Interview riddle

Question

On the Mathematics chat we were recently talking about the following problem @Chris'ssis had to solve during an interview :

$$3\times 4=8$$ $$4\times 5=50$$ $$5\times 6=30$$ $$6\times 7=49$$ $$7\times 8=?$$

We have not managed to solve it so far, all we know is the solution (which was given after we had given up) :

$224$

How do we find this solution ?

What kind of interview was this? I hate it when people use equals signs to describe relations that aren't equal. — NoName, Jul 14 '14 at 14:05
@ZachGershkoff It depends on the definition of the operation $\times$ ... that is what this problem asks for: To find a reasonable definition of $\times$. — martini, Jul 14 '14 at 14:06
@martini The problem is that we can't define "reasonable" =) — Balarka Sen, Jul 14 '14 at 14:08
"interview" and "riddle" - unless you're Tom Riddle and you're interviewing somewhere, you should never hear those two words in the same sentence. =( — corsiKa, Jul 14 '14 at 17:35
That interview question is good sign that it's time to walk out the door. — RQDQ, Jul 14 '14 at 17:48
naa, its a common practice to through such questions at the job seeker. Trust me, I have seen these kind of question. The aim, which is still debatable, is to check the logical approach of the candidate. And yes, these questions cannot be supported by a proper mathematical proof of concepts. Sigh! — MonK, Jul 14 '14 at 18:31
Did the interviewer give the answer? Or was the question 'the answer is 224, why?'. Could have been one of those questions where they just want to see what you try. Some more context would be nice! — john w., Jul 14 '14 at 18:58
To me, it would seem a better way to ask this question is to posit some function f(x,y) = z given the following examples ... determine the function. I don't see the point in using the standard multiplication symbol when determining how a candidate solves mathematical problems. — CramerTV, Jul 14 '14 at 19:23
I would probably have taken this question as a challenge to convince the interviewer that the answer isn't 224. The question is arbitrary, so it really just comes down to a battle of wills. Bonus points if you make the interviewer cry while he concedes your answer. — DanielV, Jul 14 '14 at 21:02
Btw, as a general riddle solving technique, when no pattern appears in a set of numbers, it probably involves some a posteriori information. Dates, Sports data, etc. It might be relevant what the company and job position are. — DanielV, Jul 14 '14 at 21:10
@corsiKa, It's pretty common to give a riddle at an interview not to see if you can arrive at the answer, but to see how you attempt to find an answer. — Brian S, Jul 14 '14 at 22:44
@DanielV: further bonus points if you can get the person you'll be negotiating salary with in on the discussion. "Your salary will be 56K", "excellent, and 56 = 7 * 8 = 224, so what's 224k after tax and what's the dental plan?" — Steve Jessop, Jul 14 '14 at 23:32
This is almost as bad as $\sum \limits_{n=1}^{\infty} n = -\frac{1}{12}$. :) I really abhor bad notation choices. — Ryan, Jul 15 '14 at 00:24
I think we should add the hint (even if as a spoiler) that what we need is to re-define $\times$. (I think an interviewer would answer "yes" to the question, "Does 4 still mean 4?" etc.) — Andrew Kelley, Jul 15 '14 at 02:51
I would expect a problem like this to have something to do with meta-information about the numbers like the number of letters in their words in Esperanto or something. If an employer insisted on asking questions like these I'd probably take it to mean that I don't to work there anyway. — fluffy, Jul 15 '14 at 06:22
I see a picture. What's the question? As far as I see, 7 x 8 = ?. By definition. — HostileFork says dont trust SE, Jul 15 '14 at 07:22
Take the first one, which says 12 = 8. Subtract 8 and divide by 4. You have proven that 1 = 0, so any number is equal to any other. If the interviewer claims that 7 x 8 = 224, he is right. So are you, if you claim it is any other value. — CompuChip, Jul 15 '14 at 09:18
On the one hand I am also inclined to be cynical about this question, but on the other hand, look at some of the answers it yields. It's interesting to see what patterns people discover in a seemingly irregular pattern like this one, and what (possibly arcane) methods/knowledge they apply in arriving at it. Moreover, some openly dislike the question but apparently still feel urged to act on the problem-solving itch. Speaks for them. — Roy, Jul 15 '14 at 09:59
If you're going to redefine x, why not also redefine = and/or the digits? — keshlam, Jul 15 '14 at 18:41
Simple approach: $7\times8$ is $56$. Just because the interviewer has supplied a lot of incorrect information does not mean you are obliged to supply, or consent to, an incorrect answer. — David, Jul 16 '14 at 08:04
Might be worth seeing if this would be a good fit for http://puzzling.stackexchange.com/ (I don't know if it is since although I've popped over there occasionally I've never wanted to post so haven't read their guidelines). Should have people who might be up for the challenge from a less mathematical point of view (which might be appropriate here). — Chris, Jul 16 '14 at 11:07
"There are uncountable many functions that match the given examples. Shall I enumerate them all?" — Cephalopod, Jul 16 '14 at 11:40
Just working through it, noticing the sixths. 34=124/6=8 45=20(35)/6=50 56=306/6=30 67=427/6=49 78=56(38)/6 = 224 It's increasing sixths, multiplied by 3 when the first number is 3x+1. — please delete me, Jul 16 '14 at 18:15
What job was the interview for ? So we could see if they were looking for a mathematical answer or for a "trick" based on some sort of logic answer — WizardLizard, Jul 17 '14 at 11:43
I think my answer below shows how to come to the "expected solution" 224. As it was shown here it is one of many solutions that would fit. However I think the expected solution has beauty and logic....and at the end is quite SIMPLE! I also added some more explanation to it. — user2477732, Jul 20 '14 at 09:34
I think there is a lot of negativity on this style. As long as no one said "times," I think we are O.K. We can (hopefully) all wrap our head around a well defined binary operation on $\mathbb Z$ that overloads notation that is not used elsewhere in the problem. It isn't nice, but it could be way worse. — Thoth19, Jul 21 '14 at 05:16
Interview questions such as this remind me that "communicating badly and then acting smug when you're misunderstood is not cleverness". They are also a good indicator that you don't want the job. — zzzzBov, Jul 21 '14 at 05:43

