How Can Be The Below Equation Can Be Right 29

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

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

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

Edit

Or just using that $9-1 = 10$

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


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

$$eginalign29-1&=28 xrightarrow extrounds lớn 30\14-1&=13 xrightarrowphantom ext________ 15\11-1&=10 xrightarrowphantom ext________ 10\9-1 &= phantom08 xrightarrowphantom ext________ 10\endalign$$


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This is mathematically true in $mathbbZ/2mathbbZ$, i.e. $mod 2$:

$$eginalign29 - 1 equiv 30 equiv 0 pmod 2\14 - 1 equiv 15 equiv 1 pmod 2\11 - 1 equiv 10 equiv 0 pmod 2\9 - 1 equiv 10 equiv 0 pmod 2\endalign$$


$egingroup$ +1 I had similar thoughts, thinking the arithmetic was in the field of 2 elements 0,1 $endgroup$
This is inspired by/alternative khổng lồ Olba12"s nice answer:

$$eginalign30&= 10 + 10 + 10\&= (11 - 1) + (11 - 1) + (9 - 1)\&= 31 - 3\&= 31 - 2 - 1\&= 29 - 1endalign$$

Hence: $29 - 1 = 30$.


Here"s another approach that doesn"t require redefining ‘$-$’ as a string operation, rather than a mathematical one:

$$eginalign29 - 1 ( extbase 11) &= 30 ( extbase 10)\14 - 1 ( extbase 12) &= 15 ( extbase 10)\11 - 1 ( extbase 10) &= 10 ( extbase 10)\9 - 1 ( extbase 10) &= 10 ( extbase 8)endalign$$

I lượt thích the OP"s intent better, though. It"s a clever puzzle.

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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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