# Tag Info

11

There are infinitely many solutions to this problem, all of which place the starting point either at the north pole, or very close to the south pole. Hence the bear "must" be white. Of course, real polar bears only live north of the Arctic circle, and they like to be near water, so they won't be anywhere near the north pole. Hence the puzzle is ...

10

Suppose $N>2000$ is an integer such that the period length of the (eventually) repeating $\frac1N$ equals $N$. Then in computing the decimal expansion all remainders $1,\ldots,N-1$ occur at some place. Then the fractions $\frac1N,\frac2N,\ldots, \frac{2000}N$ turn out to lead to the very same period, merely shifted. In this situation, we have ...

8

The answer is positive. Moreover, there exists a number $n$ such that additionally all numbers $2n,3n,\dots,2000n$ are obtained from $n$ by cyclic permutation of digits (naturally, we need to add some zeroes before $n$ for that). Here is this number: ...

7

I would show your dad the following table which should be easy to follow line by line: $$\begin{array}{|c|c|c|} \hline \text{Full Beers} & \text{Empty Bottles} & \text{Caps} & \text{Action}\\ \hline \color{red}{5} & 0 & 0 & \text{DRINK} \\ 0 & 5 & 5 & \text{Buy more}\\ \color{red}{3} & 1 & 1 & \text{DRINK}\\ ... 5 Executive summary: As originally posed, assuming no transactions that are not explicitly allowed by the problem statement, starting with just \10 you can drink 15 beers but no more. If you are allowed to borrow empty bottles and caps (which you must return at the end), you can drink 20 beers. If no other kinds of transaction are assumed, you must ... 4 If you have two equal disks, then as long as they're on different pegs, you can't ever move any larger disk. So it makes no sense to stay in this situation longer than necessary. The optimal solution must therefore be to keep all disks of the same size together, only splitting them up temporarily when you need to move all of them from one peg to another, ... 4 See that N+100=A^2 and N+168=B^2 for some A and B, so 68=B^2-A^2=(B-A)(B+A). As 68=17\times 2^2 it doesn't let a lot of choices for B-A and B+A, and by checking some cases you find A=16, B=18 and so N=156. 4 OPs strategy for 25 horses can be extented to 125 horses resulting in a determination of the fastest three horses within 33 races. We can obtain the following information from a race with five horses The horses at the 4^{th} and 5^{th} position can be ruled out and all other horses which are known to be slower than these. The 3^{rd} ... 3 It's interesting to try to make sense of this problem. We'll suppose there's a "real" probability that Alice is lying (either in general, or with respect to the particular statement she is making); call that q. Take a Bayesian approach, and assume a prior distribution on q. (Maybe we'll be lucky, and the answer won't depend on that prior, but, no, we ... 3 Son 3 sells 30 watermelons, then buys 20 watermelons from Son 2. Son 2 buys 10 watermelons from Son 1. 3 Each weighing can reduce the number of possibilities to 1/3 of that from before, using the same argument as in the example you linked. If you have 3^n marbles, then you need at least n weighings to find the heavier marble. If you have 3^n < N \leq 3^{n+1} marbles you should need n+1 weighings--since N is not a power of three, some weighings ... 3 This is equivalent to a variant of Nim. In this variant you have piles P_1,\ldots,P_n for some n. A move consists in choosing a pile P_k with k>1 and transferring any positive number of stones from P_k to P_{k-1}, or removing any positive number of stones from P_1. The first person who has no valid move loses. Equivalently, the person who ... 3 If you are allowed to borrow stuff, then you can get to drink 20 bottles of beer: Buy 5 bottles of beer for \10 Drink 5 bottles of beer, giving you 5 empty bottles and 5 caps Borrow 15 empty bottles and 15 caps, giving you 5+15=20 empty bottles and 5+15=20 caps in total Swap the 20 empty bottles for \frac{20}{2}=10 bottles of beer ... 3 Similar as in other answers, a state is described by a tuple (z,m,b,e,c) where z= number of bottles drunk, m= available money in dollars, b= full bottles in inventory, e= empty bottles i inventory, c= caps in inventory. The followig steps can be applied, each adding a specific tuple to the current tuple: Buy a bottle: (0,-2,+1,0,0) Drink a ... 3 You can look at the partial sums reduced modulo 3. Suppose the original sequence is a_1\ldots a_n, then define a new sequence b_0\ldots b_n by b_k = \sum_{i=0}^k a_i\pmod{3}. So basically you make the sequence of b's by starting with a 0 and then adding the next a and reducing modulo 3. Now if b_k=b_l for some i\neq j then  0 = ... 2 Hint: Keep a running total sequence mod 3. 2 If m is the number of beer bottles initially purchased and N is the total number of beer bottles that can be bought or exchanged, then we have m+\left\lfloor \frac{N-1}2\right\rfloor+\left\lfloor\frac{N-1}4\right\rfloor \geq N. That is, N\leq 4m-5 or N=4m-3. The case where N=4m-3 happens only when there are 1 empty bottle and 1 bottle cap ... 2 Take an equilateral triangle with sides of length 1, and then use each vertex as the center of a circular arc passing through the other two vertices: This is at least a contender, with area 3\cdot\frac{\pi}{6}-2\cdot\frac{\sqrt3}{4}, less than a circle of diameter 1 or a quarter-circle of radius 1, which are other convex shapes meeting the ... 2 n-1. Say n students get k, one poor unfortunate soul gets 0. The average, which I don't really even need to evaluate, is something between 0 and k. Hence, one below average, everyone else above. Similarly, we can consider the case with n-1 scoring 0, and one over achiever scoring k. By the given statement, we know that it can be no more than n-1, so it is ... 2 The bear is white, (a polar bear) since the only possible point the bear started from is the north pole. 2 Divide your marbles into three groups. Compare group 1 vs group 2. If they are the same then you know the heavier marble is in group 3. If group 1 is heavier then the heavier marble is in group 1. If group 2 is heavier then the heavier marble is in group 2. Repeat this process until you are comparing individual marbles. If you have a situation where you can ... 2 Let x be the number of problems he solved. Then 26-x is the number of problems he didn't solve. Then, from the task: 0=x*800-(26-x)*500. I am sure you can continue from here. 2 Let us assume the time taken by them to meet for A=x and for B=y. So now we know speed.time=distance so now we can generate 2 equations 7x+\frac{16y}{3}=118...(1) and B starts 1 hour late x-y=1 ..(2) solving them we get x=10,y=9 . Hope its clear.so distance travelled by A=7.t=7.10=70km 2 The wording here is really bad and ambiguous, but the answer to this puzzle is that if the son inherits 4/7 of the estate, the wife inherits 2/7 of the estate, and the daughter inherits 1/7 of the estate, the proportions between wife-son (1:2) and wife-daughter (2:1) will still hold. To find these values, translate the problem into ... 2 The subsequent smallest value of x is 1680. We need x such that$$x+1 = m^2 \text{ and }x/2+1 = n^2$$Eliminating x, we need m and n such that$$m^2 + 1 = 2n^2 \text{ or }m^2-2n^2=-1$$This is an example of Pell's equation, see here and here. The approach is to guess the smallest positive solution, which in this case is (n_1,m_1)=(1,1). All ... 1 The stock proof of this is the one you started with. Let \ell  be any line, with points P_1, \ldots, P_n (where n is at least 3). Let Q be any point not on \ell (axiom 4). Consider the lines$$ \ell_i (i = 1, \ldots, n)  where $\ell_i$ contains both $Q$ and $P_i$. These are all distinct (axiom 1), and they intersect pairwise exactly at the ...

