5
$\begingroup$

So my British friend proposed a new operator, but he wouldn't tell me how it works. He told me if I could figure it out then I could know. He's not very nice. Anyway, help me figure it out?

1 🎫 1 = 9

1 🎫 2 = 9

3 🎫 4 = 0

2 🎫 6 = 9

7 🎫 10 = 5

15 🎫 12 = 8

34 🎫 40 = 0

80 🎫 203 = 0

109 🎫 374 = 6

491 🎫 705 = ?

What's the method? What's the next answer?

He gave me a hint:

The numbers don't stay numerical the whole time


The (intended) answer has been requested, so:

There were three answers very close to the intended answer, and the closest was @puzzle-guesser's initial formula: counting out the letters in each number, multiplying those together, and then multiplying each digit of the product together until a single digit emerged. In compliance with British usage (sources: 1 2 3), my "friend" was considering the "and" as three letters in each number that required it.

Therefore 109 🎫 374 = 6 is reached by:

"one hundred and nine" (17) * "three hundred and seventy-four" (26) = 442;
4 * 4 * 2 = 16
1 * 6 = 6

Following the same method, 491 🎫 705 =

"four hundred and ninety-one" (23) * "seven hundred and five" (19) = 437
4 * 3 * 7 = 84
8 * 4 = 32
3 * 2 = 6

$\endgroup$
  • 2
    $\begingroup$ You may need to provide more examples. This isn't a lot of information to go off of. $\endgroup$ – dcfyj Jun 29 '18 at 13:43
  • $\begingroup$ @dcfyj a few more given $\endgroup$ – Blake Steel Jun 29 '18 at 13:50
  • $\begingroup$ are all numbers base 10? $\endgroup$ – sriram Jun 29 '18 at 17:45
  • $\begingroup$ All numbers are base ten $\endgroup$ – Blake Steel Jun 29 '18 at 17:47
5
$\begingroup$

Also built from QwirkyQwilfish's answer

The steps are as he suggested - spell out the numbers, count the letters in those words, and multiply them together, but the second step is a little different.
If there is only one digit after multiplication, that is the answer. However, if there is more than one digit, you simply multiply the first digit of the number by the last digit. There are not enough examples to determine if you then need to take the mod 10 of the result or if a multi-digit answer is allowed.
By this logic, the last problem will result in:
491 (operator) 705 = fourhundredninetyone (20) x sevenhundredfive (16)
20 x 16 = 320
3 x 0 = 0
Hence the answer to the last problem is 0

Another attempt using the information provided by Blake Steel:

Using the "and" is apparently important, so I took another crack and came up with the following.
As before, turn the numbers into their spelling, but when you have an "and", you break the words apart, subtracting the length of the words after the "and" from the length of the words before the "and", then discarding the "and". You then proceed to multiply the two adjusted lengths to get a result, which you then take only the last digit of (mod(10)). This works for all of the examples provided except the 15 🎫 12 one. However, it seems that by spelling 12 as "twoteen" and 15 as "fiveteen", I can get the answer to work, so I assume that 11, 12, 13, and 15 are translated as "oneteen", "twoteen", "threeteen", and "fiveteen" respectively so that the last digit is always represented by its single digit spelling (with the exception of numbers ending in 0).
So, the examples that are affected by this change would be solved as follows:
15 🎫 12 = fiveteen (8) x twoteen (7) = 56 mod 10 = 6
80 🎫 203 = eighty (6) x (twohundred (10) - three (5)) = 6 x 5 = 30 mod 10 = 0
109 🎫 374 = (one hundred (10) - nine (4)) x (three hundred (12) - seventy four (11)) = 6 x 1 = 6 mod 10 = 6
491 🎫 705 = (four hundred (11) - ninety one (9)) x (seven hundred (12) - five (4)) = 2 x 8 = 16 mod 10 = 6

This takes into account the comment by Blake indicating that some information was not taken into account and provides a different answer for the unanswered portion while still working with the other examples. If this still wrong, we might need some more examples, including one showing if the order of the operands is important (for example if 15 🎫 12 provides a different result than 12 🎫 15), as well as some examples where my method fails to work.

