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47

Believe it or not, this problem has been studied before (in the superficially different formulation of a pizza sliced into radial slices of unequal size). It turns out that the first player can only guarantee getting $4/9$ of the pizza: there are slicings of the pizza under which the second player can get $5/9$ of the pizza. See, for example, this arxiv ...


30

There's a sign error in your final equation; you want $$ a+b+998=ab $$ which simplifies to $$ (a-1)(b-1) = 999 $$ from which it should be easy to extract several integer solutions.


14

It is impossible, since the sum will always be divisible by $9$. Note that for a number to be divisible by $9$, the sum of its digits must be divisible by $9$; thus, we know that the solution must be divisble by $9$ as $1+\cdots+9=45$ is divisble by $9$, and so it can never be $100$. To comment on the problem where we can use a decimal point - we have the ...


10

Do we have an alternative to fill up with ones? Let $N=1000$, then \begin{align} \sum_{i=1}^N n_i &= \prod_{i=1}^N n_i \iff \\ N n_a &= n_g^N \end{align} where $n_a$ is the arithmetic mean and $n_g$ the geometric mean of the numbers $n_i$. Those numbers on the left and right hand side of the equation drift apart very fast. Already for $n_a = n_g = ...


9

This problem reminds me of the first problem, "Coins in a Row", in Peter Winkler's Mathematical Puzzles: A Connoisseur's Collection. There are quite a number of discussions of this problem online, for example here's a blog post that looks to maximize a linear version of this problem. Curiously, with the linear version of the problem (where players can ...


4

From $3z\equiv z\pmod{10}$ we find $2z\equiv 0\pmod {10}$, so $z=0$ (which is absurd per right hand side) or $z=5$. So we must have $xyz=\frac{555}{3}$.


3

A solution with four numbers different from 1 is: $$16 \times 4 \times 4 \times 4 \times 1^{996} = 16 + 4 + 4 + 4 + (996 \times 1) = 1024$$ How was this found? $1024 = 2^{10}$ appeared to be a promising candidate for the sum and product because it's slightly larger than 1000 and has many factors. The problem then was to find a, b, c, d such that: ...


3

Hint: $$a < a+\frac{1}{b+\frac1c} <a+1$$ Then do the same with $$b < b+\frac1c \leq b+1$$


3

Consider a square of $2 \times 2$. Plot the position of the first sentry on the horizontal axis and the second one on the vertical axis. Now the sample space is the total area of the square which is $4$. Can you draw the necessary lines (band) which represents the distance between the two is less than $\frac{1}{3}$. All you have to do is to now calculate ...


2

Hint:


2

You can't do it without trickery. This is known as the three utilities problem and the graph you are trying to draw is $K_{3,3}$ which is known to be non-planar. The types of trickery used are to draw the graph on a torus or to say they are pipelines and one can go over another.


2

$\sqrt{35+35/35}=6$ This can be extended to $6^{2n}-1$ with more square roots. $16/\sqrt{16}+\sqrt{\sqrt{16}}=6\\144/(\sqrt{144}+\sqrt{144})=6$ Other operators I have seen are concatenation, so $\sqrt {33+3}=6$ (though you probably have $3 \times 3-3=6$), decimals (so you can get $.2$ instead of $2$ and now $94/.94-94=6$), repeats (so you can do ...


2

I propose the solution is alpha-beta search. Your adjacent rule means we have n * 2(n - 2) possible searches which is well within range.


2

Clearly $A$ is even and so $A=2$ to avoid a carry in multiplication, so $E≥4$ is $8$ or $9$ but $4×9=36$ so $E=8$ (credit to the comment on the question). $B<3$ for $E=8$ . $B$ is odd since $B=4D+3$ where $3$ is the carry. So, $B=1$ and $D=2$ or $7$ If $D=2$, carry under $B$ should have been $8$ which is impossible with multiplication by $4$. So, $D=7$ ...


2

Let's represent your starting location as $0$ and the destination as $1000$. Let $f(x)$ be the greatest amount of fuel that can possibly be transported to or past $x$ miles from the starting point. For example, if you pick up $1000$ gallons, drive to $1$ (one mile), drop off $998$ gallons, drive back, repeat the trip to $1$ and back, and on the third trip ...


2

Based on the hint, a maximum of $14$ moves are required, because there are just $14$ gaps to move bricks across and you only move bricks across each gap once. If we start with all the bricks in one pile, it should be obvious that $14$ moves are necessary, so we have the answer. The challenge is proving that the hint is correct in that you can always avoid ...


