Results 1 to 10 of 10

- October 3rd 2012, 06:22 PM #1

- Joined
- Oct 2012
- From
- Zimbabwe
- Posts
- 7

- October 3rd 2012, 07:56 PM #2

- Joined
- Jun 2012
- From
- AZ
- Posts
- 616
- Thanks
- 97

- October 4th 2012, 01:10 PM #3

- Joined
- Oct 2012
- From
- Zimbabwe
- Posts
- 7

- October 4th 2012, 01:45 PM #4

- Joined
- Sep 2012
- From
- Washington DC USA
- Posts
- 525
- Thanks
- 147

## Re: Zahlenspiel

All digits are 0 is obviously one solution.

That it's the only solution follows from the rightmost column, middle digits.

The filled circle stays the same after an addition, so either the 4-quadrant circle is 9 or it's 0. But if it were 9, then AAA-BBB = 9**, which can't happen. Thus the 4 quadrant circle = 0, and also, from AAA-BBB = 0**, get A=B. From the top row, last two digits, then get filled triangle = filled circle = 0. So A = B = 6-gon = half moon. From middle column, 100's digit, get square = 0. Now the middle column prodcues a contradiction unless all the digits are 0.

Unless I made a mistake, it works out they're all zero.

- October 4th 2012, 01:55 PM #5

- Joined
- Oct 2012
- From
- Zimbabwe
- Posts
- 7

- October 4th 2012, 02:01 PM #6

- Joined
- Sep 2012
- From
- Washington DC USA
- Posts
- 525
- Thanks
- 147

## Re: Zahlenspiel

Unless I made a mistake, or there's something sneaky like the digits aren't base 10, I just gave you an explanation of why they're all *necessarily* zero. If you stipulate that they aren't zero, then that means the problem has no solution.

- October 4th 2012, 02:17 PM #7

- Joined
- Oct 2012
- From
- Zimbabwe
- Posts
- 7

## Re: Zahlenspiel

- October 4th 2012, 03:13 PM #8

- Joined
- Sep 2012
- From
- Washington DC USA
- Posts
- 525
- Thanks
- 147

## Re: Zahlenspiel

Are you sure there's a solution? I still get that it's impossible, even correcting for that trick. It's certainly possible I made a mistake somewhere. Do you have an actual solution?

Same rationale about 3rd column says that 4-quadrant circle is either 0 or 9.

Assume the 4-quadrant circle is 9.

1) Top row **9 - *9T = *99 implies the filled triangle = 0. Also, *C9 - *90 = *99 imples the solid circle = 8.

2) Bottom row: AAA - BBB = 888, so A = 9, B = 1, or A = 8, B = 0.

3) Left column: x89 + 0xy = AAA. Since can at most carry the one once, and A is 8 or 9, x is 7 or 8 or 9. Since for any of those 3, the 10's position addition will definitely result in carrying a 1, A = x+1. Thus x is 7 or 8. A = 6-gon +1, 6-gon is 7 or 8.

4) Look at column 1 again: x89 + 0xy = AAA. It's either 789+07y = 888, or 889 + 08y = 999. In the first case, 07y = 888-789 = 99.

In the second case it's 08y = 999 - 889 = 110. In neither case can it work out.

**Therefore the 4-quadrant circle is 0.**

1) Top row: one's place:**solid triangle = 0.**

2) Top row now reads: **0 - *00 = 000. Therefore fill**ed circle = 0**. And**6-gon = split-circle.**

3) Left column now reads: x00 + 0xx = AAA. So**A = 6-gon = split-circle.**

4) Bottom row reads AAA - BBB = *0*, so**A = B.**

5) From A = B and bottom row, right column now reads: 000 + y0* = 000. Thus y =**split box = 0.**

6) Middle column now reads: B00 - 0BB = BBB. Thus**B = 0.**

With that, everything is 0.

- October 4th 2012, 04:09 PM #9

- Joined
- Oct 2012
- From
- Zimbabwe
- Posts
- 7

- October 4th 2012, 04:17 PM #10

- Joined
- Sep 2012
- From
- Washington DC USA
- Posts
- 525
- Thanks
- 147

## Re: Zahlenspiel

Yes, I understand that AAA is a 3 digit number.

I've produced a solution - unless you can find an error in my reasoning (which is certanly possible, given how convoluted the problem is). The only solution is that everything in sight is the digit 0.