Denis · Accepted Answer · 2014-07-18T16:34:44.760

109

These interview problems are sometimes weird, where notations are bad, rules are arbitrary, and they expect only one answer where several could fit.

Here is one, which could be the expected one, but probably not:

To compute $a \times b$, take the numerator of $\dfrac{ab^2}{6}$ after simplification of the fraction.

I don't see how they could argue it is wrong.

edited Jul 18 '14 at 16:34

answered Jul 14 '14 at 14:56

Denis

6,945
1
21
22

Um, the numerator of $\frac{n(n-1)}{6} \implies k = (n)(n-1)$. Thus, you say $(n-1)\times n = n^2(n-1)$, which doesn't match $3\times 4$ from the OP. – apnorton Jul 14 '14 at 17:41
@anorton, no it is the numerator of the fraction after simplification, i.e. if 2 or 3 divide it they are removed from the numerator – Denis Jul 14 '14 at 18:31
Ah! ok. ${}{}{}$ – apnorton Jul 14 '14 at 18:33
24

Holy **** how can you come up with this answer?? – Hoàng Long Jul 15 '14 at 03:31
I don't think this is the answer because it requires disregarding part of the result. Let's look for where the 6 in the denominator comes from on the results that simplify completely ... my guess is it's calculated as opposed to defined. We're probably looking for a formula where each answer is returned exact. – MrCMedlin Jul 15 '14 at 03:43
39

+1 "I also don't like these problems, where notations are bad, rules are arbitrary, and they expect only one answer where several could fit." – Sawarnik Jul 15 '14 at 08:44
5

@HoàngLong similarly to anorton's answer: I entered the sequence of quotients 2,10,5,7,28 in OEIS, then people helped to simplify it in this particular case (see the comment of Joseph Geipel) – Denis Jul 15 '14 at 15:25
1

@Denis Hahahah! I did the same thing! Go OEIS :) – Albert Renshaw Jul 16 '14 at 05:04
1

https://oeis.org/. Then again, I assume you could have found that yourself ;) – CompuChip Jul 16 '14 at 06:38
How is it apply for the second one? $\frac{4*5^2}{6}$ != 50 – Himanshu Jul 16 '14 at 11:26
2

