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"The two sides of this equation appear to be different, don't they?" asked Grandpa

I looked at the numbers he had written on a piece of paper and nodded in agreement.

2011 = 1012

"Can you prove to me that the equation is right?" He asked. "No silly thing like moving the digits!"

This time I knew what he was talking about. Do you?

Although @Stiv has given a great answer, Grandpa had a different logic and that also holds true for

211 = 112

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    $\begingroup$ "Does this equation make sense?" No. No it doesn't. Please stop talking nonsense again, Grandpa. Did you take your pills this morning? $\endgroup$ – Vilx- Feb 17 at 13:21
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Revised answer:
Grandpa's words are important here - we mustn't move the digits... or in fact do anything with the digits at all. Because what he wants us to do is:

focus on the letters that spell these numbers! Both equations now included in the puzzle have the property that the numbers either side of the equals sign can be spelled out using the same letters (i.e. they are anagrams of each other).

See as follows:

TWO THOUSAND AND ELEVEN = ONE THOUSAND AND TWELVE

TWO HUNDRED AND ELEVEN = ONE HUNDRED AND TWELVE

The key thing to spot here is that TWO/ELEVEN and ONE/TWELVE use the exact same letter set! Of course, this also holds true if we don't spell out the 'AND' in these numbers, like some people may pronounce them.


Initial answer (revised after OP post clarification at +4):
The two halves of this equation are equivalent if we write them as:

concatenated Roman numerals.

Specifically:

If we split each side into smaller chunks like so:
20 1 1 = 10 12

Then converting these smaller chunks into Roman numerals gives:

20 1 1 = XX I I
10 12 = X XII

i.e. without moving the digits around (merely by converting them into a different numeral system), both sides can be written as XXII!

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  • $\begingroup$ Wow. I did not think of this path. There is another logic to this @Stiv $\endgroup$ – DrD Feb 15 at 15:38
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    $\begingroup$ @DrD You know what, I think I've just spotted what you're after...! $\endgroup$ – Stiv Feb 15 at 15:39
  • $\begingroup$ Absolutely nailed it! $\endgroup$ – DrD Feb 15 at 15:46
  • $\begingroup$ Avpr. Fb gurl ner nyfb rdhny gb 'ryrira gubhfnaq naq gjb' naq 'gjryir gubhfnaq naq bar' $\endgroup$ – Eric Duminil Feb 16 at 14:07
  • $\begingroup$ @EricDuminil Exactly - that middle section can be anything you like, as long as it's the same for both! :) $\endgroup$ – Stiv Feb 16 at 14:11
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Another way that this can work is if we interpret the numbers as

Dates

In particular, we can read them as

2011 = 2/01/1, or, in European reading, 2nd of January of the year 1, and
1012 = 1/01/2, or, in Japanese reading, year 1, January 2.

In reading them this way, it is seen that the two numbers can be interpreted as the same.

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  • $\begingroup$ Does this also work for the second example "equality" given? $\endgroup$ – bobble Feb 16 at 1:36
  • $\begingroup$ @bobble - you can just as easily interpret the month as a single digit, the same as the day and the year. $\endgroup$ – Glen O Feb 16 at 2:25
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One observation is that in both sets, after the first digit the second digit is the same. Also the last two digits in both sets are the same: 11,12. In addition to that, 2+0+1+1=4 and 1+0+1+2=4 also 2+1+1=4 and 1+1+2=4.

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  • $\begingroup$ Yeah, I think to avoid the obvious suggestions here (sum the digits, or count the number of occurrences of each digit), there should be some false examples. Like 12 ≠ 21 or 1102 ≠ 2011, etc. $\endgroup$ – Darrel Hoffman Feb 16 at 14:50
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It’s plain, ordinary math! Nothing weird going on!

Juxtaposition of numbers is multiplication, i.e., ab is a × b. So 2011 = 2 × 0 × 1 × 1 = 0, as is 1012 = 1 × 0 × 1 × 2 = 0. This also holds for the second equation: 211 = 2 × 1 × 1 = 2 and 112 = 1 × 1 × 2 = 2. Grandpa decided to save himself some time and not write parentheses around the digits to disambiguate.

But what do you want, the man is old!

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Another option...

It could be shorthand for a modular equation. So saying a = b means ax = b mod 5. And it turns out that 211x = 112 mod 5 and 2011x = 1012 mod 5 both have the same solution: 2 + 5c for c = 0, 1, 2, 3, ....

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  • $\begingroup$ Or, more simply, gurl'er nyy rdhny zbqhyb avar. $\endgroup$ – Gordon Davisson Feb 18 at 8:24
  • $\begingroup$ @GordonDavisson Good point! You should post that as an answer. $\endgroup$ – bob Feb 18 at 14:28
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For each equation, assume that the two sides of the equation are numbers in different bases. Noting that the highest digit in use is '2', so the lowest base that can be used is base 3. For example, for the equation "2011=1012" if we set the LHS to base 4 then the RHS must be in base 5 (or base 5.01312, to be precise):

2(4^3) + 4^1 + 4^0 = n^3 + n^1 + 2(n^0) ∴ n≈5

You will need to solve a cubic equation to derive that solution.

For the second equation, apply the same technique. I.e. select some base (3 or above) for the LHS and then solve for the RHS; this equation will involve solving a quadratic.

It turns out that there are an infinite number of solutions when using pairs of bases in this way.

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