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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

bobble
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Amruth A
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  • You mean if I guess the answer I am.... imaginary. – Marius Jun 10 '17 at 09:42
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    I think the hint is confusing. – jamesdlin Jun 10 '17 at 12:28
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    @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. – Alex Hall Jun 10 '17 at 15:04
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    How about how = 2? – Frozn Jun 12 '17 at 11:33
  • Instead of "-1" it should have been "-i". It would have led to interesting proofs and been more correct. – EvSunWoodard Jun 12 '17 at 13:50
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    @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. – jamesdlin Jun 13 '17 at 06:25
  • Athena and Minerva are both counterparts of each other ..one in Rome other in Greek .. @jamesdlin – Amruth A Jun 13 '17 at 06:27
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    @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. – jamesdlin Jun 13 '17 at 07:49
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    @jamesdlin Maybe not two jumps. I think you can go "Greek Numerals => Roman Numerals" on the off chance one doesn't know who Minerva is. Greek numerals, and games you can play with them, are almost if not more popular and famous than Roman ones. – Lan Jun 13 '17 at 12:44
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    @AmruthA why is there a new open bounty on this post when there's already an accepted answer? – El-Guest Jul 20 '18 at 14:13
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    @El-Guest That is what I was wondering too – PotatoLatte Jul 22 '18 at 02:32
  • @ZaniXu i wanted this question to be highest viewed question.. – Amruth A Jul 22 '18 at 10:03

8 Answers8

143

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

ffao
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    If this is actually the answer, then the hint is misleading - it should have been about Minerva. :) – fluffy Jun 11 '17 at 19:26
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    @fluffy " like Athena " not Athena ..counterpart for Minerva in Greek .. – Amruth A Jun 12 '17 at 10:33
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    @AmruthA but you're saying the answerer will be like Athena, or at least that's what the statement implies. – TheWanderer Jun 12 '17 at 11:34
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    @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.) – fluffy Jun 12 '17 at 20:55
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    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... – mr23ceec Jun 13 '17 at 11:08
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    Good luck on getting 8 more upvotes! – Beastly Gerbil Jun 13 '17 at 20:01
46

