# Tag Info

29

First consider choosing a prime $p$ in the range $[2,N]$. You lose only if the computer chooses a multiple of $p$ or a number smaller than $p$, which occurs with probability $$\frac{(\lfloor{N/p}\rfloor-1)+(p-2)}{N-1}=\frac{\lfloor{p+N/p}\rfloor-3}{N-1}.$$ The term inside the floor function has derivative $$1-\frac{N}{p^2},$$ so it increases for $p\le ... 12 To understand the formula, it would be easiest to explain how it works conceptually before we derive it. Let's simplify the problem and say there are only 3 bags each with 2 coins in them. 2 of those bags have the 1 gram coins and one has the 1.01 gram gold coins. Let's denote the bags arbitrarily as$Bag_0$,$Bag_1$, and$Bag_2$. Similarly to your ... 8 I feel like all of the other responses are somewhat more complicated and hard to follow, so I hope to provide something more straight-forward. I'm always impressed with the beautiful formatting on MSE, though. First, any number$n$you choose will have some set of prime factors$(p1, p2, ...)$. For each prime factor$p_i$, there are$x_i =\lfloor ...

8

I used Javascript and brute-forced the solution. It wasn't terribly fast, but it worked. 29 or 31 are the best numbers to pick. function find_primes(max) { // returns all primes less than or equal to max for (var primes = [], i = 2; i <= max; ++ i) { for (var j = 0; j < primes.length; ++ j) if (i % primes[j] == 0) break; // stop if it's ...

7

Let's assume that you stick with picking prime numbers first. In this case, to win you should choose $i$ to maximize the probability $$p_{\text{win}}(i)\approx p_{\text{common}}(i)p_{\text{larger}}(i)+p_{\text{no common}}(i)p_{\text{smaller}}(i)\approx\frac{1}{i}\frac{i}{N}+\frac{i-1}{i}\frac{N-i}{N}$$ where $N$ is the largest available number for picking. ...

6

Fermat's theorem: A number N is expressible as a sum of $2$ squares if and only if in the prime factorization of N, every prime of the form $(4k+3)$ occurs an even number of times Knowing this, you start trying $a$ values. $0,1,2$ are not ok, because they give numbers that do not respect Fermat's condition. $a=3$ gives you $b^2 + c^2 = 2014-81=1933$ that ...

6

He doesn't even need his special device to find the road in a finite time. He just needs to walk in an expanding spiral, in straight-line steps shorter than two units, until he hits the road. The spiral can be any scale, but if he knows that he is 1000 units from the origin, it is enough to go forward 1000 units, left 1000 units, left 2000 units, left 2000 ...

6

$14=2^{2^2}-2$and$18=2^{2^2}+2$$13=\frac{22}2+2$$24=\frac{(2^2)!}{\frac22}$$20=\sqrt{{22}^2}-2=(2^2)!-2^2$$22=\sqrt{\left[(2^2)!-2\right]^2}$

5

Though I suspect this may be pushing common functions.. $2^2 + 2 + \Gamma(2) = 7$ $22/{\sqrt{2}^2} = 11$ $\int_{2/2}^{22}dx =21$ $\int_{2-2}^{22}dx =22$ $(2+2)! + \sqrt{2+2} = 26$

5

I got a bigger one : $87531\times 9642=843973902$ What I did was start with two blanks. Then I started putting the greater digits one by one, keeping the sum of digits of both the number somewhat balanced. The greater digits are given more preference, by putting them at a greater place value.

5

Note that if $a\gt b$ and $c\gt 0$ then $(10a+b)c\gt (10b+a)c$ - so we can organise the two numbers so that the largest digits are in the places of highest significance. If the two numbers we are multiplying have $m$ and $n$ digits their product is less than $10^m\times 10^n=10^{m+n}=10^9$ so the product must have at most nine digits, and since nine digits ...

4

Be systematic. Label all your points. Choose a unique label for each triangle, e.g. by listing point labels in alphabetic order. Enumerate triangles in some order so you can check for them one at a time. I'd suggest lexicographic order. Exploit symmetry. Many triangles will occur four times, or even eight times, throughout the figure in rotated and posibly ...

4

Most of these answers take an empirical approach; I would rather take an analytical approach, although I'm not the most qualified to do so. There are two kinds of numbers that could be good choices: Large numbers that share common factors with many smaller numbers, and Small numbers that share common factors with few larger numbers Large Numbers The ...

4

According to the given values, the gold coins have essentially the same weight (down to a single percent) as the base ones. Since gold is heavy this means that each gold coin is significantly smaller. Forget about weighing and just take the bag whose coins are much smaller than the coins in the other bags.