I'd love to discover the answer that Blake had in mind when the puzzle was created.

$\endgroup$
  • $\begingroup$ Your method wasn't expected and did not lead to the intended answer, but it does work with all the available info.. That being said, if you can figure out an answer involving the rot13("naq"), it would be most correct for the riddle, rot13( tvira vapyhqvat naq vf cneg bs Oevgvfu ahzoref-gb-grkg pbairagvbaf ). +1 for now, accepted soon $\endgroup$ – Blake Steel Jun 30 '18 at 4:03
  • 2
    $\begingroup$ @BlakeSteel I'm no Puzzling Expert, but if an answer fulfils all the given examples, I think that has to be the/a correct answer regardless of what the original maker intended. Then again, I'm a student who often has to deal with teachers giving vague questions with multiple answers, so maybe I'm a bit biased :P $\endgroup$ – Aryaman Jun 30 '18 at 5:29
  • $\begingroup$ This was where I started solving, but couldn't get the last one to match. $\endgroup$ – Herb Wolfe Jun 30 '18 at 14:14
  • 1
    $\begingroup$ @BlakeSteel sorry about the silence, I don't usually browse these forums on the weekends. I took another crack at it considering your first comment. I'd like to get the solution you intended, so please provide some more feedback (and maybe a couple more clues/examples) if this doesn't work. $\endgroup$ – cpcodes Jul 3 '18 at 0:58
  • $\begingroup$ @cpcodes I did not expect so many different answers to the question with what I thought was a pretty solid and unique method. I'll just have to be more careful next time ;) $\endgroup$ – Blake Steel Jul 3 '18 at 13:52
6
$\begingroup$

I believe it is

Length of word-version of numbers multiplied together and then the digits of that answer multiplied together repeatedly until it is a single digit.

1 🎫 1 = 9

one one 3*3=9

1 🎫 2 = 9

one two 3*3=9

3 🎫 4 = 0

three four 5*4=20 2*0=0

2 🎫 6 = 9

two six 3*3=9

7 🎫 10 = 5

seven ten 5*3=15 1*5=5

15 🎫 12 = 8

fifteen twelve 7*6=42 4*2=8

34 🎫 40 = 0

thirtyfour forty 10*5=50 5*0=0

80 🎫 203 = 0

eighty twohundredthree 6*15=90 9*0=0

109 🎫 374 = 6

onehundrednine threeseventyfour 14*16=224 2*2*4=16 1*6=6

491 🎫 705 = ?

fourninetyone sevenhundredfive 13*16=208 2*0*8=0

Credit to QwirkyQwilfish for the first step

$\endgroup$
  • 1
    $\begingroup$ In the 109 🎫 374 = 6, you rot13( pbhagrq gur jbeq "uhaqerq" bapr, ohg abg gur frpbaq gvzr, juvyr guvf znxrf lbhe zngu pbeerpg, vg vfa'g pbafvfgrag ). Pretty close though! $\endgroup$ – Blake Steel Jun 30 '18 at 3:45
5
$\begingroup$

Is it

number of letters in the numbers multiplied together modulus 10?

It doesn't seem to hold up for the latter half though.

$\endgroup$
  • $\begingroup$ Npghnyyl, vg ubyqf sbe nyy ohg 15b12 naq 109b374, juvpu znxrf zr jbaqre vs vg jnf n zvfgnxr (znlor zvffcryyvat 15 nf "svirgrra"? Be znlor gung vf gur vaqraqrq fcryyvat, fhpu gung guvegrra jbhyq or "guerrgrra" sbe gur checbfrf bs guvf bcrengbe. Fgvyy oernxf ba gur bgure, gubhtu. $\endgroup$ – cpcodes Jun 29 '18 at 21:50
  • $\begingroup$ You got the first half right, but the second needs some work $\endgroup$ – Blake Steel Jun 29 '18 at 22:29
2
$\begingroup$

I have skipped some basic math steps to try and make my answer as short as I can. If you do not understand, comment down below and I will make myself a bit more clear.