@hims056 $\frac{4*5^2}{6}=\frac{50}{3}$ so the numerator is $50$. – Denis Jul 16 '14 at 11:28
@Denis - Ohh got it. – Himanshu Jul 16 '14 at 11:30
Us old people can't read the tiny font under ab2. Is that the number 6? – Reactgular Jul 18 '14 at 16:22
yes it is 6, I just increased the size. – Denis Jul 18 '14 at 16:35
1

@MathewFoscarini, this meta post might help with that. – Antonio Vargas Jul 18 '14 at 16:43
It seems this is the answer because 7x8 = 224 is correct. – acegs Jul 20 '14 at 11:08
1

@xuwicha Obviously it is because Denis is the interviewer. – ministic2001 May 10 '15 at 05:16

score 96 · Answer 2 · edited Jul 17 '14 at 00:29

96

Easy, just define

$$\begin{array}{rcl}a \times b &=& \hspace{10.5pt}(a-4)(b-5)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/72 + \\&& 25(a-3)(b-4)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/18 + \\&& 15(a-3)(b-4)(a-4)(b-5)(a-6)(b-7)(a-7)(b-8)/8 \hspace{5.25pt}+ \\&& 49(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-7)(b-8)/36 + \\&&\hspace{5.5pt}7(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-6)(b-7)/18\end{array}$$

edited Jul 17 '14 at 00:29

Alexander Gruber

26,963

answered Jul 15 '14 at 08:14

CaptainCodeman

1,639

53

You mean $\frac{181 a ^{4} b ^{4} - 4318 a ^{4} b ^{3} + 37880 a ^{4} b ^{2} - 144623 a ^{4} b + 202440 a ^{4} + -3594 a ^{3} b ^{4} + 85893 a ^{3} b ^{3} - 754833 a ^{3} b ^{2} + 2886774 a ^{3} b - 4047120 a ^{3} + 26012 a ^{2} b ^{4} - 622766 a ^{2} b ^{3} + 5482627 a ^{2} b ^{2} - 21003793 a ^{2} b + 29493240 a ^{2} + -81093 a b ^{4} + 1944783 a b ^{3} - 17150580 a b ^{2} + 65813730 a b - 92559600 a + 91560 b ^{4} - 2199120 b ^{3} + 19423320 b ^{2} - 74648280 b + 105134400}{36} $? – Axel Kemper Jul 15 '14 at 08:30
19

@AxelKemper Multiplication has never looked so epic! – CaptainCodeman Jul 15 '14 at 08:34
9

Holy Mother God! – MonK Jul 15 '14 at 18:09
6

Hilarous $\phantom{}$ – Newb Jul 15 '14 at 19:41
3

This is obviously the answer. How can it be anything else. – Lost1 Jul 21 '14 at 13:55
@CaptainCodeman how did you get that ? – ParaH2 Nov 05 '16 at 14:12
Did someone even check whether all the pairs give correct answers? – Truth-seek Jun 28 '17 at 17:34
1

@Hexacoordinate-C For every value of a,b all of the expressions evaluate to zero except one. Then you just need to change the coefficient for that expression to make it be whatever you want it to be. – CaptainCodeman Jul 02 '17 at 03:30

Omran Kouba · Answer 3 · 2014-07-14T16:56:58.477

69

This might be a possible solution. For a positive integer $n$, let $\nu_2(n)$ be the largest $k$ such that $2^k|n$, and similarly, let $\nu_3(n)$ be the largest $k$ such that $3^k|n$. Finally let $$h(n)=\frac{n}{3^{\nu_3(n)}2^{1+4\lfloor \nu_2(n)/4\rfloor}}$$ If we consider $$ a\times ~ b {\buildrel \rm def\over =}~b h(ab) $$ then $(k-1)\times k$ coincides with the proposed results for $k=4,5,6,7$ and yields $224$ for $k=8$.

edited Jul 14 '14 at 16:56

answered Jul 14 '14 at 16:25

Omran Kouba

28,772

44

Your answer made me laugh just trying to imagine an interviewer who could comprehend such a rule let alone expect anyone to come up with it on the spot during an interview. – heropup Jul 14 '14 at 17:20
12

@heropup, So, you understand my feelings while trying to to write it down! :) – Omran Kouba Jul 14 '14 at 17:57
39

How in the world did you come up with this? – Balarka Sen Jul 14 '14 at 18:38
I don't even... – Amal Murali Jul 15 '14 at 13:03
54

