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Math Help - Rule for multiplying by 17

  1. #1
    Bar0n janvdl's Avatar
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    Rule for multiplying by 17

    Apparently there has to be a rule for multiplying by 17, just like the one for multiplying by 11.

    Example: 32 x 11...

    Digit 1 = 3 x 1 = 3
    Digit 3 = 2 x 1 = 2
    Digit 2 = Digit 1 + Digit 3 = 5

    Therefore 32 x 11 = 352


    Does anyone know what the rule for 17 is? Thanks!
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  2. #2
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    Hello, janvdl!

    Apparently there has to be a rule for multiplying by 17,
    just like the one for multiplying by 11.

    We could create one, but why would we want one?

    Example: 32 x 11...

    Digit 1 = 3 x 1 = 3
    Digit 3 = 2 x 1 = 2
    Digit 2 = Digit 1 + Digit 3 = 5

    Therefore 32 x 11 = 352

    There is a lot more to the 11-rule . . .

    What about 59 x 11 ?
    The answer is 649 . . . How does your rule give that result?


    To multiply 7856 times 11, append a zero to both ends: . 078560

    Starting at the right, add consecutive pairs of digits.
    . . Perform the "carrying" as you work.


    . \begin{array}{cccccccccccccccccccccccc}<br />
0 & & & &7 & & & &8 & & & &5 & & & &6 & & & &0 \\<br /> <br />
& \searrow && \swarrow && \searrow && \swarrow && \searrow && \swarrow && \searrow && \swarrow && \searrow && \swarrow \\<br /> <br />
& & 7 & & & & ^15 & & & & ^13 & & & & ^11 & & & & 6 \\<br /> <br />
& & \downarrow & & & & \downarrow &&& & \downarrow &&& & \downarrow &&& & \downarrow \\<br />
\end{array}
    . - - - . 8 \qquad\qquad\qquad\: 6 \qquad\qquad\qquad\; 4 \qquad\qquad\qquad\:\, 1 \qquad\qquad\qquad\: 6


    . . . Answer: . 8\ 6\ 4\ 1\ 6

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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, janvdl!
    ...
    Soroban, thank you for the clarification of the 11-rule.

    You asked why I would want a rule for the number 17:
    It was actually on my university's maths notice board. As a challenging problem. And i was wondering what the answer would be. So that's why I would want to know the rule for the number 17.
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  4. #4
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    Hi, janvdl!

    Of course, such a rule would be messier than the 11-rule.


    Let multiply abcde by 17 and see what happens.


    . . . \begin{array}{ccccccc} & a & b & c & d & e \\<br />
& & & & 1 & 7 \\ \hline<br />
& 7A & 7b & 7c & 7d & 7e \\<br />
a & b & c & d & e \\ \hline<br />
& 7a & 7b & 7c & 7d & 7e \\<br />
+a & +b & +c & +d & +e<br />
\end{array}


    The digits are: . [a]\;\;[7a+b]\;\;[7b+c]\;\;[7c+d]\;\;[7d+e]\;\;[7e]

    . . with the appropriate "carry" performed.



    The basic step seems to be: "Starting at the right, multiply each digit by 7
    . . and add the preceding digit (and carry)."



    Example: . 234 \times 17

    Append a zero (0) to both ends: . 0\,2\,3\,4\,0


    . . \begin{array}{ccccc}0 & 2 & 3 & 4 & \quad0\\<br />
\downarrow & \downarrow & \downarrow & \downarrow \\<br />
7(0)+2 & 7(2)+3 & 7(3) + 4 & 7(4) + 0 \\<br />
\downarrow & \downarrow & \downarrow & \downarrow \\<br />
2 & ^17 & ^25 & ^28 \\<br />
\downarrow & \downarrow & \downarrow & \downarrow \\<br />
3 & 9 & 7 & 8 <br />
\end{array}


    Therefore: . 234 \times 17 \;=\;3978

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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Soroban View Post
    Hi, janvdl!

    Of course, such a rule would be messier than the 11-rule.


    Let multiply abcde by 17 and see what happens.


    . . . \begin{array}{ccccccc} & a & b & c & d & e \\<br />
& & & & 1 & 7 \\ \hline<br />
& 7A & 7b & 7c & 7d & 7e \\<br />
a & b & c & d & e \\ \hline<br />
& 7a & 7b & 7c & 7d & 7e \\<br />
+a & +b & +c & +d & +e<br />
\end{array}


    The digits are: . [a]\;\;[7a+b]\;\;[7b+c]\;\;[7c+d]\;\;[7d+e]\;\;[7e]

    . . with the appropriate "carry" performed.



    The basic step seems to be: "Starting at the right, multiply each digit by 7
    . . and add the preceding digit (and carry)."



    Example: . 234 \times 17

    Append a zero (0) to both ends: . 0\,2\,3\,4\,0


    . . \begin{array}{ccccc}0 & 2 & 3 & 4 & \quad0\\<br />
\downarrow & \downarrow & \downarrow & \downarrow \\<br />
7(0)+2 & 7(2)+3 & 7(3) + 4 & 7(4) + 0 \\<br />
\downarrow & \downarrow & \downarrow & \downarrow \\<br />
2 & ^17 & ^25 & ^28 \\<br />
\downarrow & \downarrow & \downarrow & \downarrow \\<br />
3 & 9 & 7 & 8 <br />
\end{array}


    Therefore: . 234 \times 17 \;=\;3978

    That's amazing! Thanks Soroban!
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  6. #6
    Bar0n janvdl's Avatar
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    By the way, I was working on it as well and got the following:

    1 x 17 = 17
    2 x 17 = 34
    3 x 17 = 51

    Etc...

    In short:
    17k = x ; Where k is an element of all positive integers.

    But:

    All the seperate digits of x added together plus k = 9

    So if we have 34:
    (3 + 4) + 2 = 9.

    If we have 68

    (6 + 8) + 4 = 18 <- Seperate into 1 and 8.
    1 + 8 = 9


    ---

    That was basically what I found. Do you have any comment on this, Soroban?
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  7. #7
    Member Henderson's Avatar
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    I'll comment on it:

    If you move the addition of k before you start summing digits, you'll realize that you've been summing digits of

    17k + k = 18k, which obviously is divisible by nine, leading to the sum equaling nine.
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