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Math Help - multiplying on fingers also work with bases other than ten, why

  1. #1
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    multiplying on fingers also work with bases other than ten, why

    At school I was shown how to multiply two intergers, both > 5 and <11 using the fingers on both hands.
    Each hand used to represent one of the two numbers.
    It works like this e.g 7x8= 56.
    On the left hand count from 6 to 7 by folding down two fingers so that two are bend and three are not.
    Then on the right hand count from 6 to 8 so that three are bend and two are not.
    Count the number of straight fingers on each hand 3x2=6. this represents 6 units.
    Count number of bent fingers on both hands and this represent the number of tens, five tens, or fifty.
    So the answer is fifty plus six
    The question is:
    How come this works if we didnt have ten fingers and we counted for example in base 8 and had 8 fingers? If we counted in base 8 we would have 4 fingers on each hand and could multipy numbers from 5 to 8 using this method.
    Attached Thumbnails Attached Thumbnails multiplying on fingers also work with bases other than ten, why-base-10.bmp  
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  2. #2
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    How come this works if we didnt have ten fingers and we counted for example in base 8 and had 8 fingers?
    Why does it work for ten fingers? It is probably easy to generalize to other bases.
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  3. #3
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    Hello, misterchinnery!

    Wow! . . . a fascinating trick!


    We have two sets of five fingers, each set numbered from 6 to 10.

    . . \begin{array}{ccccccccccccccccccc}<br />
| & | & | & | & | &&& |&|&|&|&| \\<br />
6&7&8&9&10 &&& 6&7&8&9&10  \end{array}



    To multiply 7 \times 9:

    On the left hand, bend fingers 6 and 7.
    On the right hand, bend fingers 6, 7, 8 and 9.

    . . \begin{array}{ccccccccccccccccccc}<br />
\times & \times & | & | & | &&& \times & \times & \times & \times &| \\<br />
6&7&8&9&10 &&& 6&7&8&9&10  \end{array}


    . . . \begin{array}{c}\text{Left hand} \\ \hline<br />
\text{bent: 2} \\ \text{straight: 3} \end{array} \qquad\quad \begin{array}{c}\text{Right hand} \\ \hline \text{bent: 4}\\ \text{straight: 1}\end{array}


    \begin{array}{ccccccc}\text{Ten's digit:} & \text{(bent)} + \text{(bent)} &=& 2 + 4 &=& 6 \\<br /> <br />
\text{Unit's digit:} & \text{(straight)} \times \text{(straight)} &=& 3 \times 1 &=&3 \end{array}


    . . Therefore: . 7 \times 9 \;=\;63


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I tried a general proof (for base-ten).

    Let \,a and \,b be the two numbers: . 6 \,\le\, a,b\,\le\,10


    . . . \begin{array}{c}\text{Left hand} \\ \hline<br />
\text{bent: }a-5 \\ \text{straight: }10-a \end{array} \qquad\quad \begin{array}{c}\text{Right hand} \\ \hline \text{bent: }b-5\\ \text{straight: }10-b\end{array}


    \begin{array}{cccccccccc}\text{Ten's digit:} & \text{(bent)} + \text{(bent)} &=&(a-5) +(b-5) &=& a + b - 10 & (T)\\<br /> <br />
\text{Unit's digit:} & \text{(straight)} \times \text{(straight)} &=& (10-a)(10-b)&=& 100 - 10a - 10b + ab & (U) \end{array}


    The product is: . 10T + U \;=\;10(a+b-10) + (100-10a - 10b + ab)

    . . . . . . . . . . . . . . . . . . =\; 10a + 10b - 100 + 100 - 10a - 10b + ab

    . . . . . . . . . . . . . . . . . . =\quad ab

    Hey, it works!
    Last edited by Soroban; September 19th 2010 at 09:10 AM.
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  4. #4
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    dont know how to edit your solution so that the 10s and the 5s you use will represent the base used. Obviously I can do this by hand but I dont know the notation or type of maths which can do this.
    Did you see my other post involving triangles?
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