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19

How about this?$$\binom{3!}{3}$$


13

The object you are interested in is called a De Bruijn graph. Construct a graph where each node is one 4-digit sequence. Then put a directed edge from $a$ to $b$ if the last $3$ digits of $a$ are the same as the first $3$ digits of $b$. See: http://en.wikipedia.org/wiki/De_Bruijn_graph A directed Hamiltonian path will tell you what sequence to enter the ...


9

You can do it with $10,003$ letters, and I believe that this is the shortest possible string. We start by creating a De Bruijn sequence of the $4$ letter words over the alphabet $\{0,1,\dots,9\}$. What this is is a cyclic sequence (meaning that when you get to the end, you start reading from the beginning again) which contains every possible $4$ letter word ...


9

You can show them that they are adding the digits of multiples of $9$ between $18$ and $90$ and then just list them. $18 \implies 1+8=9$ $27\implies 2+7=9$ $36\implies 3+6=9$ $45\implies 4+5=9$ $54\implies 5+4=9$ $63\implies 6+3=9$ $72\implies 7+2=9$ $81\implies 8+1=9$ $90\implies 9+0=9$ While this approach is not sophisticated, it might be ...


9

For any integer $k$, $3(3k+3)=9(k+1)$ is a multiple of $9$, and one can prove that such a number has digital root $9$.


8

We are allowed to use any mathematical sign. Ok, let $f(0)=20$, and $$ f(3-3)=20 $$ It's really stupid, I think.


4

Answer 1: $\sigma_2(3) \times \sigma_0(3)$ Where $\sigma_2(x)$ is the sum of the squares of the positive divisiors of $x$, $\sigma_0$ is the number of positive divisiors of $x$. Answer 2: $\varphi(33)$ $\varphi(x)$ is the Euler-phi function. Answer 3: $g(3 \times 3)$ $g(x)$ is Landau's function. Lazy Answer: $S(S(3 \times 3!))$ With ...


4

Googling comes up with the paper Erik D. Demaine and Martin L. Demaine. Jigsaw Puzzles, Edge Matching, and Polyomino Packing: Connections and Complexity. Graph. Comb. 23, 1 (February 2007), 195-208. where corollary 3 states that It is NP-complete to decide whether $n$ given polyomino pieces, each fitting within an $\Theta(\log n) \times \Theta(\log ...


4

It's $(40 + 10) = 50$. Understand it like this: The man stole the money and gave it back to the shop owner in order to purchase the product, so this cancels each other. But the shop owner gave him the bread which costs $\$40$ and $\$10$ change. So, this is the extra money which he loses.


4

We know that two operational symbols have to exist in the red places : $$2\color{red}{\circ}\circ\color{red}{\circ}\circ=2015.$$ Then, we have $$\text{LHS}\le 2\times 9\times 9=162\lt 2015.$$


4

HINT: $$222(a+b+c)-(cba)=2536$$ Try with multiples of $222$ nearing $2536$ OR $$\iff220a+202b+22c=2536\iff202b+22c=2536-220a$$


3

One elementary fact is key to making these problems tractable: If you pass through a given square, you must cross its northeast/southwest diagonal in exactly one place. So, consider the upper-left problem. The number of allowed paths is equal to the total number of paths (on the original grid), minus the number of paths that pass through either of ...


3

The following pattern works with $31$ weighings: $${0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, \ 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, \ 0, 1, 1, 1}.$$ This beats the $35$-weighing solution found earlier. $${1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, \ 1, 1, 1, 0, 0, 0, 1, 0, ...


3

This is an elementary approach, but for larger numbers, it is more computable than the total approach given by Tad. So probably this is far from being optimal: If you first weigh the coins $0$, $12$ and $25$ it takes 38 weighings according to the rule, so you can tell all coins apart. Proof: Set $w_i$ the weigh of the $i$'th coin, and $M_j$ the result of ...


3

This answer is a proposition for a partial answer if you suppose that liars can tell the truth. Fix $A$ in the population. If you ask $2 \times 27 + 1 = 55$ people that you select randomly (not $A$), you can make them "vote" to say if $A$ is a liar or if he tells the truth. If A tells the truth you've found one. Otherwise, $A$ is a liar so you know there ...


2

I ask you for additional explanation. Meanwhile I'll post here another approach. Denote by $\tau_i^5$ the random variable that counts the time required to get five heads starting from $i$ heads, ok? What we want is exactly $E[\tau_0^5]$, right? Now, you can evaluate $E[\tau_0^5]$ conditioning at the first step. $$ E[\tau_0^5] = \frac{E[\tau_0^5]}{2} + ...