Plot twist: @OmranKouba was the interviewer all along – yuritsuki Jul 16 '14 at 05:41

score 36 · Answer 4 · answered Jul 14 '14 at 20:35

36

The left-hand-side input and the right-hand-side output can be imagined as binary numbers in a kind of truth table:

enter image description here

All eight output bits can be calculated from the seven input bits evaluating simple Boolean expressions.

answered Jul 14 '14 at 20:35

Axel Kemper

4,943

21

I figured binary boolean logic was involved. However, without knowing the final answer to the question, you wouldn't come up with the right answer, because without knowing from 7x8=224 that h=i=d, and also not knowing that results larger than 6 bits exist, your method would give you an answer of just 100000 (32). – KeithS Jul 14 '14 at 20:59
+1 for the analysis though. – KeithS Jul 14 '14 at 21:04
2

All the operations are very arbitrary, though. Given just the first 4 and no solution, you could do many distinct operations for each digit and get many different solutions. You could make the "answers" to the "equations" just about any 5 numbers and come up with binary operations – Khan Jul 15 '14 at 16:56
1

where is 'L' and 'O' ? – user3459110 Jul 16 '14 at 05:25
2

@Awal: Developers often skip l and o to avoid confusions. Now, I seem to have confused you? – Axel Kemper Jul 16 '14 at 06:34
Awesome. Bits h, i, and j are completely arbitrary. So, you could set h = 0, i = 0, and j = g and still be consistent with the four given clues. Then, insist that $7 \times 8 = 0$ until you get the job. – 200_success Jul 20 '14 at 05:54

score 30 · Answer 5 · answered Jul 14 '14 at 18:19

30

Spoiler Alert: (I use the answer given above in the response below. If you don't want to see it, you may want to skip this answer...)

I'm replacing $\times$ by $\circ$, as the latter is more commonly used with unknown operations. I hate it when people redefine a common symbol, then "$=$" to describe a relationship.

Note that $$\begin{align}3\circ4 &= 4\cdot 2\\ 4\circ 5 &= 5\cdot 10\\ 5\circ 6 &= 6\cdot 5\\ 6\circ 7 &= 7\cdot 7 \\ 7\circ 8 &= 8\cdot 28 \\ \end{align}$$

Thus, we can define: $$a\circ b\quad{\buildrel \rm def\over =}\quad b\cdot x_a$$ Where $x_n$ is some sequence. OEIS yields three possible sequences: $$x_n = \frac{\binom{n+2}{2}\gcd(n,3)}{3},\quad n \ge 0$$ (A234041) $$x_n = \text{denominatorOf}\left(\frac{(n-2)(n+3)}{(n)(n+1)}\right)\quad n \ge 3$$ (A027626: GCD of $n$-th and $(n+1)$st tetrahedral numbers, offset by me for this problem)

The last sequence from OEIS is A145911 which is not promising at all. (It's a combination of, what appears to be, $3$ other sequences.)

answered Jul 14 '14 at 18:19

apnorton

17,706

Note: I may have goofed up in how I offset the sequence, but I think I did it right... – apnorton Jul 14 '14 at 18:21
64

To me, this confirms that the interviewer is smoking crack. – heropup Jul 14 '14 at 18:24
I agree with @heropup, although it does depend what the job is for. I mean, if this is a job ad for some mathematics specialist, then ... maaaybe ... – GreenAsJade Jul 17 '14 at 01:05
If you search the first 4 terms of $x_a$ on OEIS, the 5th term in the first few results is 28, so that's a way this could plausibly be "solved". – Jakob Weisblat Jul 20 '14 at 15:36

score 21 · Answer 6 · answered Jul 15 '14 at 03:23

The answer is $42$.

$69$ is also the answer.

"Purple feelings" is also an answer.

The truth of each of these is, of course, vacuous. :)

If the question is posed as something other than multiplication, then it is the fault of the questioner for miscommunicating.

