# Tag Info

7

Did I miss anything from the question ?

5

I assume all the variables are single digits, but do not assume that each digit is represented once in the grid. It is clear that you are not supposed to respect the usual order of operations, because $a$ must be at least $5$, so if you read the first column $a+\frac cf=4$ there is no solution. It must be $\frac{a+c}f=4$ $e$ and $g$ must be $1,2,$ or ...

4

This is a very broad question. $P(n,n)$ and $P(n,3)$ are known, and this allows $P(5,4)$ to be deduced. I'm not aware that any other values are known. For $k=n$, the answer is given in the dissertation Probability that $n$ random points are in convex position (Pavel Valtr, 1994): $$P(n,n)=\left(\frac{\binom{2n-2}{n-1}}{n!}\right)^2\;.$$ For $k=3$, the ...

4

With the limited examples, $\left \lceil \frac{a+b+c}{3} \right\rceil -1$, where $\left \lceil \cdot \right\rceil$ denotes the ceil or ceiling function seems to work. As the saying goes: one man's ceiling is another man's floor

3

In all cases the output is less than the average of the inputs. If the average is a whole number the output is one less than the average. if the average is not a whole number the output is the greatest integer less than the average of the inputs. The greatest integer strictly less than the average works for all cases.

3

Just for kicks, define $$"a+b=c"\quad\Longleftrightarrow\quad a=\left\lfloor \frac{c}{b} \right\rfloor$$ and I propose $88$ as a result :)

3

A possible law is $$a+b=ab+a$$ (this is valid for the three first sums) so an answer is $$8+11=8\cdot11+8=96$$ It could be possible maybe another law satisfying the three first sums.

3

We use a now-standard technique called Vieta jumping, which has a fascinating and surprisingly recent history. Suppose $a^2 + a + b^2 + b = kab$ for some integer $k$. Then, writing this as a polynomial in $a$ with integer coefficients: $$a^2 - (kb-1) a + (b^2+b) = 0,$$ we see that if $(a,b)$ is one of the roots of the above quadratic, then the other one ...

2

Here is an example for $k=7$: \begin{array}{ccccccccc} . & . & . & . & . & . & 7 & . & .\\ 9 & . & 7 & 1 & 2 & 3 & . & 5 & 6\\ 4 & . & 6 & . & 7 & . & . & . & 3\\ 3 & 1 & 2 & 8 & . & 5 & . & 6 & .\\ . & . & . & 3 & ...

2

If the ages are constrained to integer values only, then prime factorize $756$ as: $$756=2\times2\times3\times3\times3\times7$$ Thus we find: $$(a+b+c)(abc) = 756$$ Then simply use trial and error, keeping in mind that $(a+b+c)$ must be less than $abc$. I found at least one solution:

2

Doing in the same method, another solution is $(18,2,1)$ $(18×2×1)(18+2+1) = 36×21 = 756$

2

I got 40... But I guess there could be any number, because you can use the Lagrange Interpolation Polynomial

2

This is a form of the Gambler's Ruin problem. Here is a pdf with an analysis of the problem. If you go to the bottom of page 2, you will see that the probability of surviving indefinitely, given that $p > 0.5$, is $2-1/p$.

