Thread: Rule for multiplying by 17

1. 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!

2. 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}
0 & & & &7 & & & &8 & & & &5 & & & &6 & & & &0 \\

& \searrow && \swarrow && \searrow && \swarrow && \searrow && \swarrow && \searrow && \swarrow && \searrow && \swarrow \\

& & 7 & & & & ^15 & & & & ^13 & & & & ^11 & & & & 6 \\

& & \downarrow & & & & \downarrow &&& & \downarrow &&& & \downarrow &&& & \downarrow \\
\end{array}$

. - - - . $8 \qquad\qquad\qquad\: 6 \qquad\qquad\qquad\; 4 \qquad\qquad\qquad\:\, 1 \qquad\qquad\qquad\: 6$

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

3. Originally Posted by Soroban
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.

4. Hi, janvdl!

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

Let multiply $abcde$ by 17 and see what happens.

. . . $\begin{array}{ccccccc} & a & b & c & d & e \\
& & & & 1 & 7 \\ \hline
& 7A & 7b & 7c & 7d & 7e \\
a & b & c & d & e \\ \hline
& 7a & 7b & 7c & 7d & 7e \\
+a & +b & +c & +d & +e
\end{array}$

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: . $0\,2\,3\,4\,0$

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

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

5. Originally Posted by Soroban
Hi, janvdl!

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

Let multiply $abcde$ by 17 and see what happens.

. . . $\begin{array}{ccccccc} & a & b & c & d & e \\
& & & & 1 & 7 \\ \hline
& 7A & 7b & 7c & 7d & 7e \\
a & b & c & d & e \\ \hline
& 7a & 7b & 7c & 7d & 7e \\
+a & +b & +c & +d & +e
\end{array}$

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: . $0\,2\,3\,4\,0$

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

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

That's amazing! Thanks Soroban!

6. 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?

7. 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.