- 27 Jan '15 22:52 / 1 editGuys and girls in this forum that know more than me in number theory please help me out on this since it is driving me crazy.

I received this simple problem via whatsapp (I'm paraphrasing):

Follow the instructions below to see what of these movies is more related to you:

think of a number between 1 and 9

Multiply it by 3

Sum 3 to the result

Multiply the result by 3

Edit: Add the final digits and check your movie!

1 - Matrix

2 - My big fat wedding

...

9 . The pleasure of taking it in the a$$

...

18 - Rambo

I thought of 3 and got 9. Well I obviously wouldn't tell people what I've got so I tried again with 2 and also got 9. Hmmmmm... What's going in here?! Let me try 6... It is also 9!!! Gasp! adam's warlock conjecture: The result is always 9 regardless of the initial value one chooses.

Let me write in full the steps we are given:

x is the original value. Then by following the instructions it is

3*(3x+3)

Which is

9*(x+1)

Well this is just the multiplication table of 9 which can be written like this. The right most digits are

9

8

7

6

5

4

3

2

1

0

and the left most digits are

0

1

2

3

4

5

6

7

8

9

Putting the left most and right most digits together it is

09

18

27

36

45

54

63

72

81

90

which "explains" why the sum must be 9.

Well of course this doesn't explain anything at all since the "rule" that we used to write the multiplication table of 9 is what needs explanation in the first place.

So I can see that the result of summing the digits is 9 because one of the factors is 9 and we are using the decimal base as our number system (if the expression is 4*(x+1) and you are using base 5 the sum of the digits will always be 4 and you can proceed analogously from there) but I can't prove it in a rigorous way...

:'(:'(:'(

Can someone just give me hint on how to do it? - 27 Jan '15 23:11 / 1 editYou're missing some steps/instructions out.

What you have written makes no sense.

9(x+1) where x is an integer between 1 and 9 inclusive gives;

18

27

36

45

54

63

72

81

90

As the set of answers... Not 9.

Adding the first and last digit together from each number does indeed give

the answer 9 for each value of x... But nowhere do you specify that step.

This result is not universal. summing the digits of n(x+1) does not result in

a consistent result for all [or many] values of n.

EDIT: missed your bit, about base 5 ... I still think you need to be clearer about

what your problem actually is. - 27 Jan '15 23:50

Edited and inserted the final step which is to add the digits of the final result as you deduced that it should be.*Originally posted by googlefudge***You're missing some steps/instructions out.**

What you have written makes no sense.

9(x+1) where x is an integer between 1 and 9 inclusive gives;

18

27

36

45

54

63

72

81

90

As the set of answers... Not 9.

Adding the first and last digit together from each number does indeed give

the answer 9 for each value of x... But nowhere do yo ...[text shortened]... it, about base 5 ... I still think you need to be clearer about

what your problem actually is.

As for the base 5 comments.

If you work in base 5 the available digits are: 0,1,2,3,4.

Hence the analogous expression to 9*(x+1) would be 4*(x+1).

Remember that in base 5 4+1=0 and you should carry 1 just like in base 10 (our familiar decimal system) it is 9+1=0 and you should carry 1.

So for instance pick x=2 in our base 5 system. The expression is:

4*(2+1) = 4*3

If you're not familiarized with multiplying integer numbers in other bases other than 10, the trick is just to see multiplication as a repeated addition.

Hence 4*3 is the same as 3+3+3+3. Remember our initial rule that 4+1=0 and carry 1 so that the result is

3+3= 1 carry 1

1+3=4 carry 0

4+3=2 carry 1

Hence the right most digit is 2 and the left most digit is the sum of the numbers you need to carry: 1+0+1=2 (the exact same method you use to sum integers in base 10!).

Thus the final result is 4*(2+1)=4*3=22 in base 4.

Finally 2+2=4.

If you try it out in base 6 and write out a similar expression for the problem you'll see that the sum of final digits will always be 5 (6-1).

Did that made my comments about base 5 more intelligible?