$$9 - 1 = 10 = 11 - 1,$$ thus

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

Edit

Or just using that $9-1 = 10$

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

Olba12
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    Why did you(?) vote for delete? – Olba12 Jun 11 '17 at 23:06
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    This is probably the best thing I've seen all day. I can't believe it only has 2 upvotes. – Kröw Jun 12 '17 at 00:50
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    I was surpriced to see that no one else had posted it. Thanks! @Kröw – Olba12 Jun 12 '17 at 00:56
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    @Olba12- Good answer! – ksp585 Jun 12 '17 at 04:36
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    @M.Herzkamp It is given right there in the question that 11 - 1 = 10. – Masked Man Jun 12 '17 at 14:01
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    Can you make the assumption that 29 = 20 + 9? – UKMonkey Jun 12 '17 at 15:17
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    Imagine That you have set $X$ with two operations, ${+,-1}$ think of the $+$ as we normally do, and $-1$ as an operator who is defined by the last digit in a number. $29-1 = (2\cdot 10^1 + 9\cdot 10^0)-1 = 2\cdot 10^1 +(9-1)$. Now $(9-1)$ is defined to be $10$. Thus $-1$ maps it to $10$ and $2\cdot10^1$ maps to itself. Then for instance, what is $28-1$? Well, $(28)-1=(10 + 9+9)-1 = 10 + (9 + 9)-1 = 10 + 9-1 + 9-1 = 30$. You asked a really good question. I tried to answer it.It is far from perfect. Obviously I didnt think of this, when I came up with the answer to the question. @UKMonkey – Olba12 Jun 13 '17 at 00:25
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    @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). – Masked Man Jun 13 '17 at 05:38
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    Now you can. ;)@M.Herzkamp – Olba12 Jun 13 '17 at 10:18
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    Okay, before the edit, there was still the issue of 31-1 = 30, which was definitely not provided ;-) After the edit you use associativity of + and - (i.e. (x+y)-1 = x + (y-1) ) of which I am not convinced yet... – M.Herzkamp Jun 13 '17 at 11:02
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    I decieded that $-1$ would be a linear operator. I also defined it to map $20$ to $20$ and $9$ to $10$. If you look in my comment above tagged to UKMonkey. So its basically linearity. @M.Herzkamp – Olba12 Jun 13 '17 at 11:18
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    You cannot build a consistent operation that way which fulfills the three equations above. Your assumption is (x+y)-1 = x-1 + y-1. Using 9-1 = 10 and 11-1 = 10 you get 2-1 = 0, which leads to 1-1 + 1-1 = 0, so 1-1 should be zero as well. But then any number must be mapped to 0, which contradicts the equations given in the puzzle. – M.Herzkamp Jun 13 '17 at 11:41
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    How did you get $2-1 = 0$? If you wrote it as $2-1=(11-9)-1 = 11-1 - 9-1 = 0$ Then I dont agree with you. @M.Herzkamp – Olba12 Jun 13 '17 at 11:44
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    10 = 11-1 = (9+2)-1 = 9-1 + 2-1 = 10 + 2-1. Since the "+" is our usual plus, and both sides need to be identical, it follows that 2-1 = 0 – M.Herzkamp Jun 13 '17 at 11:56
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    @M.Herzkamp Yes, that is correct, and then one (I) must find a better way to define -1. – Olba12 Jun 13 '17 at 12:23
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    Lets say you have an $x$ then we look at the expressions, $x - (10\cdot k + 9), x-(10\cdot k + 11) , x - (10\cdot k+ 14)$ where $k$ is natural number, and $-$ in this expression is normal minus. Then let $y$ be the smallest, must be positive or $0$, of the three expressions. Given $x$ assume for clarity that $x-(10\cdot k +9)$ was the smallest. Then we define $x-1$ as $x-1 = (x-9) + 9-1 = (x-9) + 10 = x + 10.$ Then $28-1 = (28-14) + 14-1 = 29$. @M.Herzkamp – Olba12 Jun 13 '17 at 22:10
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    To be honest, I don't understand your definition... But you need to keep in mind, that it must also be valid for the three cases given in the question! – M.Herzkamp Jun 14 '17 at 10:24
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    It is, because $14-1 = (14-14) +14-1 = 0+15= 15$, $9-1= (9-9)+ 9-1= 0 + 10=10$, $11-1=(11-11)+11-1$=0+10=10$. You have to seperate normal minus from the operator $-1$.@M.Herzkamp – Olba12 Jun 14 '17 at 10:52
23

I suppose you could always round the answer to the nearest multiple of 5, although that has nothing to do with Athena.

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

grg
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Amadeus
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    yup, that's what I thought it was. The puzzle isn't clear enough to eliminate this possibility. – NH. Jun 12 '17 at 15:13
19

This is mathematically true in $\mathbb{Z}/2\mathbb{Z}$, i.e. $\bmod 2$:

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

boboquack
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ClementWalter
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    +1 I had similar thoughts, thinking the arithmetic was in the field of 2 elements {0,1} –  Jun 13 '17 at 14:28
5

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: $29 - 1 = 30$.

grg
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Masked Man
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    Like Olba12, you cannot mix the "$-$" in the puzzle with our usual minus sign. They behave rather differently! – M.Herzkamp Jun 14 '17 at 10:23
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    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. – Masked Man Jun 14 '17 at 13:02
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    By your definition you have 10 = 9 - 1 = 10 - 1 - 1 = 10 - 2 = 8, which is a contradiction. – M.Herzkamp Jun 19 '17 at 12:05
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    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. – Masked Man Jun 19 '17 at 12:37
4

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.

grg
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Mark Kreitler
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    How does this work? Do we just choose any random base at each step? – Wen1now Jun 12 '17 at 08:10
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    not a random base, but just cherrypick the one that fits. Arbitrary ≠ random. – NH. Jun 12 '17 at 15:16
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    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.

     – Mark Kreitler Jun 14 '17 at 06:03
2

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

SteamCode
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Lawrence
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0

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

Shaleen Jain
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