I do not mean to steal the answer of @QwirkyQwilfish as he/she was so close, but I believe the operator is

If you have two numbers each with $m$ and $n$ digits respectively, with $x$ letters and $y$ letters respectively, then the operator is equal to
$n\times (m+x-1)$ if $m>n$
Unknown if $m=n$
$m\times (n+y-1)$ if $m<n$


Supporting Examples:

$1$ 🎫 $1 = \underbrace{\text{one}}_{3}\times \underbrace{\text{one}}_{3} = 3\times 3 = 9$.

$1$ 🎫 $2=\underbrace{\text{one}}_{3}\times \underbrace{\text{two}}_{3} = 3\times 3 = 9$.

$3$ 🎫 $4 = \underbrace{\text{three}}_{5}\times \underbrace{\text{four}}_{4}=5\times 4 = 20 \to\require{cancel}{\cancel2}0=0$. $\qquad m=5, n=4$.

$2$ 🎫 $6 = \underbrace{\text{two}}_{3}\times \underbrace{\text{six}}_{3}=3\times 3 = 9$.

$7$ 🎫 $10 = \underbrace{\text{seven}}_{5}\times \underbrace{\text{ten}}_{3} = 5\times 3 = 15 \to \cancel{1}5=5$. $\qquad m=5, n=3$.

$10$ is two digits, but we disregard that because it does not have the greatest letter-amount.


Now, regarding the second half, it is a bit different because the numbers have two digits instead of $1$. But we can still use the same rule.


Supporting Examples

$15$ 🎫 $12=\underbrace{\text{fifteen}}_{7}\times \underbrace{\text{twelve}}_{6}=(7+1)\times 6 = 8\times 6 = 48\to \cancel{4}8 = 8$. $\; m=7, n=6$.

$34$ 🎫 $40=\underbrace{\text{thirty-four}}_{11}\times \underbrace{\text{forty}}_{5} = (11+1)\times 5 = 12\times 5 = 60\to \cancel{6}0=0.$

As you can see above, the dash/hyphen "$-$" counts as a letter. Moving on,

$80$ 🎫 $203= \underbrace{\text{eighty}}_{6}\times \underbrace{\text{two-hundred-and-three}}_{18}=(18+2)\times 6 = 20\times 6=120\to \cancel{12}0=0$.

In this example, $203$ has three digits, so we add $2$ to the greatest number, and we disregard $80$ with two digits because its letter-amount is less than the letter-amount of $203$, i.e. we do not add $1$ to $6$ because it is not the greatest number. Overall, the general formula is stated in the first sandbox above.



The last answer is:

$$\begin{align}491 \, 🎫 \, 705&=\underbrace{\text{four-hundred-and-ninety-one}}_{27}\times \underbrace{\text{seven-hundred-and-one}}_{21} \\ \\ &= (27+3-1)\times 21\qquad \binom{\text{since } 27>21 \text{ and}}{491 \text{ has } 3 \text{ digits }} \\ \\ &= (27+2)\times 21 \\ \\ &= 29\times 21 \\ \\ &= 609\to \cancel{60}9=9.\end{align}$$

$\endgroup$
  • 1
    $\begingroup$ I am really impressed with this answer, but the hyphens aren't spelled out all the time, here's a source on hyphens in English numbers. (I did not count hyphens though). $\endgroup$ – Blake Steel Jun 30 '18 at 3:58
  • $\begingroup$ @BlakeSteel ok, thank you. Did I get the last answer, though? :) $\endgroup$ – Mr Pie Jun 30 '18 at 4:51
  • 1
    $\begingroup$ no, unfortunately. I wish your method had worked though. $\endgroup$ – Blake Steel Jun 30 '18 at 13:14
  • $\begingroup$ @BlakeSteel ah dang. Oh well :) $\endgroup$ – Mr Pie Jun 30 '18 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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