p) * 2^(q - p)] + [2^(p) * 2^(r - p)] = 42
    2^(p).[1 + 2^(q - p) + 2^(r - p)] = 42 ← there is an odd number in the second bracket
    2^(p).[1 + 2^(q - p) + 2^(r - p)] = 2 * 21
    2^(p).[1 + 2^(q - p) + 2^(r - p)] = 2^(1) * 21 → you can deduce that:
    2^(p) = 2^(1) → p = 1
    [1 + 2^(q - p) + 2^(r - p)] = 21
    2^(q - p) + 2^(r - p) = 20 → recall: p = 1
    2^(q - 1) + 2^(r - 1) = 20
    [2^(q) * 2^(- 1)] + [2^(r) * 2^(- 1)] = 20 → recall: x^(- a) = 1/x^(a) → then: 2^(- 1) = 1/2^(1) = 1/2
    [2^(q) * 1/2] + [2^(r) * 1/2] = 20
    (1/2).[2^(q) + 2^(r)] = 20
    2^(q) + 2^(r) = 40
    2^(q) + 2^(r + q - q) = 40
    2^(q) + 2^(q + r - q) = 40
    2^(q) + [2^(q) * 2^(r - q)] = 40
    2^(q).[1 + 2^(r - q)] = 40 ← there is an odd number in the second bracket
    2^(q).[1 + 2^(r - q)] = 8 * 5
    2^(q).[1 + 2^(r - q)] = 2^(3) * 5 → you can deduce that:
    2^(q) = 2^(3) → q = 3
    [1 + 2^(r - q)] = 5
    2^(r - q) = 4
    2^(r - q) = 2^(2)
    r - q = 2
    r = 2 + q → recall: q = 3
    → r = 5
    Summarize
    p = 1
    q = 3
    r = 5

  • @saltydog584
    @saltydog584 День тому +1

    Just write it as a binary number. 32=10101 then convert back to base 10. 10101=32 + 8 + 2 = 2^5 + 2^3+2^1so P is 5, Q is 3 and r =1 (only works when P,Q&R are whole numbers)

    • @pijusuaac.chattopadhakjm2676
      @pijusuaac.chattopadhakjm2676 7 годин тому

      How is 32(base 10)=10101(base 2)?

    • @oahuhawaii2141
      @oahuhawaii2141 4 години тому

      @pijusuaac.chattopadhakjm2676: OP is very sloppy. I would show the conversion as:
      42 = 2Ah = 101010b = 2⁵ + 2³ + 2¹

  • @TheNizzer
    @TheNizzer 14 годин тому

    32,16,8,4,2,1,0,…
    Which 3 sum to 42?
    32,8,2
    P=1, q=3, r=5.
    This is supposed to be difficult? Given non integer powers of 2 are irrational….

    • @oahuhawaii2141
      @oahuhawaii2141 3 години тому

      Untrue, as 2 raised to a negative integer is a fraction, which isn't irrational. Example: 2⁻⁵ = 1/32 = 0.03125 .

    • @TheNizzer
      @TheNizzer Годину тому

      @ and three negative powers add up to more than 1? They aren’t under consideration, for obvious reasons.