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## Question

An arbitrary number could be expressed as

A1A2 A3 ....An-1An

Each digit in the number is unknown and the number is also unknown.

Applying a right shift to the number, the new number becomes

An A1 A2 A3 .... An-1

(a) It was found that the new number is 1.5 times larger than the original number.

What will be the original number?

(b) If the original number is 1.5 times larger than the new number.

What will be the original number?

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## 4 answers to this question

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(a) 285714 or 571428

(b) 5294117647058823

Don't ask me how I got these, it was very hodgepodge

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(a) 285714 or 571428

(b) 5294117647058823

Don't ask me how I got these, it was very hodgepodge

Other possibilities for part 1 other than those mentioned by Tuckleton are:

285 714 285 714

571 428 571 428

There are still infinitely many more solutions possible; the next 2 solution would contain 18 digits, next 2 would have 24 digits and so on...

Will look into part 2 later!!

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I thought I knew math fairly well but WOW. Seriously, how do you guys figure that stuff out?

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I thought I knew math fairly well but WOW. Seriously, how do you guys figure that stuff out?

This is probably overly complicated and I apologize in advance if my explanation is very confusing.

So a bit of an introduction to the method I used. Say I have:

```
AB

* 1.5

-----

??

```
As long as A and B are both even then we can multiply the way we did in elementary school. B*1.5 into the units column and A*1.5 in the tens column (and carry the one when necessary.
```
42

* 1.5

-----

3

6

-----

63

```
But say A was odd it would look like this:
```
52

* 1.5

-----

3

75

-----

78

```
Having to go back again to add the 5 when a digit was odd was problematic so instead of looking at something like 52 as 50+2, I looked at it like 40+12:
```
4 12

* 1.5

-----

1  8

6

-----

7  8

```
Anyway, back to the problem. So let's start with (a). To make getting started easier I'm going to call The original number ABCDE and the new number EABCD. There may be more or less than 5 digits. I just chose 5 to help me get a good visualization of the problem. We've got:
```
ABCDE

* 1.5

-----

EABCD

```
Looking at this, E must be even so the product is an integer. E can't be zero since it is the first digit of the product and it's supposed to be bigger than the original number. From here I made a little lookup table: 0=0;15 2=3;18 4=6;21 6=9;24 8=12;27 The first number is just *1.5. The second is if we brought over a 1 because the next number is odd. So I just picked a number for E and saw where it took me. If E=2 then D=3 if the next number is even or 8 if the next number is odd. But since D is the next number. This doesn't work. So E is not 2. E is not 6 for the same reason. So now try E=4. D would be 6 if D was even or D would be 1 (with a carryover of 2) if D was odd. No contradiction so far. But if D=6 we run into the same problem as before. So D must be 1:
```
(2)

ABC14

* 1.5

-----

4ABC1

```
Since we took 1 away from D earlier (to get 14 * 1.5 = 21) we treat it as 0. Remembering to carry the 2, C would be 2 if C was even or C would be 7 (with a carryover of 1) if C was odd. No contradiction but if C were 2 we'd have a familiar problem. So C must be 7:
```
(1)

AB714

* 1.5

-----

4AB71

```
We took a 1 away from C so treat it as a 6. The reason we don't run into a problem with that is because of the carryover of 1. B(even) gives B=0 (Carryover of 1). B(odd) gives B=5 (Carryover of 2). Because of the carryover of 1 when B is even, We'd run into the same contradiction we run into when we have a 2 or 6 with no carryover. So B must be 5. We're starting to reach the limit of our letters so we'll just expand it a bit:
```
(2)

...X5714

*    1.5

--------

4...X571

```
Treat 5 as 4, X(even)=8, X(odd)=3(Carry 2). 3 would act like 2 (carry 2) which doesn't work so X=8:
```
(0)

...X85714

*     1.5

---------

4...X8571

```
X(even)=2(Carry 1), X(odd)=7(Carry 2). 7 would act like 6 (carry 2) which doesn't work so X=2:
```
(1)

...X285714

*      1.5

----------

4...X28571

```
X(even)=4, X(odd)=9(Carry 1). 9 acts like 8 (carry 1) which doesn't work so X=4. Since there is no carryover here and the 2 numbers satisfy the condition, we can stop! Or we could keep going, repeating the same logic every 6 digits as DeeGee pointed out. This gives one answer of 285714. And, instead of doing it all over again for E=8. We can notice that when we got 8 before it had no carryover. Which means starting at 8 will look like just chopping off what came before it in the previous calculations. and since 285714 repeats we can immediately arrive at the other family of answers: 571428. Part (b) works much the same. Start with:
```
Z..XY

* 1.5

-----

..XYZ

```
Y must be even again. So let's choose a number. When I did it I chose 2. Let's do 0: If Y is zero, then Z is either 0 or 5. Since Z is the start of one of the numbers it can't be 0. So it must be 5 with a carryover of 1 and X must be odd (I denote this with a +):
```
+

1

5..X0

* 1.5

-----

..X05

```
Now in order to make that zero X needs to be a number that produces a 9 so that when we add the carryover we get 0. Referring to our table from before, that number is 6 when the next number is even. Also X needs to be odd so we turn 6 into 7 (We go up since we turned that first 0 into 10 in order to get that 5 at the end of the product). So X=7 with a carryover of 1 and the next number is even:
```
+

11

5..X70

*  1.5

------

..X705

```
In order to get 7, X must produce a 6. Which is 4 where the next number is even:
```
+

11

5..X470

*   1.5

-------

..X4705

```
In order to get 4, X must produce a 4. Which is 6 where the next number is odd (and we'll have a carryover of 2). Continue this reasoning until you get a number that satisfies the condition. It's pretty big so I'll just jump to the end:
```
+   ++ + +++  +

02100211201220110

5882352941176470

*            1.5

-----------------

8823529411764705

```
Now this number repeats just like the number from part (a). Which means that anyplace along here that has no carryover and is even, is a place where we can cut the number and append the cut part to the end to make a new number:
```
|+  | |++ + |+++ | +

0|210|0|21120|1220|110

|588|2|35294|1176|470

|*  | |     |    |1.5

-|---|-|-----|----|---

|882|3|52941|1764|705

```

So we get:

8823529411764705

3529411764705882

5294117647058823

1764705882352941

7058823529411764

Which all work and repeat.

...phew...

P.S: Too lazy to go back over my work so please excuse any small errors I may have made.

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