3

If all coins would have equal weight ($1.00$ gram), then the total weight of taking $0$ coins from bag $0$, $1$ coin from bag $1$, etc. would be $4950$ grams. Verify: $1 + 2 + \cdots + 99 = 4950$. To work with the example: if you take $25$ coins from bag $25$, then the total offset in weight is $0.25$ grams. Not that I know of. When the rest of the rules ...

3

For #1: imagine for a moment that all the coins are fake. If we took 0 coins from bag 0, 1 coin from bag 1, 2 coins from bag 2... we'd have $99\times100/2=4,950$ coins, and those 4,950 coins would weigh a total of 4,950 grams. But now, say that bag 25 were the one with real coins that are slightly heavier: 0.01 grams heavier, in fact. So the total weight ...

3

Since the problem statement says "he," may we assume the mathematician is not female? That rules out one obvious answer that would get the mathematician back on the road in one step. The initial reading tells the mathematician that the origin of the plane is somewhere in an annulus with inner radius $999.5,$ outer radius $1000.5,$ and center at his current ...

3

The de Bruijn sequence gives you approximately 1428 possibilities (10000/7 plus or minus one or two due to rounding): shift a window of size 10 through it, each time shifting by 3 positions. Edit (sorry, I also missed the part that you want permutations). This might bring you a bit closer to what you need: "Universal cycles of k-subsets and k-permutations" ...

3

The following is a complete list of the rationals obtainable by the five binary operations you gave, from a list of four twos. There are numerous expressions for many of the results and several invalid expressions (like $(2-2)^{2-2}$), but only the first valid expression my program found is shown below. $$\begin{array}{ll} -4194302 & 2-2^{22} \\ -482 ... 3 0=2+2-2-2 uses four twos not three. 10=\cfrac {22-2}2 3 \left\lfloor \exp\left(2^2\right)\right\rfloor+2+2=11 3 The following Haskell program: import Control.Monad; remove1 x [] = [] remove1 x (h:t) = if (x == h) then t else h : remove1 x t remove xs ys = foldr remove1 ys xs digits = [0..9] numeral = foldl (\a b -> a * 10 + b ) 0 solutions = do s <- remove [0] digits; e <- remove [s,0] digits; v <- remove ... 3 I believe that it would be the number of faces of the tetrahedron (4), plus the number of faces of the pyramid (5), minus the two faces that got glued together since they will not be on the outside surface of the resulting polyhedron, so 7 faces total. 3 Let ABCD be the base square of the pyramid, S its tip, and M the midpoint of BS. The segment BS is a hinge connecting two equilateral triangles; therefore the plane of the triangle \triangle:=AMC intersects BS orthogonally. It follows that the angle \alpha:=\angle(AMC) is the angle between two adjacent walls of the pyramid. Using the cosine ... 3 This is a classic riddle, whose solution surprisingly requires the axiom of choice as discussed here. The solution, however, is short and clever. Encode the colors into 0 and 1, and define the equivalence relation on 2^\Bbb N, \langle x_i\rangle\sim\langle y_i\rangle if and only if there is some k such that for all n\geq k, x_n=y_n. Using the ... 2 A slightly more creative way for 12 and 17: |2+2\sqrt{-2}|^2 = 12 |(2+2)!+\sqrt{-2}|/{\sqrt{2}} = 17 2 We don't seem to have 15 yet: {2\cdot 2 + 2 \choose 2} = 15 If we're going to use \Gamma and the like, we can get 30 too: 2 \cdot {\Gamma(2+2) \choose 2} = 30 No 19 or 27 or 29 yet either (though this keeps getting more absurd): \lfloor{2+\sqrt{2}}\rfloor^{\lfloor{2+\sqrt{2}}\rfloor} = 27 \lfloor \mathrm{Exp}(2)\rfloor \cdot ... 2 I'm pretty sure that 7 is impossible. In fact, 11,13,15,17,19 and every number from 20 to 30 is imposssible, too. A=Possible results with two twos: 0,1,4 B=New possible results with three twos: 3,6,8,16 C=New results of an operation with a two and an element of B: 5,9,10,12,14,18,32,36,64,256 Operating two elements of A gives ... 2 Clearly, 2 and 5 cannot divide the number. Hence, let the number be of the form 3^a7^b. We have$$3^a \equiv 1,3,9,7 \pmod{20} \text{ and }7^b \equiv 1,7,9,3\pmod{20}$$Hence,$$3^a7^b \equiv 1,3,7,9\pmod{20} However, the number ending in $11$ is $\equiv 11\pmod{20}$.

2

Instead of iterating between 10-digit numbers, make a list of all the possible 4-digit numbers (which are $10*9*8*7=5040$ and merge 7 4-digit numbers at a time to obtain 10-digit numbers. It is a dynamic programming question and you will have to backtrack to get already used 4-digit numbers into 10-digit numbers that require them the most. Using this ...

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