Thread: 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!

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: .

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


.
. - - - .


. . . Answer: .

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.

Hi, janvdl!

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


Let multiply by 17 and see what happens.


. . .


The digits are: .

. . 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: .

Append a zero (0) to both ends: .


. .


Therefore: .

Originally Posted by Soroban

Hi, janvdl!

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


Let multiply by 17 and see what happens.


. . .


The digits are: .

. . 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: .

Append a zero (0) to both ends: .


. .


Therefore: .

That's amazing! Thanks Soroban!

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?

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.