6
$\begingroup$

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

Improve this question
$\endgroup$
8
  • $\begingroup$ Related: this $\endgroup$
    – dcfyj
    Oct 30, 2017 at 19:36
  • 1
    $\begingroup$ Do those 'formulas' each stand by themselves or do they inter-relate in some manner that is crucial to the solution? $\endgroup$
    – Dr t
    Oct 30, 2017 at 20:02
  • $\begingroup$ my answer would be :things i've never seen. $\endgroup$
    – Jason V
    Oct 30, 2017 at 20:07
  • 1
    $\begingroup$ Apple mathematics? :) macrumors.com/2017/10/24/ios-11-calculator-animation-bug $\endgroup$
    – Bojan B
    Oct 30, 2017 at 21:34
  • $\begingroup$ @Drt They all use the same principals, if that helps, $\endgroup$
    – Chowzen
    Oct 30, 2017 at 21:40

4 Answers 4

Reset to default
10
$\begingroup$

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

Improve this answer
$\endgroup$
4
  • 3
    $\begingroup$ Excellent and great to see answering again. Small typo maybe EIGHT > 5 $\endgroup$
    – Tom
    Oct 31, 2017 at 16:26
  • 2
    $\begingroup$ 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" $\endgroup$
    – Sconibulus
    Oct 31, 2017 at 16:55
  • $\begingroup$ Correct, Levieux, and correct as well, @Sconibulus, nice job of lateral-thinking. $\endgroup$
    – Chowzen
    Oct 31, 2017 at 17:49
  • $\begingroup$ I figured it was ASCII based, but failed to see the pattern. $\endgroup$
    – Octopus
    Oct 31, 2017 at 21:33
2
$\begingroup$

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

Improve this answer
$\endgroup$
2
$\begingroup$

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): $$ add(a, b) = \frac{(10+a)\cdot(10+b)}{3}$$
$$ mul(a, b) = add(a, b) - 1$$
$$ exp(a, b) = add(a, b) + 1$$
$$ div(a, b) = mul(a, b) + \frac{add(a, b)}{100}$$

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ 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! $\endgroup$
    – boboquack
    Oct 31, 2017 at 10:20
  • 1
    $\begingroup$ @boboquack hahaha - yeah I was going to state right-hand-sides use normal operators! Guess I'll edit... $\endgroup$
    – Paul Evans
    Oct 31, 2017 at 10:42
1
$\begingroup$

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

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Though that does seem to work, that is not the method that I used to derive my answer. $\endgroup$
    – Chowzen
    Oct 30, 2017 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.