Although, one could arguably blame the person trying to solve this problem for not doing enough to extract enough requirements from the 'customer' to be able to provide a solution. In some settings, this is an extremely important skill.

maddog2k · Answer 7 · 2014-07-15T03:23:09.197

16

56 Did the question explicitly say there was a pattern to be found or is it just like you've presented it here? The symbols for multiplication(x) and equality(=) have well defined mathematical meaning and therefore 7 x 8 = 56 regardless of what misleading noise was written before. It may just be a test of the ability to avoid presumption.

edited Jul 15 '14 at 03:23

answered Jul 15 '14 at 03:12

maddog2k

177

1

That is one approach, but there are a large class of problems like this (although I hate them all) that redefine either $=$ or $\times$. This is, most likely, one of these problems. – apnorton Jul 15 '14 at 03:24
3

+1. If you want to redefine standard mathematical operations, you should say so. Otherwise, stick to mathematical convention. If you want me to solve a "mathematics" puzzle and can't manage to bring it across properly, I don't expect we'll be able to communicate very well and this is probably not a person I want to work for :) – CompuChip Jul 15 '14 at 09:20

score 14 · Answer 8 · edited Apr 13 '17 at 12:21

$$p(x)=$$

$$-\frac{1486263915627335609976345925580307452480}{198824918770116952269605821139049374259}-\frac{23535858736574459335924875719051524464677 x}{1789424268931052570426452390251444368331}+\frac{1532186339457747628597246965489647712097745599 x^2}{742539494635629574624160683858739355082631760}-\frac{5300973178829466500668773673899060773511329723 x^3}{62373317549392884268429497444134105826941067840}+\frac{425139989729581169917246837619141657974952401 x^4}{374239905296357305610576984664804634961646407040}-\frac{15160892592292573821061148160317799661783 x^5}{7128379148502043916391942565043897808793264896}+\frac{2379833487879115598578638026951579913181 x^6}{1496959621185429222442307938659218539846585628160}-\frac{133849478325585275186149006837381343 x^7}{249493270197571537073717989776536423307764271360}+\frac{9291465647310545015926219743101 x^8}{136087238289584474767482539878110776349689602560}$$

Then
$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x & 12\color{grey}{(3\times 4)} & 20\color{grey}{(4\times 5)} & 30\color{grey}{(5\times 6)} &42\color{grey}{(6\times 7)}&56\color{grey}{(7\times 8)}& \color{grey}{1729}&\color{grey}{2014}&\color{grey}{2015}&\color{grey}{2016}\\ \hline p(x)& 8 & 50 & 30 &49&\color{red}{224}& \color{grey}{1729}&\color{grey}{2014}&\color{grey}{2015}&\color{grey}{2016} \\ \hline \end{array}$$

To learn to play this 'game', read me.

I was tempted to write a similar solution :) an interpolating polynomial. — Lost1, Jul 21 '14 at 14:03

Asinomás · Answer 9 · 2014-07-14T18:13:26.313

9

Here is something I did which lead me to an incorrect result, but it is still pretty close.

Since all the values we are given are of the form $a\times (a+1)$, I decided to make the function $f(a)=a\times (a+1)$. Assuming $f$ is a polynomial of grade $4$ or less we obtain $f$ is equal to $\frac{101 x^3}{6}-233 x^2+\frac{6301 x}{6}-1500$ using interpolation.

This function gives us $f(7)=208$, which comes close, but is still not correct.

edited Jul 14 '14 at 18:13

answered Jul 14 '14 at 16:41

Asinomás

105,651

The f function in not incorrect, although it may yield defferent result from that in the "supposed" answer. – Cthulhu Jul 16 '14 at 07:50
Haha, well I guess you are right, I don't think anyone would have arrived at the "correct" solution if they hadn't provided it. – Asinomás Jul 16 '14 at 13:22
1

"Correct" in the context of this interview question seems to mean someone thinking the same way they do. My alternative answer is (101 b^3)/6 - (567 b^2)/2 + (4700 b)/3 - 280, which gives the same answer as yours. Therefore, our answers are correct because we think the same way – Daniel Gibson Aug 26 '23 at 12:43

score 4 · Answer 10 · answered Jul 14 '14 at 18:54

This is what I have so far, it seems a bit more intuitive than Omran's solution.

Based on the flip-flopping numbers, I figured the answer has to rely on the prime factorization of the numbers in question. So in particular, we see:

$$3 \times 2^2 \Rightarrow 2$$ $$2^2 \times 5 \Rightarrow 2*5$$ $$5 \times 2 * 3 \Rightarrow 5$$ $$2 * 3 \times 7 \Rightarrow 7$$ $$7 \times 2^3 \Rightarrow 2^2*7$$

So my initial hypothesis, which is that you took the highest prime and any primes with power greater than $1$ fails for the first equation. But it does look like a promising lead.