PS: Even though the problem only states for one to chose numbers between 1 to 9 it is obvious that one can choose from 0 to 9. Not so obvious but also possible is for one to choose numbers greater than 9. The thing is that you have to sum the final digits more than one time... - 28 Jan '15 06:19

Nice! That's actually a smart way of solving the problem.*Originally posted by wolfgang59***for 0 < x < 10**[/b]

9x = 10(x-1) + (10-x)

So the digits in the product are (x-1) and (10-x)

add them

(x-1) + (10-x) = x -1 +10 - x = [b]9

When dealing with a base n number system it is

(n-1)x=n(x-1)+(n-x)

Adding the digits is x-1+n-x=n-1

Which is the result I stated but didn't prove.

PS: The only thing that bothers me is that the formula 9x = 10(x-1) + (10-x) only works for 0 < x < 10 when I already saw that the method also works for 0 and for numbers above 10. But that is something that a little change of variables and a little reworking of your formula will solve. - 28 Jan '15 13:18

(n-1)x = nx-x = nx + n-n -x = nx-n + n-x = n(x-1) + (n-x)*Originally posted by adam warlock***Nice! That's actually a smart way of solving the problem.**

When dealing with a base n number system it is

(n-1)x=n(x-1)+(n-x)

Adding the digits is x-1+n-x=n-1

Which is the result I stated but didn't prove.

PS: The only thing that bothers me is that the formula 9x = 10(x-1) + (10-x) only works for 0 < x < 10 when I already saw that the met ...[text shortened]... s something that a little change of variables and a little reworking of your formula will solve.

This is true for any n and any x, regardless of base. Therefore, it is certainly also true when n is your base, for any x; x doesn't have to be 0 <= x < n for that. It's just as true when x = 15n+8, or even when x = -3.45n+0.339. That answers your last problem.

It also answers the original one, because when it*is*true that 1 <= x < n, and n is the number base, it's easy to see that n(x-1) + (n-x) specifies a two-digit number (with 0 allowed for the first digit!). So all digits multiply by the highest digit to such a two-digit number; it follows from the pigeonhole principle that these are also*all*two-digit numbers of that form higher than 00. The final step is realising that in this form, the first digit is x-1, the second is n-x, and the sum of the digits is therefore always n-1, the highest digit in the n-ary system. QED.

It even works in binary - 28 Jan '15 15:36

Have you considered the possibility that you just like taking it in the ass?*Originally posted by adam warlock***Guys and girls in this forum that know more than me in number theory please help me out on this since it is driving me crazy.**

I received this simple problem via whatsapp (I'm paraphrasing):

Follow the instructions below to see what of these movies is more related to you:

think of a number between 1 and 9

Multiply it by 3

Sum 3 to the result

Mu ...[text shortened]... n't prove it in a rigorous way...

:'(:'(:'(

Can someone just give me hint on how to do it? - 28 Jan '15 23:31 / 1 edit
**Well of course this doesn't explain anything at all since the "rule" that we used to write the multiplication table of 9 is what needs explanation in the first place.**

Why is it that the tens increase by one, but the ones decrease by one each time you add 9?

That's because our numbers are based on 10. Each time you add 10 you increase the 10's digit by one, but 9 is (10-1) which means you are decreasing the ones digit by one as you increase the ten's digit by one.

I hope that helps.

Your initial number of 9 is a 0 in the ten's digit and 9 in the ones digit and of course 9 + 0 = 9. Just saying - 29 Jan '15 06:39

Very rigorous. .... NOT*Originally posted by Eladar***[b]Well of course this doesn't explain anything at all since the "rule" that we used to write the multiplication table of 9 is what needs explanation in the first place.**

Why is it that the tens increase by one, but the ones decrease by one each time you add 9?

That's because our numbers are based on 10. Each time you add 10 you increase the 10 ...[text shortened]... ber of 9 is a 0 in the ten's digit and 9 in the ones digit and of course 9 + 0 = 9. Just saying[/b] - 29 Jan '15 07:32 / 1 edit

It doesn't.*Originally posted by Eladar***I hope that helps.**

First it isn't rigorous. Second I know about positional notation and how different numeral systems work (so what you said didn't bring anything new to me).

What I wanted was a rigorous proof of why this particular formula (and its generalizations) always gets n-1 as its final result.