Interview riddle

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 $k=8$.

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

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

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.)

Easy, just define

$a \times b = (a-4)(b-5)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/72 + 25(a-3)(b-4)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/18 + 15(a-3)(b-4)(a-4)(b-5)(a-6)(b-7)(a-7)(b-8)/8 + 49(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-7)(b-8)/36 + 7(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-6)(b-7)/18$

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.

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.

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.

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.

$$p(x)=$$

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

Then
$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x & 12\color{grey}{(3\times 4)} & 20\color{grey}{(4\times 5)} & 30\color{grey}{(5\times 6)} &42\color{grey}{(6\times 7)}&48\color{grey}{(7\times 8)}& \color{grey}{1729}&\color{grey}{2014}&\color{grey}{2015}&\color{grey}{2016}\\ \hline p(x)& 8 & 50 & 30 &49&\color{red}{224}& \color{grey}{1729}&\color{grey}{2014}&\color{grey}{2015}&\color{grey}{2016} \\ \hline \end{array}$$

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

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.

The answer stares you right in the face.

7 x 8 is a question mark.

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

1. cm := cm + 1
2. cm := fm * cm
3. cm := fm - 1
4. cm := fm
5. cm := cm * fm / 2
6. 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

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