I reached the same conclusion (and honestly reached it, unlike my comment to boywholived where I clearly lied through my teeth). Prime factorization plays a role somehow. — corsiKa, Jul 16 '14 at 17:57

vuur · Answer 11 · 2014-07-15T13:26:25.077

One solution is to define the operation $\times$ between two integers as

$m \times n = n \cdot \left\{ \begin{array}{ll} \frac{1}{3}\sum_{k=1}^m k &\mbox{if } 3 \mid \sum_{k=1}^m k \\ \sum_{k=1}^m k &\mbox{otherwise.} \end{array} \right.$

The point is, that what remains of the RHS after dividing by $n$ can be recognized as the sum of the first $m$ integers, divided by $3$ should that be possible.

user2477732 · Answer 12 · 2014-07-20T09:28:57.087

numbers sequence

3 . 4 = 4 X 2 = 8

4 . 5 = 5 X 10 = 50

5 . 6 = 6 X 5 = 30

6 . 7 = 7 X 7 = 49

7 . 8 = 8 X 28 = 224

8 . 9 = 9 X 4 = 36

9 . 10 = 10 X 5 = 50

10 . 11 = 11 X 55 = 605

11 . 12 = 12 X 11 = 132

12 . 13 = 13 X 13 = 169

13 . 14 = 14 X 91 = 1274

14 . 15 = 15 X 7 = 105

15 . 16 = 16 X 8 = 128

16 . 17 = 17 X 136 = 2312

17 . 18 = 18 X 17 = 306

18 . 19 = 19 X 19 = 361

19 . 20 = 20 X 190 = 3800

20 . 21 = 21 X 10 = 210

a.(a+1) = (a+1)x((a+1)/2) - even number/2

b.( b+1) = (b+1)x(((a+1)/2 )xc)) - middle of the sandwich

c.(c+1) = (c+1)xc

d.(d+1) = (d+1)x(d+1) - odd number - copy

e.(e+1) = (e+1)x((d+1)x(f/2))

f.(f+1) = (f+1)x(f/2)

a+1=b, b+1=c,…

The problem is more about finding the patterns and relations between numbers and given equations.

3×4=8
4×5=50
5×6=30
6×7=49
7×8=?

There are some assumptions that have to be made:
1) look at the given equations as a sequences of numbers (sequences is plural - not just one sequence)
2) results on the right side can always by divided by the second number on the left side 8:4=2, 50:5=10, 30:6=5, 49:7=7 => the result of the last equations is therefore multiple of 8 => 7x8=8x?=???
(Note: Why did they use "x" when multiplication is clearly not what is done with those numbers? Why not better use symbol ∘ for unknown operations? My guess is - it's also a hint.... multiplication is necessary in the answer. ....so don't try to come up with solutions that are more complex than that ;-) But that's just my guess)
3) we can write down what we assume so far:
3 ∘4 = 4 X 2 = 8
4 ∘ 5 = 5 X 10 = 50
5 ∘ 6 = 6 X 5 = 30
6 ∘ 7 = 7 X 7 = 49
7 ∘8 = 8 X ?=???

we can also say that after 7∘8=8x???

some other equations should follow and the patter we know so far is is:

8 ∘9 = 9 X ?=???
9 ∘10 = 10 X ?=???

5) now look at the numbers sequence (fourth number in each equation): 2, 10, 5, 7, ... there are of course many things we can do (2+8=10, 10-5=5, 5+2=7,etc.)... but we also have a possible relation to 3.4, 4.5, 5.6,6.7
6) the easy patter would be "sandwich"- second number = first*third
7) how to define first and third number? - check the relation with 3.4 and 5.6 and first number also has a relation to 7.7
..the rest I already explained in the comment section below ;-)

i made excel spreadsheet with formulas to calculate the fourth number. — user2477732, Jul 15 '14 at 14:12
the fourth number sequence is 2, 10, 5, 7, 28, 4, 5, 55, 11, 13, 91, 7,….etc — user2477732, Jul 15 '14 at 14:15

I. J. Kennedy · Answer 13 · 2019-11-17T19:41:30.253

1

Surely even this interviewer meant for $\times$ to be commutative, so I propose $$a \times b = \left( \binom{a+1}{3},\binom{b+1}{3} \right) \cdot \max(a,b)$$ where the inner brackets are binomial coefficients, and the outer brackets indicate the gcd.

edited Nov 17 '19 at 19:41

answered Nov 23 '14 at 19:42

I. J. Kennedy

4,048

score 1 · Answer 14 · answered Jul 15 '14 at 13:49

The first multiplicant is given. So the open question is "what is the second multiplicant"?

