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 ?
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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 $k=8$.
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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.
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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.)
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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.
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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.
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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.
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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.
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Easy, just define
$$\begin{array}{rcl}a \times b &=& \hspace{10.5pt}(a-4)(b-5)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/72 + \\&& 25(a-3)(b-4)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/18 + \\&& 15(a-3)(b-4)(a-4)(b-5)(a-6)(b-7)(a-7)(b-8)/8 \hspace{5.25pt}+ \\&& 49(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-7)(b-8)/36 + \\&&\hspace{5.5pt}7(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-6)(b-7)/18\end{array}$$
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$$p(x)=$$
$$-\frac{1486263915627335609976345925580307452480}{198824918770116952269605821139049374259}-\frac{23535858736574459335924875719051524464677 x}{1789424268931052570426452390251444368331}+\frac{1532186339457747628597246965489647712097745599 x^2}{742539494635629574624160683858739355082631760}-\frac{5300973178829466500668773673899060773511329723 x^3}{62373317549392884268429497444134105826941067840}+\frac{425139989729581169917246837619141657974952401 x^4}{374239905296357305610576984664804634961646407040}-\frac{15160892592292573821061148160317799661783 x^5}{7128379148502043916391942565043897808793264896}+\frac{2379833487879115598578638026951579913181 x^6}{1496959621185429222442307938659218539846585628160}-\frac{133849478325585275186149006837381343 x^7}{249493270197571537073717989776536423307764271360}+\frac{9291465647310545015926219743101 x^8}{136087238289584474767482539878110776349689602560}$$
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.
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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.
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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 ?=???
5) 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 ;-)
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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.
4. Increment fm by one on each rowresult = fm * cm. You only change cm from row to rowThese are the rules for the six rows
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
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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).
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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.
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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 $c_1,c_4$.
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.