Birds on a Tree
There are 10 birds on a tree. One hunter shoots at them and hits 2. How many birds are left on the tree?
SOLUTION
Zero – the remaining 8 birds get scared and fly away.
Category: Puzzles
We do not know where this puzzle originated from. If you have any information, please let us know via email.
Penrose Railway Maze 3
Get from A to B, moving only over smooth lines, without reversing and making sharp turns.
SOLUTION
The solution is shown below.
Traveling Salesmen
Between every pair of major cities in Russia, there is a fixed travel cost for going from either city to the other. Traveling salesman Alexei Frugal starts in Moscow and visits all cities exactly once, choosing every time the cheaper flight to a city he has not visited so far. Salesman Boris Lavish starts in St Petersburg and visits all cities exactly once, choosing every time the most expensive flight to a city he has not visited so far. Can Alexei end up spending more money than Boris after they end their journeys?
SOLUTION
No, it is impossible. For every trip of Alexei we will choose a trip of Boris, which costs at least as much.
Let the number of cities is n and Alexei visits them in order 1 -> 2 -> … -> n.
If Boris visits city n-1 before city n, then pair Alexei’s trip (n-1, n) with Boris’ trip (n-1, #). Notice that C(n-1, n) < C(n-1, #).
If Boris also visits city n-2 before city n, then pair Alexei’s trip (n-2, n-1) with Boris’ trip (n-2, #). Notice that C(n-2, n-1) < C(n-2, n) < C(n-2, #).
Continue like this until get to a city k which Boris visits after city n. Then pair Alexei’s trip (k, k+1) with Boris’ trip (n, #). Notice that C(k, k+1) < C(k, n) < C(n, #).
Next, check whether Boris visits city k – 1 before city k, and pair Alexei’s trip (k-1, k) with either (k-1, #) or (k, #). Continue like this, until pair all of Alexei trips with more expensive Boris trips.
Chess Aggression
Starting from this position, can you make 39 consecutive checks – 20 from White and 19 from Black?
SOLUTION
The sequence is as follows:
1. Nh2+ f1N+
2. Rxf1+ gxf1N+
3. Ngxf1+ Bg5+
4. Qxg5+ Bg2+
5. Nf3+ exf3+
6. Kd3+ Nc5+
7. Qxc5+ Re3+
8. Nxe3+ c1N+
9. Qxc1+ d1Q+
10. Qxd1+ e1N+
11. Qxe1+ Bf1+
12. Nxf1+ f2+
13. Ne3+ f1Q+
14. Qxf1+ Qxf1+
15. Nxf1+ Re3+
16. Nxe3+ b1Q+
17. Rxb1+ axb1Q+
18. Nc2+ Nf2+
19. Bxf2+
NASA and the Meteor
NASA locates a meteor in outer space and concludes that it has either a cubical or spherical shape. In order to determine the exact shape, NASA lands a spacecraft on the meteor and lets a rover travel from the spacecraft to the opposite point on the planet. By measuring the relative position of the rover with respect to the spacecraft throughout its travel on the planet (in 3D coordinates), can NASA always determine the shape, no matter the route taken by the rover?
SOLUTION
The answer is NO.
The question is equivalent to analyzing the intersection of a cube and a sphere which share a common center. Thus the question gets reduced to figuring out whether such intersection, which is a curve, can connect two opposite points on the sphere/cube.
Let the edge of the cube has length 1. If you pick the radius of the sphere equal to √2/2, the intersection will consist of 6 circles inscribed in the sides of the cube. Then the rover can just move along these circles from one point to its opposite and NASA won’t be able to figure out the exact shape.
Remark: It is not hard to see that 2:√2 is the only edge-radius ratio, for which NASA can’t figure out the shape.
The Answer to This Riddle Is a Number
The key to this riddle is only for you
Below are instructions, above this the clue
First strike the one near the head of a year
Then remove he who begins a cheer
Next take away the end of a tunnel
Then let us go to solve this puzzle
What you first took you must now take again
With three you are left, but fret not dear friend
Fattest to front and thinnest to rear
Add in two “eyes” and all becomes clear
Remark: The instructions in this riddle are quite literal.
SOLUTION
The answer is XVIII, or 18 in Roman numerals.
The key is “only for you” – EXCLUSIVE. Remove “the one near the head of a year” – E. Remove “he who begins a cheer” – C. Remove “the end of the tunnel” – L. Remove “us” – U and S. Remove again E. Now you are left with X, I, V. Arrange them “fattest to front and thinnest to rear” – X > V > I. Add “two ‘eyes'” – I and I, so you get XVIII=18.
Source:
The Poisoned Glass
You are given 4 identical glasses, completely filled with transparent, odorless liquids. Three of the liquids are pure water, and the fourth is poison, which is slightly heavier. If the water glasses weigh 250 grams each, and the poisoned glass weighs 260 grams, how can you figure out which one is which, using a measuring scale just once?
SOLUTION
Empty the first glass, fill around 1/4th of it with liquid from the second glass, and the rest 3/4ths with liquid from the third glass. Then, measure the first and fourth glasses simultaneously. If their total weight is:
– 500 grams -> the first glass is (was) the poisoned one
– between 500 and 505 grams -> the second glass is the poisoned one
– between 505 and 510 grams -> the third glass is the poisoned one
– 510 grams -> the fourth glass is the poisoned one
Source:
The Dark Bridge
Four friends are trying to cross a bridge in complete darkness, but have only one flashlight. They need respectively 1, 2, 7, and 10 minutes to cross the bridge, and if any three of them step on the bridge at the same time, it will collapse. How many minutes do they need at least in order
SOLUTION
They need 17 minutes. Label the friends A (1min), B (2min), C (7min), D (10min). A and B cross the bridge, then A returns back with the flashlight. C and D cross the bridge, then B returns back with the flashlight. Finally, A and B cross the bridge.
In order to see that this is optimal, notice that when D crosses, he needs at least 10 minutes. If C crosses separately, this will make already 17 minutes in total. Therefore C and D must cross together, and A and B must be at that time on the two opposite sides of the bridge. From here it is easy to conclude that the friends indeed need at least 17 minutes.