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How does this work?:

$$\begin{array}\ &2+3 = 52 \\ &3 \times 2 = 51 \\ &2^3 = 53\\ &\frac{3}2 = 51.52 \\ \end{array} $$ Hint:

The "5" in the first position of the answers is not "fixed," per se, yet, coincidentally, I can't conceive of an answer that would not start with 5.

More equations:

7 x 2 = 56
9 + 8 = 57
3 - 2 = 51
3 - 1 = 51

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  • $\begingroup$ Related: this $\endgroup$ Commented Oct 30, 2017 at 19:36
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    $\begingroup$ Do those 'formulas' each stand by themselves or do they inter-relate in some manner that is crucial to the solution? $\endgroup$ Commented Oct 30, 2017 at 20:02
  • $\begingroup$ my answer would be :things i've never seen. $\endgroup$ Commented Oct 30, 2017 at 20:07
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    $\begingroup$ Apple mathematics? :) macrumors.com/2017/10/24/ios-11-calculator-animation-bug $\endgroup$ Commented Oct 30, 2017 at 21:34
  • $\begingroup$ @Drt They all use the same principals, if that helps, $\endgroup$ Commented Oct 30, 2017 at 21:40

4 Answers 4

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I think the answer is:

The ASCII code of the number of characters in the written version of the answer. ASCII codes for numbers are in the range [48-57] (for [0-9]).

So

2+3 = 5 > FIVE > 4 > 52
3*2 = 6 > SIX > 3 > 51
2^3 = 8 > EIGHT > 5 > 52
3/2 = 1.5 > ONE.FIVE > 3.4 > 51.52
7x2 = 14 > FOURTEEN > 8 > 56
9+8 = 17 > SEVENTEEN > 9 > 57
3-2 = 1 > ONE > 3 > 51
3-1 = 2 > TWO > 3 > 51

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    $\begingroup$ Excellent and great to see answering again. Small typo maybe EIGHT > 5 $\endgroup$ Commented Oct 31, 2017 at 16:26
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    $\begingroup$ If this is the answer, sqrt(2-3) should give i, which would be 49. I don't know if that falls afoul of "can't concieve of an answer that would not start with 5" $\endgroup$ Commented Oct 31, 2017 at 16:55
  • $\begingroup$ Correct, Levieux, and correct as well, @Sconibulus, nice job of lateral-thinking. $\endgroup$ Commented Oct 31, 2017 at 17:49
  • $\begingroup$ I figured it was ASCII based, but failed to see the pattern. $\endgroup$ Commented Oct 31, 2017 at 21:33
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I think the rule is:

Calculate the answer: if number 5 append with 2: e.x. 2+3=5 => 52; if number >5 then final answer is 5 appended with the difference of number and 5: e.g. 3*2 =6 => 51 i.e.(5(6-5)); if number <5 then answer 5 appended with number itself e.g 1 is <5 so 51 so 3/2 = 1.5 => 51.52

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Could be an infinite number of things, but one is:

Redefine addition, multiplication, exponentiation and division as $add(a ,b)$, $mul(a, b)$, $exp(a, b)$, and $div(a, b)$ such that (using normal operators on the right-hand-side): $$ add(a, b) = \frac{(10+a)\cdot(10+b)}{3}$$
$$ mul(a, b) = add(a, b) - 1$$
$$ exp(a, b) = add(a, b) + 1$$
$$ div(a, b) = mul(a, b) + \frac{add(a, b)}{100}$$

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    $\begingroup$ Recursive definition alert: $a+b=\displaystyle\frac{\left(\frac{20\cdot\left(\frac{20\cdot\dots}{3}\right)}{3}\right)\cdot\left(\frac{20\cdot\left(\frac{20\cdot\dots}{3}\right)}{3}\right)}{3}$, and that's not even taking into account multiplication! $\endgroup$ Commented Oct 31, 2017 at 10:20
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    $\begingroup$ @boboquack hahaha - yeah I was going to state right-hand-sides use normal operators! Guess I'll edit... $\endgroup$ Commented Oct 31, 2017 at 10:42
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I think it follows the following rules

Subtract 5 and prepend the number 5 to the remainder. i.e. 8 becomes 53, five, because its remainder is 0, is replaced by a 5

AND

If the number ends in "5", append 2, so 5 becomes 52 and 1.5 becomes 1.52

Thus

2+3 = 5, becomes 5 after the first rule is applied and 52 after the second is applied
3x2 = 6 becomes 51 after the first rule is applied, second rule doesn't apply
2^3 = 8 becomes 53 after the first rule is applied, second rule doesn't apply
3/2 = 1.5, which has a remainder of 1.5, becomes 51.5 after the first rule and 51.52 after the second

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    $\begingroup$ Though that does seem to work, that is not the method that I used to derive my answer. $\endgroup$ Commented Oct 30, 2017 at 22:26

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