1

The smallest i could get was $1680$. ($1681=41^2, 841=29^2$).

1

The answer according to me should be $33$. EXPLANATION: Let us mark the horses as $X_i$ where $i=1,2,3,..,125$ First we divide the $125$ horses into $25$ groups of $5$ each. Say we divide the horses in such a way that the $n^{\text{th}}$ group contains horses $X_{5n-4}$ to $X_{5n}$. For each group, there is a group race where all the $5$ horses of the ...

1

I am changing my answer because it appears you are talking about a given $B$, rather than trying to undo $A$ as I originally supposed. In this case, unless $d = \pm 1$, it will never be $I_3$. Because $\det A = -d$ and $\det B = d^2$, we see that the determinant of any finite product of them is $\pm d^k$ for some $k$. As $\det I_3 = 1$, the only possible ...

1

Let $a$ be the number on the chocolate flavoured jellybean. Let $b$ be the number on the strawberry flavoured jellybean. Let $c$ be the number on the vanilla flavoured jellybean. Let $d$ be the number on the peppermint flavoured jellybean. Calculate $h=1008a+336b+1344c$. Calculate $i=672a+1680b+336d$. Calculate $j=336a+672c+1680d$. Calculate $k=b-2c$. ...

Only top voted, non community-wiki answers of a minimum length are eligible