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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8 Answers
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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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32$\begingroup$ If this is actually the answer, then the hint is misleading - it should have been about Minerva. :) $\endgroup$– fluffyCommented Jun 11, 2017 at 19:26
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5$\begingroup$ @fluffy " like Athena " not Athena ..counterpart for Minerva in Greek .. $\endgroup$– Amruth ACommented Jun 12, 2017 at 10:33
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4$\begingroup$ @AmruthA but you're saying the answerer will be like Athena, or at least that's what the statement implies. $\endgroup$ Commented Jun 12, 2017 at 11:34
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3$\begingroup$ @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.) $\endgroup$– fluffyCommented Jun 12, 2017 at 20:55
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5$\begingroup$ 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... $\endgroup$– mr23ceecCommented Jun 13, 2017 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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4$\begingroup$ I was surpriced to see that no one else had posted it. Thanks! @Kröw $\endgroup$– Olba12Commented Jun 12, 2017 at 0:56
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2$\begingroup$ @M.Herzkamp It is given right there in the question that 11 - 1 = 10. $\endgroup$ Commented Jun 12, 2017 at 14:01
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4$\begingroup$ Can you make the assumption that 29 = 20 + 9? $\endgroup$– UKMonkeyCommented Jun 12, 2017 at 15:17
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2$\begingroup$ @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). $\endgroup$ Commented Jun 13, 2017 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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3$\begingroup$ yup, that's what I thought it was. The puzzle isn't clear enough to eliminate this possibility. $\endgroup$– NH.Commented Jun 12, 2017 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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2$\begingroup$ +1 I had similar thoughts, thinking the arithmetic was in the field of 2 elements {0,1} $\endgroup$– user19988Commented Jun 13, 2017 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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1$\begingroup$ Like Olba12, you cannot mix the "$-$" in the puzzle with our usual minus sign. They behave rather differently! $\endgroup$ Commented Jun 14, 2017 at 10:23
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1$\begingroup$ 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. $\endgroup$ Commented Jun 14, 2017 at 13:02
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1$\begingroup$ By your definition you have 10 = 9 - 1 = 10 - 1 - 1 = 10 - 2 = 8, which is a contradiction. $\endgroup$ Commented Jun 19, 2017 at 12:05
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1$\begingroup$ 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. $\endgroup$ Commented Jun 19, 2017 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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4$\begingroup$ How does this work? Do we just choose any random base at each step? $\endgroup$– Wen1nowCommented Jun 12, 2017 at 8:10
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6$\begingroup$ not a random base, but just cherrypick the one that fits. Arbitrary ≠ random. $\endgroup$– NH.Commented Jun 12, 2017 at 15:16
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2$\begingroup$ 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. $\endgroup$ Commented Jun 14, 2017 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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4$\begingroup$ '
11-1=10' does not work here:11+1=12, not10. $\endgroup$ Commented Jun 22, 2017 at 4:41
how = 2? $\endgroup$