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Posers and Puzzles

Posers and Puzzles

  1. 21 Mar '06 10:03
    I was wondering if anyone might contribute some interesting ways to perform mental arithmacy quickly. For instance, for squaring numbers ending in five, you take the numbers infront of the five and multiply it by that plus one more, and then put 25 at the end. Proof:
    Let n denote the numbers after the 5,
    (10n+5)^2 = 100n^2 + 100n + 25
    = 100n (n+1) + 25

    I find this to be a good game when in the cars. When I see a licence plate I square the number. But I find it difficult to square four digit numbers without developing a head ache. Any suggestions?
  2. Standard member XanthosNZ
    Cancerous Bus Crash
    21 Mar '06 10:40 / 1 edit
    Originally posted by Conrau K
    I was wondering if anyone might contribute some interesting ways to perform mental arithmacy quickly. For instance, for squaring numbers ending in five, you take the numbers infront of the five and multiply it by that plus one more, and then put 25 at the end. Proof:
    Let n denote the numbers after the 5,
    (10n+5)^2 = 100n^2 + 100n + 25
    = 100n (n+1) + 25 ...[text shortened]... find it difficult to square four digit numbers without developing a head ache. Any suggestions?
    If you want to know (n+1)^2 and know n^2 you can just add 2*n - 1

    So if you know that 16^2 = 256 then you can quickly say that 17^2 = 256+34-1 = 289.

    This combined with your trick for squaring numbers ending in 5 (and the obvious one for squaring numbers ending in zero) allows you to quickly (compared to actually doing the squaring) find the square of any number ending in 4,5,6,9,0 or 1 (you can find the square of (n-1) if you know the square of n by reversing the above process).
  3. 21 Mar '06 11:53
    Multiplying two number that is equally far from a easy square is easily done:

    Say 27*33 = (30-3)(30+3)=30^2-3^2=900-9=891.
    Using (x-y)(x+y)=x^2-y^2.
  4. 22 Mar '06 02:44
    Originally posted by FabianFnas
    Multiplying two number that is equally far from a easy square is easily done:

    Say 27*33 = (30-3)(30+3)=30^2-3^2=900-9=891.
    Using (x-y)(x+y)=x^2-y^2.
    To multiply 27 times 33 you could multiply 27 by 3 and then by 11. A shortcut to multiplying by 11, would be to add up the digits and insert the sum in the middle of the original number and also carrying the remainder. i.e. 3*27*11 = 81*11: 8+1 = 9, hence, 11*81 = 891.
  5. 22 Mar '06 08:43 / 1 edit
    Originally posted by Conrau K
    To multiply 27 times 33 you could multiply 27 by 3 and then by 11.
    I was posting a general rule with certain conditions.
    Your method is specific for 27*33 only.

    Try my method for 91*109, 44*56, 19*31 etc and you'll notice how easily the calculation is done by thinking alone.

    General methods are generally better. Specific methods are helpful when you can't find any general method to apply.
  6. 22 Mar '06 10:02
    Originally posted by FabianFnas
    I was posting a general rule with certain conditions.
    Your method is specific for 27*33 only.

    Try my method for 91*109, 44*56, 19*31 etc and you'll notice how easily the calculation is done by thinking alone.

    General methods are generally better. Specific methods are helpful when you can't find any general method to apply.
    I completely agree. I suppose specific methods are only number tricks that don't work with other math languages. Where as general methods always work. Heres one for multiplying (not to sure how it works):
    1. decide a reference number close to the two numbers
    2. subtract this reference number from one of the numbers and add to the other.
    3. Multiply this number by the reference number.
    4. Multipily the subtractions of the two numbers from the reference number, together
    5. Add answer from step 4 to step 3.

    i.e.
    23*18
    1. Reference number= 20
    2. 23-20 = 3, 18-20 = -2, -2 + 23 = 21 (3+18 will also work because theyre communicable)
    3. 21*20 = 420
    4. -2*3=-6
    5. 420+ -6 = 414
    23*18 = 414
  7. 22 Mar '06 11:39 / 1 edit
    Nice!