2

$a_n + b_n = c_n$ should in fact be interpreted as $a_n \cdot b_n + n = c_n$ with $b_n = a_n + 3$ and the following recursive relation for $a_n$ $a_n = a_{n-1} \cdot a_{n-2} + (n-2),\:\:$ $n \geq 3$ $a_0 = 0, \:\:\:\:a_1 = 1, \:\:\:\:a_2 =2$ $a_1 \cdot b_1 + 1 = c_1$ $a_{2} \cdot b_{2} + 2 = c_{2}$ $a_{3} \cdot b_{3} + 3 = c_{3}$ $a_{4} \cdot ... 2 If you number your ‘equations’ as$a_n+b_n =c_n$, you have the relation$c_n=a_n+b_n+c_{n-1}. Hence \begin{align*} 4+7&=32,\\5+8&=45,\\6+9&=60,\\7+10&=77,\\8+11&=96.\end{align*} 2 I agree with your reasoning. Let B denote an unfired round and X a fired round. The event which occurred was either XBB, BXB or BBX, all equally likely. In two of these events, namely XBB and BXB the second round will also be fired so the odds are\frac{2}{3}$. When the second round is fired you know that the event was either XXB or BXX, both equally ... 2 I believe your answer is correct yes. For two bullets to be consecutively fired there are only two positions you could have originally landed on. Either you landed on the first bullet out of the three consecutively lined up or you landed on the second bullet. Landing on any other position would mean two consecutive shots could not have been fired. From ... 2 Let$X_k=1$if the$k$-th person to arrive needs a new table and$0$otherwise; clearly $$\Bbb E[X_k]=\Bbb P[X_k=1]=(1-p)^{k-1}\;,$$ since the$k$-th arrival needs a new table if and only if he or she is not friends with any of the previous$k-1$arrivals. Let$X=\sum_{k=1}^NX_k$; by linearity of expectation $$\Bbb E[X]=\sum_{k=1}^N\Bbb ... 2 I have absolutely no idea how to solve this problem by hand. Here's the code of a Python program to solve the problem: def calc(a, b, c, d, e, f, g, h): ''' This function checks whether or not all six equations work. If one of them fails, it says False. If all of them work, it says True. ''' if a+b-9 != 4: return False if (c-d)*e ... 2 If you are allowed to rotate the assembled cube freely there is exactly one way to assemble it. Proof. Let \{0,1,2,3\} be the set of colors. Color each vertex of the assembled 2-cube with the color of the 1-cube it belongs to. Vertices of the assembled cube having the same color are necessarily space-diagonally opposite. Assume that a vertex having ... 1 This answer looks at Question 2, specifically the generalization to n=3. Consider the problem of assembling a 3\times 3\times 3 cube using a collection of 3^3=27 cubies of 3^2=9 colors, with 3 cubies of each color, so that each color appears once on each face. There are four roles a cubie may play: a corner cubie contributes its color to three ... 1 Heuristically, it might be reasonable to expect that 1 + p_1\cdots p_n behaves like any other number of its size that is known to be coprime to p_1, p_2, \ldots, p_n. The magnitude of 1 + p_1\cdots p_n is roughly e^{p_n}, so the density of primes in that size range is about 1/p_n \sim 1/(n\log n). To account for the avoidance of the first n ... 1 It depends on how you define different ways to assemble the cube. If there are two constructions of the cube such that I can get from one to the next just by rotating the cube without disassembling and assembling, is that considered two different cubes or one? If you look at the large cube as 8 defined spots for the 8 small cubes, and a cube created by ... 1 345-12=333 does the job nicely. Your statement about combining 2,3,4,5 to make 332 is actually wrong. Consider this example: we can combine 1 and 2 to make 1, but can we use 2 by itself to get 0? 1 Three answers:$$345-12=333\\ \frac{(3!)!-5-1}{2} - 4!=333\\ 3\cdot 5! - 4!-1-2=333$$1 Also$$ \left\lfloor \frac{(2\cdot 5)^{4-1}}{3}\right\rfloor = 333. $$1 (a) Let the distance travelled be FB mi. Then front tyre wears out at B/mi, and back one wears out at F/mi. For max. dist, each tyre must cover 0.5(B+F), thus D_a = \frac{FB}{0.5(F+B)}, and the tyres need to be rotated at half this distance. (b) 3 tyres are now sharing the load, so D_b = 1.5D_a Changes should be made at ... 1 Let's assume you could divide everything into quadrilaterals. Now, let's count the number of edges in this division; call this number E (in your example, I count E=15, but I might have miscounted). Now, each quadrilateral has 4 edges, but with the exception of the 3 edges on the border of the triangle, each edge belongs to 2 quadrilaterals. Therefore, if ... 1 96 For each line add the two numbers and store it in D. Add D to the C from the previous step. A+B=C D 1+4=5 5 (4+1 gives 5) 2+5=12 7 (5+2 is 7. Add 7 to 5 = 12 from previous step) 3+6=21 9 (6+3 is 9. Add 9 to 12 = 21 from previous step) 4+7=32 11 (4+7 is 11. Add 11 to 21 = 32 from previous step) 5+8=45 13 6+9=60 15 ... 1 I would start going backwards - what is the integer factorization of 756?$$756=2^2\times 3^3\times 7$$Now we should try to distribute these prime numbers between x and y s.t. the first two conditions apply. y should have at least 3 prime numbers, since it is the result of$a\times b\times c\$. Also, since a solution with 1's, doesn't work, notice that y>x ...

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