Not a Nickel
I have two coins, worth 55 cents in total. If one of them is not a nickel, what are the coins?
SOLUTION
One of them is not a nickel, but the other one is. The two coins are a 50 cents and a nickel.
Category: Brain Teasers
A collection of Math, Chess, Detective, Lateral, Insight, Science, Practical, and Deduction puzzles, carefully curated by Puzzle Prime.
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Seven Dwarfs
In one house deep in the forest
SOLUTION
The seventh dwarf is also playing chess; 2 people are needed for this.
Wizards with Hats
There are 2 wizards and each of them has infinitely many hats on his head. Every hat has 50-50 chance to be white or black, and the wizards can see the hats of the other person, but not their own. Each wizard is asked to identify a black hat on his head without looking, and they win if both succeed to guess correctly. If the wizards are allowed to devise a strategy in advance, can they increase their chance of winning to more than 25%?
SOLUTION
Each wizard guesses the position of the lowest black hat on the head of the other wizard. Then the chance of winning becomes 1/4 + 1/16 + 1/64 + … = 1/3. It can be shown that this is an optimal strategy as well.
Rubik’s Chess
Last week we found out that Puzzle Pranks Co. have invented a new type of puzzle – Rubik’s Chess. The goal is simple – you get a scrambled cube with various chess pieces on its sides, and you must unscramble it so that on each side there is one mated King, assuming the kings
We are usually good with this type of puzzles, but we spent our entire weekend trying to solve this one without any success. We even started wondering if it can be actually solved, so decided to share it with you and see if you can help us figure that out.
Below you can see the way the cube looks when seen from 8 different angles:
Remark: The orientations of the pieces are irrelevant to the final solution, i.e. they don’t need to be consistent on each side.
SOLUTION
The Rubik’s Chess puzzle cannot be solved. You can see a detailed solution HERE.
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.
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 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.