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Bored in math class again, so I made this

  1. Oct 25, 2005 #1
    If B - A = 1,
    Then
    A^2 + A + B = B^2.

    Good for figuring out an exponent next to one you know.
    Example:
    You know 50^2 is 2500, but need 49^2.

    B^2(2500) - B(50) - A(49) = B^2(2401).

    Make sense?
    Useful, useless, w/e?
     
  2. jcsd
  3. Oct 25, 2005 #2

    TD

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    Sure:

    [tex]a^2 + a + b = b^2 \Leftrightarrow a^2 - b^2 = - \left( {a + b} \right) \Leftrightarrow \left( {a - b} \right)\left( {a + b} \right) = - \left( {a + b} \right) \Leftrightarrow \left( {a - b} \right) = - 1 \Leftrightarrow b - a = 1[/tex]

    I suppose so, but I doubt it's "new" :smile:
     
  4. Oct 25, 2005 #3
    Probably not, but I need to know that the crap I think up in math class isn't junk! ^_^
    Ty ^_^
     
  5. Oct 25, 2005 #4

    Galileo

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    Knowing and applying 'tricks' like these is usually how I am able to multiply big numbersin my head. Note: Big means 2 digits.
    For example, using (a+b)(a-b)=a^2-b^2, calculating 47 times 53 is easy:
    [tex]47 \cdot 53=(50+3)(50-3)=50^2-3^2=2500-9=2491[/tex]
     
  6. Oct 25, 2005 #5
    To find roots near 50, use (50+/-x)^2 = 2500 +/- 100x + x^2. In other words, to find 47^2 just subtract 3 from 25 to get 22 and square 3 to get 09, so 47^2=2209.

    Read it from one of Feynman's books.

    - Kamataat
     
  7. Oct 25, 2005 #6
    Unfortunately, these sorts of tricks alone will not get you very far.
     
  8. Oct 26, 2005 #7

    Galileo

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    They always work like a charm for me. Then again, I really suck at mental calculations without 'tricks'.

    To square a number a ending with a five quickly:
    Write [itex]a = 10b+5[/itex]. (division by 10 with remainder 5). b is simply the number you get after dropping the 5 mentally.
    [tex]a^2=(10b+5)^2=100b^2+100b+25=100b(b+1)+25[/tex]
    So you simply take b, multiply with the next integer and glue 25 at the end.

    25^2: 2 times 3 is 6. 'add' 25 to get 625
    85^2: 8 times 9 is 72. 'add' 25 to get 7225
    etc.
     
  9. Oct 26, 2005 #8

    quasar987

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    I think this trick given to Feynman by Hans Bethe while they were at Los Alamos!
     
  10. Oct 26, 2005 #9
    mental calculations as you coined it really depend on your memory.
     
  11. Oct 26, 2005 #10
    Another one I have found is this.

    5^2= 25 0=n
    15^2= 225 2=n
    25^2= 625 6=n
    35^2=1225 12=n
    45^2=2025 20=n
    55^2=3025 30=n

    How would I express that algebracly?
     
  12. Oct 26, 2005 #11

    shmoe

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    Try expanding (m*10+5)^2
     
  13. Oct 26, 2005 #12
    Yes, it was indeed!

    - Kamataat
     
  14. Oct 26, 2005 #13
    My achievements in math: a^2=(a+1)*(a-1)+1 for {a>N/a>0}

    in other words 49^2=50*48

    so 49^2=2400
     
  15. Oct 26, 2005 #14

    You forgot to add 1.
     
  16. Oct 31, 2005 #15
    A power has the same factors as its rational roots.
     
  17. Nov 1, 2005 #16
    49^2 = 2401.

    Anyway, if it's a^2=(a+1)*(a-1)+1
    That becomes
    49^2=(50)*(48)+1
    reduces to
    2401 = 2401.

    Just clearing that up. ^_^

    Much more simplistic version of my equation, nice Robo! ^_^''
     
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