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Mental calculation of squares

  1. May 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Find A:

    A2 = (4/5)(18292 + 12982)

    in about 3min, because this comes at the end of a rather difficult geometry problem with 6 min for the entire question. (edit: Yes, calculators weren't allowed because it was a competition. I have verified that everyone comes to this step, so I didn't make any mistakes before that.)

    2. Relevant equations

    None.

    3. The attempt at a solution

    I'm quite sure there's some algebraic manipulation to help you with this one, but I can't think of it. I did the 18292 and 12982 the 'brute force' way, then added them together for the sum. This step took the longest - more than 3min for me.

    Then I converted this to standard form so as to mentally estimate the common logarithm to 3 decimal points. I also know common lg 4 and lg 5 by memory to 3 decimal points - then I took a 'guess' at A:

    lg A = (lg 4 + lg(18292+12982) - lg 5)/2

    Comes around to 2 * 103~ rather quickly, correct a bit and 'try' a few values, and I found 2006 eventually (edit: , which is the correct answer.)

    Also did think of putting it in (1829+1298)2 - 2*1829*1298, but it still takes about as long. Alternatively, I had a suggestion to use 18002 + 292 + 2*1800*29, and similar for 12002 + 982 + 2*1200*98, which takes about as long for me.

    Does anyone see an algebraic manipulation to make this easier? Thanks in advance ;)
     
    Last edited: May 30, 2009
  2. jcsd
  3. May 30, 2009 #2

    Mark44

    Staff: Mentor

    Not really algebraic manipulation -- just factoring.
    A2 = 4/5[18292 + 12982]
    = 4/5[592 * 312 + 542 * 112 * 22]
    = 4/5[592(312 + 112 * 4)]
    = 4/5[592(961 + 484)]
    = 4/5[592(1445)] = 4/5[592 * (5 * 289)]
    = 4[592 * 172]

    ==> A = 2*59*17 = 2006
     
  4. May 30, 2009 #3
    Oh, I didn't see 59^2 and 17^2 (don't come across these primes often)! Thanks very much. I wish you a good weekend.
     
  5. May 31, 2009 #4

    robphy

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    Science Advisor
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    Would 1829=1830-1 be useful?
     
  6. May 31, 2009 #5

    Hurkyl

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    What was the problem? Working it differently might have lead to something with simpler arithmetic at the end.


    If you're really serious about speed arithmetic, then you might want to practice multiplication; you should be able to compute (4/5)(1829^2 + 1298^2) in under three minutes the direct way... and even faster with shortcuts (e.g. computing 1830^2 and 1300^2 first, and adjusting to get the correct answer)


    Also, I assume you're practicing from three-year old problems? It is common to have problems involve the year somehow -- so if you know there's an answer around 2000, the very first thing you should guess is the year. :smile: Maybe even memorize other interesting things -- such as its prime factorization.


    You can use modular arithmetic to pick the answer out of a narrow range. e.g.

    A^2 = (4/5)(1829^2 + 1298^2) = (4*2) (2^2 + 2^2) = 8*8 = 1 (mod 9)
    A = 1 or 8 (mod 9)

    A^2 = (4/5)(1829^2 + 1298^2) = (4*3) (2^2 + 3^2) = 5 * 6 = 2 (mod 7)
    A = 3 or 4 (mod 7)

    Based on the modulo 9 criterion, the possibilties nearest 2000 are
    ..., 1990, 1997, 1999, 2006, 2008, ...

    And modulo 7, they are
    ..., 1991, 1992, 1998, 1999, 2005, 2006, ...
     
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