2

If you had $43+46=618198$ then I can see the following pattern: $$\color{red}{4}\color{blue}{3}+\color{red}{4}\color{blue}{6}=\color{red}{\text{reverse(4*4)}}\color{blue}{\text{reverse(3*6)}}\color{green}{\text{reverse(43+46)}}=\color{red}{61}\color{blue}{81}\color{green}{98}$$


2

Please check your puzzle again if $43+46$ is actually $618198$ If not then expanding on @Daniel's comment $$22+22=(\text{reverse digits}(2*2))(\text{reverse digits}(2*2))[22+22+(2*2-2*2)]=4444$$ $$43+46=(\text{reverse digits}(4*4))(\text{reverse digits}(3*6))[(43+46+\left(6*3-4*4\right))]=618191$$ So the number of ...


2

WLOG the top block is in the canonical order, because we can just relabel, as in 5 -> 1, 6 -> 2, … in your example. I'm not going to make that replacement through the rest of your grid, but will just pretend you started off like that. +-------+-------+-------+ | 1 2 3 | · · · | · · · | | 4 5 6 | · · · | · · · | | 7 8 9 | · · · | · · · | ...


2

Presumably you are allowed to dump coal along the way and pick it up later. You will leave $A$ three times, so will come back to $A$ twice. The first depot should be where five trips burn $3000$ coal so you can leave twice going forward. That means the first depot is $600$ km in and when you leave from there outbound the first time there is $3000$ on the ...


2

You have to know that in the multiples of 11 the sum of even digits and odd digits is the same or differ for a multiple of 11. So,you can write in both the cases: $K+N+A+O=A+G+R+O+K11$ And so you can reduce to $K+N=G+R+K11$ Now you have to put the numbers in order to have the maximum and the minimum respecting the boundary conditions. You have that $K$ ...


1

I worked this out on paper by drawing a sudoku grid and putting down all the numbers for each individual digit place, column and row it could and couldn't be. As long as the programme you're using is 100% reliable, there'll will be at least one right solution to each 27 numbers in the diagonal. Sorry I can't give you a formula for proof, this sort of stuff ...


1

Ross Milikan's answer can be improved as follows. Definitions and notations A non-empty, finite sequence of non-negative integers $x = \left(x_1, x_2, \dots, x_n\right)$ shall be referred to as a block (of length $n$). The length of a block $x$ shall be denoted $\#x$. Given a block $x$ and an index $j \in \left\{1, 2, \dots, \#x\right\}$, define $$ ...


1

Not sure this is the best, but I would: load jeep up, travel 200mi, dump 600 gallons, go back to base repeat step 1) load jeep, go 200 mi. There is now 2000 gallons, 800mi from destination travel 333 1/3mi, dump 333 1/3gallons, go back to 800mi mark. load jeep, travel 333 1/3 mi. There's now 1000 gallons, 466 2/3 miles from destination travel 466 2/3 ...


1

After considering all the suggested methods and some calculations I came up with this recursive call: We all know that when there are no blocks of similar width we use the call : $T_{n+1} = 2T_{n}+1$ So some insight will clearly tell you that in case the block is of same width as the previous it's call would be $T_{n+1} = T_{n}+1$


1

Since each square touches itself and up to four others, the number required cannot be less than $\frac{nm}{5}$ and this is in a handwaving sense a possible limit when $n$ and $m \to \infty$ . The only problem is the the edges, as shown here, where the "existing squares" are red. The problem might go away on a particular torus.


1

$\begin{eqnarray*} \frac{25}{19} &=& 1+\frac{6}{19} \\[5pt] &=& 1+\frac{1}{\frac{19}{6}} \\[5pt] &=& 1+\frac{1}{3+\frac{1}{6}} \\ \end{eqnarray*}$ $\therefore a+b+c=1+3+6=10$ P.S. The above work shows the existence of $a,b,c$. For the uniqueness see also Ex. 1 on p.3 of ...


1

We have the following figure We know that for the numbers $5$ and $6$, the sum of the numbers in the neighboring cells is equal to 13. When $c_1=5$ or $6$ then it would have to hold that $1+2+c_3=13 \Rightarrow c_3=10$. This cannot be true since the greatest number is $9$. When $c_5=5$ or $6$ then it would have to hold that $4+3+c_3=13 \Rightarrow ...


1

First of all, they are all numbers.


1

The probability of success on a single throw is $\dfrac{n 5^{n-1}}{6^n}$ making the expected number of throws until success $\dfrac{6^n}{n 5^{n-1}}$ and the expected profit from a game played until success $p-\dfrac{c6^n}{n 5^{n-1}}$. This will be maximised when $\dfrac{6^n}{n 5^{n-1}}$ is minimised, which for integer $n$ is when $n=5$ or $6$. In those ...



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