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I need to figure out what number base makes the expression 32 + 12 = 28 true. It looks like the second number is signed, I've tried converting to decimal and using 10's complement but haven't gotten an answer that makes sense.

afc2020
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    Are these numbers all in the same base, or is 28 meant to be base 10? – tgdavies Sep 02 '21 at 00:55
  • Show us what you've tried so far. Please read the [Open letter to students with homework problems](https://softwareengineering.meta.stackexchange.com/questions/6166/open-letter-to-students-with-homework-problems) – Alexander Sep 02 '21 at 00:55
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    This seems more like a math problem rather than a request for help with code, so this would be more appropriate for the math stackexchange. That said, this seems impossible if every number is the same base. 32 > 28 in any base, so 32 + 12 > 28 + 12. – BatWannaBe Sep 02 '21 at 00:58
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    If you can freely assign base, there are multiple solutions. For example 6/4/9, 4/10/9, 7/3/10 would all be valid. The only limitation is that the first cannot go lower than base 4 (it contains a 3), the second cannot go lower than base 3 (it contains a 2) and the sum cannot go lower than base 9 (it contains a 8). – 3limin4t0r Sep 02 '21 at 01:16
  • They're all supposed to be the same base. – afc2020 Sep 02 '21 at 01:18

2 Answers2

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If this is a positional number system, then you can replace the numerals with an expression showing how their digits multiply by the meaning of their columns.

That is,

  "32 + 12 = 28"
≣ (3*R^1 + 2*R^0) + (1*R^1 + 2*R^0) = (2*R^1 + 8*R^0)
≣ (3*R + 2*1) + (1*R + 2*1) = (2*R + 8*1)
≣ (3*R + 2) + (1*R + 2) = (2*R + 8)
≣ 4*R + 4 = 2*R + 8
≣ 2*R = 4
≣ R = 2

For some radix R.

You can solve this equation to get the value of R. If you attempt to solve it, you'll see that you get R = 2, which clearly can't be the case, since you have a digit 3 (a base-2 system would only have ... 2 digits. 0, and 1).

Thus, there is no radix R that would allows "32 + 12 = 28" to encode a valid equation.

Alexander
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On the face of it, it looks impossible. Adding positive integers in any base should result in a larger integer; if the second integer is negative, you end up with 20 = 28 which is clearly impossible. Considered modulo arithmetic? 32 mod 4 is 0, 12 mod 4 is 0, 28 mod 4 is 0, 0 + 0 = 0.

tsc_chazz
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