# Stranger Math Things

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

• Related: this – dcfyj Oct 30 '17 at 19:36
• Do those 'formulas' each stand by themselves or do they inter-relate in some manner that is crucial to the solution? – Dr t Oct 30 '17 at 20:02
• my answer would be :things i've never seen. – Jason V Oct 30 '17 at 20:07
• Apple mathematics? :) macrumors.com/2017/10/24/ios-11-calculator-animation-bug – Bojan B Oct 30 '17 at 21:34
• @Drt They all use the same principals, if that helps, – Chowzen Oct 30 '17 at 21:40

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

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

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

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

• 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! – boboquack Oct 31 '17 at 10:20
• @boboquack hahaha - yeah I was going to state right-hand-sides use normal operators! Guess I'll edit... – Paul Evans Oct 31 '17 at 10:42

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

• Though that does seem to work, that is not the method that I used to derive my answer. – Chowzen Oct 30 '17 at 22:26