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How is $29 - 1 = 30$?

If also

$14 - 1 = 15$

$11 - 1 = 10$

$9 - 1 =10$.

Hint:

Guess the answer and be like Minerva

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    $\begingroup$ I think the hint is confusing. $\endgroup$ – jamesdlin Jun 10 '17 at 12:28
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    $\begingroup$ @AlbertMasclans your comment (particularly the word before 'equivalent') is kind of a spoiler. It made me realise the answer, and I hadn't even looked at the hint. $\endgroup$ – Alex Hall Jun 10 '17 at 15:04
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    $\begingroup$ How about how = 2? $\endgroup$ – Frozn Jun 12 '17 at 11:33
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    $\begingroup$ @AmruthA "like Athena" might as well be "like Buddha" or "like Sophocles" or something. The point is that it's like something that is like Athena, not like Athena herself, and that's confusing. $\endgroup$ – jamesdlin Jun 13 '17 at 6:25
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    $\begingroup$ @AmruthA I am very much aware of that. The point remains that your hint requires two leaps (Athena => Minerva => Roman) which is misleading, and there's no basis for making that jump. The question would be better without the hint. $\endgroup$ – jamesdlin Jun 13 '17 at 7:49
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Explanation:

If you're using Roman numerals, you can remove I (one) from the representations of the numbers before the minus sign to get the representation of the numbers in the right.

29 - 1 = 30

XXIX - I = XXX

14 - 1 = 15

XIV - I = XV

11 - 1 = 10

XI - I = X

9 - 1 =10

IX - I = X

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    $\begingroup$ If this is actually the answer, then the hint is misleading - it should have been about Minerva. :) $\endgroup$ – fluffy Jun 11 '17 at 19:26
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    $\begingroup$ @fluffy " like Athena " not Athena ..counterpart for Minerva in Greek .. $\endgroup$ – Amruth A Jun 12 '17 at 10:33
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    $\begingroup$ @AmruthA but you're saying the answerer will be like Athena, or at least that's what the statement implies. $\endgroup$ – TheWanderer Jun 12 '17 at 11:34
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    $\begingroup$ @AmruthA My point being that the ancient Greeks didn't use Roman numerals. (Arguably Romans didn't use Roman numerals the same way we do either, though.) $\endgroup$ – fluffy Jun 12 '17 at 20:55
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    $\begingroup$ Well, Athena was born by splitting Zeus' head. So maybe it's something like "you'll facepalm so hard when the answer comes to you". Not much of a hint, though, if it is... $\endgroup$ – mr23ceec Jun 13 '17 at 11:08
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$$9 - 1 = 10 = 11 - 1,$$ thus

$$29 - 1 = 20 + 9 - 1 =\\ = 20 + 11 - 1 = 31 - 1 = 30.$$

Edit

Or just using that $9-1 = 10$

$$ 29 - 1 = 20 + 9-1 = 20 + 10 = 30.$$

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    $\begingroup$ Why did you(?) vote for delete? $\endgroup$ – Olba12 Jun 11 '17 at 23:06
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    $\begingroup$ I was surpriced to see that no one else had posted it. Thanks! @Kröw $\endgroup$ – Olba12 Jun 12 '17 at 0:56
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    $\begingroup$ @M.Herzkamp It is given right there in the question that 11 - 1 = 10. $\endgroup$ – Masked Man Jun 12 '17 at 14:01
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    $\begingroup$ Can you make the assumption that 29 = 20 + 9? $\endgroup$ – UKMonkey Jun 12 '17 at 15:17
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    $\begingroup$ @UKMonkey Given that the question is tagged 'mathematics', I think it should be fine to assume common mathematical conventions (except, of course, when it contradicts the puzzle itself, for example 29 - 1 = 30). $\endgroup$ – Masked Man Jun 13 '17 at 5:38
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I suppose you could always round the answer to the nearest multiple of 5, although that has nothing to do with Athena.

$$\begin{align} 29-1&=28 \xrightarrow{\text{rounds to }}30\\ 14-1&=13 \xrightarrow{\phantom{\text{________}}} 15\\ 11-1&=10 \xrightarrow{\phantom{\text{________}}} 10\\ 9-1 &= \phantom08 \xrightarrow{\phantom{\text{________}}} 10\\ \end{align}$$