2

$$\lfloor e^3\rfloor$$ $$(3?)?-\text{sgn}(3)$$ $$\frac{\Gamma(3!)}{3!}$$ $$(3-\cos(\pi))!-(3-\cos(\pi))$$ $$\frac{3^{\bar3}}{3!}$$


2

The sum of all six numbers is 222c + 222b + 222a If we omit cba = (100c + 10b + a), what is left is N = 122c + 212b + 221a which is what we are given. Observing 221 = 13*17, we have N = 5c + 4b mod 13 and N = 3c + 8b mod 17 (We could have used factors of 122 or 212, but 2*61 and 2*2*53 are not nearly so convenient as 13 and 17.) For N = 2536, we have ...


2

She lives on floor 9. Divide 36 by 4 (ratio 3:1) = 9. Because it is more likely to go up, it means there are more levels above than below. Therefore the answer is floor 9.


1

Suggestion regarding $\textbf{hesitation}$: The ostensible reason for the hesitation is that each is waiting to see if one of the others knows the color. So I would interpret the puzzle as saying that if any two do not know the color of their own hats, then the third one's hat is white. This version of the puzzle, with the quantifier "if any two do not" ...


1

Number the seats 1 through 8 clockwise, with the odd seats being the corners. We might as well put P at 1. Then S is in 6 and Y can only go in 3 or 4. Assuming Y is in 4, R must go in 7 and T conflicts with Y, so Y is in 3. V must then go in 7, R in 5, T in 2, Q in 4, W in 8. So the order around the table is PTYQRSVU


1

Suppose you have an optimal scheme for solving the problem. It'll involve transferring fuel at various times; let's call those $t_0, t_1, \ldots$, with $t_0 = 0$ being the moment when all the trucks are fueled up and start driving. I'll call the times $t_i$ the "transfer points." I'm going to assume trucks get 1 mile per gallon, and hence have 100 gallon ...


1

According to @5xum just add $(7-7)$... $$43 = (7-7)+ (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + (6-5) + ...


1

A somewhat forgotten problem from Paul Erdős is proving his following conjecture: Let $f(r)$ equal the maximum of the sum of the side lengths of $r$ squares inscribed in a unit square such that they have no interior points in common. Erdős conjectured: For any positive integer $k$, $$f(k^2+1)=k.$$ As far as I know, this is not proven.


1

In the following I shall treat the continuous version of the game: The rolling of the die is modeled by the drawing of a real number uniformly distributed in $[0,1]$. Only the bettor can have a strategy, and this strategy is completely characterized by some number $\xi\in\ ]0,1[\> $. It reads as follows: Draw once more if the current sum $x$ is ...


1

I would approach this by essentially working backwards. First, you can figure out what the dealer's probabilities are if the bettor outcome is fixed. Let $p_w(m,n)$, $p_d(m,n)$, $p_l(m,n)$ be the probabilities that the dealer wins, draws or loses respectively if his current total is $n$ and bettor's total is $m$. (Assume the bettor hasn't already lost and $m ...


1

Since $cab + bac + bca + abc + acb = 2536$ and we know that the sum of digits are Ones digit: $2c + 2b + a$ Tens digit: $2c + b + 2a$ Hundreds digit: $c + 2b + 2a$ So now we have $\quad \quad 100(c + 2b + 2a) + 10(2c + b + 2a) + (2c + 2b + a) = 2536$ $\implies 122c + 212b + 221a = 2536$


1

Recursively color the board in white and black such that for any legal move, there is an odd number of black squares that are flipped. In your case the coloring will look like an inverted serpinski pattern shifted diagonally : Then consider the parity of the number of $1$s on black squares. At every move you invert an odd number of black squares, so the ...


1

We have to take .care of the fact that (a) more than one ball can go into a box (b) arrangements through stars and bars are not equi-probable. (c) probabilities (including box n) must sum to 1. I believe we shall have to apply PIE, as illustrated below for n = 6, k = 10 Let $m_k$ denote the # of ways k boxes can be filled. Applying PIE, we get $m_1 = ...


1

I don't see a general closed form way for many deletions of cells. You can just build a Pascal's triangle starting from A. There is 1 way to reach A, then 1 way to reach each neighboring cell, and keep going. The missing cells have 0 in them, as you can't get there at all. Each non-zero cell has the sum of the one above and the one to the left.



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