The list can be grouped in sixes. So lines 1-6 is one group, the rules count for each group.

Define fm as the given first multiplicant of the row. Start with 4. Increment fm by one on each row
Set cm (current multiplicant) to 1
The result for each row is result = fm * cm. You only change cm from row to row

These are the rules for the six rows

cm := cm + 1
cm := fm * cm
cm := fm - 1
cm := fm
cm := cm * fm / 2
cm := (fm - 1) / 2

The sequence of cm would be

2, 10, 5, 7, 28, 4, 5, 55, 11, 13, 91, 7, 8, 136, 17, 19, 190, 10

I think you can continue like that

In order to add to @user164475's result I give the result for 21 . 22 = 22 X 11 = 242 — devnull69, Jul 15 '14 at 13:59

score 1 · Answer 15 · answered Jul 17 '14 at 22:45

The answer stares you right in the face.

7 x 8 is a question mark.

Now I should add that one moderator apparently believed this was a joke answer. It should be obvious that it isn't. If I wanted to make a joke, I would have added a comment. Mathematics is about the manipulation of symbols, and this is an example of symbol manipulation creating a riddle with the answer hidden in plain sight.

The riddle equates various symbols resembling products with other symbols in a rather pointless way. The question of the riddle is what the last symbol "7 x 8" equates. It obviously is meant to equate whatever symbol is to the right of the "=" sign.

There is one answer here by maddog2k that I would consider better (that 7 x 8 = 56, since we shouldn't care about all the wrong answers given in the riddle but just give the correct answer).

I don't think this is a joke answer. But I'm not a mod, either. — Almo, Jul 18 '14 at 13:58
In it's present form, this answer is clearly not a joke. In it's prior form, I thought that it was certainly one. Currently, though, I think this (and maddog2k's) answer are stubborn responses to the format of the question: instead of trying to solve the presented problem, these responses pick at the method through which the problem is conveyed. The ? and the x characters are artifacts of how the question is presented, not part of the problem itself. (My opinion, yours may differ.) — apnorton, Jul 19 '14 at 21:29

Kal S. · Answer 16 · 2021-08-26T12:54:53.603

We rewrite the riddle using prime decomposition.

Notation: $2^{x_1}3^{x_2}5^{x_3}7^{x_4}\leftrightarrow (x_1,x_2,x_3,x_4)$ We have: $$ (a_1,a_2,a_3,a_4) \odot (b_1,b_2,b_3,b_4) = (c_1,c_2,c_3,c_4)\\ (0,1,0,0) \odot (2,0,0,0) = (3,0,0,0)\\ (2,0,0,0) \odot (0,0,1,0) = (1,0,2,0)\\ (0,0,1,0) \odot (1,1,0,0) = (1,1,1,0)\\ (1,1,0,0) \odot (0,0,0,1) = (0,0,0,2)\\ (0,0,0,1) \odot (3,0,0,0) = ? $$ We note that the following definition of $(c_1,c_2,c_3,c_4)$ satisfy the given relations: $$ c_1=a_2+a_3+b_1+b_3-b_2-b_4+2a_4\\ c_2=a_3\\ c_3=a_1+b_2-b_4\\ c_4=a_4+2b_4\\ $$

We then calculate: $$ (0,0,0,1) \odot (3,0,0,0)=(5,0,0,1) \leftrightarrow 2^57=224 $$

We note though that $a_4$ can appear anywhere and satisfy the given relations namely the riddle can have multiple answers. Knowing the answer is $224$ we can include $a_4$ in the defining relation of $c_1,c_4$.

score 0 · Answer 17 · edited Jul 21 '14 at 13:43

0

$A\cdot B = C$

$\dfrac{AB²}{\gcd(AB²,6)} = C$ or $\dfrac{\text{lcm}(AB²,6)}6 = C$

edited Jul 21 '14 at 13:43

Hippalectryon

7,730

answered Jul 21 '14 at 11:05

WizardLizard

125

Interview riddle

17 Answers17

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