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    $\begingroup$ yup, that's what I thought it was. The puzzle isn't clear enough to eliminate this possibility. $\endgroup$ – NH. Jun 12 '17 at 15:13
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This is mathematically true in $\mathbb{Z}/2\mathbb{Z}$, i.e. $\bmod 2$:

$$\begin{align} 29 - 1 \equiv 30 \equiv 0 \pmod 2\\ 14 - 1 \equiv 15 \equiv 1 \pmod 2\\ 11 - 1 \equiv 10 \equiv 0 \pmod 2\\ 9 - 1 \equiv 10 \equiv 0 \pmod 2\\ \end{align}$$

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    $\begingroup$ +1 I had similar thoughts, thinking the arithmetic was in the field of 2 elements {0,1} $\endgroup$ – Bad_Bishop Jun 13 '17 at 14:28
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This is inspired by/alternative to Olba12's nice answer:

$$\begin{align} 30 &= 10 + 10 + 10\\ &= (11 - 1) + (11 - 1) + (9 - 1)\\ &= 31 - 3\\ &= 31 - 2 - 1\\ &= 29 - 1 \end{align}$$

Hence: $29 - 1 = 30$.

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    $\begingroup$ Like Olba12, you cannot mix the "$-$" in the puzzle with our usual minus sign. They behave rather differently! $\endgroup$ – M.Herzkamp Jun 14 '17 at 10:23
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    $\begingroup$ The minus sign in my solution is not at all contradicting the minus sign in the puzzle. I take the minus sign in the puzzle to define a new operator, so that in the cases given in the question, it gives those results. In other cases, it behaves like the usual minus sign. The plus operator works as usual, since the question doesn't say anything about it. The puzzle is tagged mathematics, so we can assume basic mathematical conventions and rules, except when the puzzle explicitly overrides them. $\endgroup$ – Masked Man Jun 14 '17 at 13:02
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    $\begingroup$ By your definition you have 10 = 9 - 1 = 10 - 1 - 1 = 10 - 2 = 8, which is a contradiction. $\endgroup$ – M.Herzkamp Jun 19 '17 at 12:05
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    $\begingroup$ You should first question the OP who came up with this stupid puzzle and tagged it Mathematics. 10 = 9 - 1 is already a contradiction, everything else you "deduce" from it will be likely to be a contradiction anyway. $\endgroup$ – Masked Man Jun 19 '17 at 12:37
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Here's another approach that doesn't require redefining ‘$-$’ as a string operation, rather than a mathematical one:

$$\begin{align} 29 - 1\ (\text{base}\ 11) &= 30\ (\text{base}\ 10)\\ 14 - 1\ (\text{base}\ 12) &= 15\ (\text{base}\ 10)\\ 11 - 1\ (\text{base}\ 10) &= 10\ (\text{base}\ 10)\\ 9 - 1\ (\text{base}\ 10) &= 10\ (\text{base}\ 8) \end{align}$$

I like the OP's intent better, though. It's a clever puzzle.

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    $\begingroup$ How does this work? Do we just choose any random base at each step? $\endgroup$ – Wen1now Jun 12 '17 at 8:10
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    $\begingroup$ not a random base, but just cherrypick the one that fits. Arbitrary ≠ random. $\endgroup$ – NH. Jun 12 '17 at 15:16
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    $\begingroup$ Unfortunately, yes, this requires an arbitrary base at each step -- which is why I like the OP's solution better, despite the requirement that one redefine the meaning of '-'. In my defense, I based this solution off the title of the puzzle: "29 - 1 = 30, how?" Which implied a single case. When I saw multiple cases, I realized the solution wasn't good. Now, if the cases had been presented like this: 14 - 1 = 15 29 - 1 = 30 11 - 1 = 10 the 'different bases' solution would make more sense. $\endgroup$ – Mark Kreitler Jun 14 '17 at 6:03
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This is a cheeky answer.

Assume the usual rules of arithmetic and logic.

We are told that $9 - 1 = 10$, that is, $8 = 10$, which is a contradiction.

By the rules of logic, since we have derived a contradiction, we can now derive anything else (like 'magic', cf the storybook character Minerva), such as, that

$29 - 1 = 30$.

QED

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It is just the '-' operator has changed its semantics to compute sum of its operands.

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    $\begingroup$ '11-1=10' does not work here: 11+1=12, not 10. $\endgroup$ – boboquack Jun 22 '17 at 4:41

protected by Beastly Gerbil Jun 12 '17 at 20:12

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