Ø Three people check into a hotel. They pay £30 to the manager and go to their room. The manager suddenly remembers that the room rate is £25 and gives £5 to the bellboy to return to the people. On the way to the room the bellboy reasons that £5 would be difficult to share among three people so he pockets £2 and gives £1 to each person. Now each person paid £10 and got back £1. So they paid £9 each, totalling £27. The bellboy has £2, totalling £29. Where is the missing £1?
Answer:We have to be careful what we are adding together. Originally, they paid £30, they each received back £1, thus they now have only paid £27. Of this £27, £25 went to the manager for the room and £2 went to the bellboy.
There are three houses, and three utilities: gas, electricity and water. Your task is to connect each house to all three utilities. Therefore each house will have three lines and each utility will also have three lines. However, you cannot cross lines. You cannot pass lines through houses or utilities. You cannot share lines. Can you draw the 9 lines required?
This puzzle is a classic one which has no solution in 2D. However, if you place the items on a doughnut shape in 3D you can solve it. In the picture below, E is linked to 3 by going over the top and re-entering through the hole in the middle.
Think of words ending in -GRY. Angry and hungry are two of them. There are only three words in the English language. What is the third word? The word is something that everyone uses every day. If you have listened carefully, I have already told you what it is.
There are only two: angry and hungry. The rec.puzzles archive offers a large collection of words that end in -GRY, but none of them could be considered even remotely common. There are many generally unsatisfying "trick" answers to the problem, which depend on a specific wording of the question or that the question be spoken instead of written. There seems to be no agreement among puzzle historians about which form is the original, or even the age of the problem. In any event, it is apparent that the frequent mutations of the puzzle statement over the years have erased whatever answer was intended by the original author. The usual trick is to play on the expression "the English Language", you are then asked for the third word - which is of course Language! QED.
Below is a very special grid, around each shaded number are 8 white squares. However, each white square should have a number from 1 to 7. Once filled in, these 8 numbers will sum to the shaded number. In addition, once completed correctly, no row nor column will contain a duplicate number within a white square. For example, the top row may be 5 6 4 2 3 1 7, etc. This is BrainBashers most difficult puzzle, but is solvable without the aid of a computer.
There are 5 houses in 5 different colours. In each house lives a person of a different nationality. The 5 owners drink a certain type of beverage, smoke a certain brand of cigar, and keep a certain pet. Using the clues below can you determine who owns the fish?
The Brit lives in a red house.
The Swede keeps dogs as pets.
The Dane drinks tea.
The green house is on the immediate left of the white house.
The green house owner drinks coffee.
The person who smokes Pall Mall rears birds.
The owner of the yellow house smokes Dunhill.
The man living in the house right in the middle drinks milk.
The Norwegian lives in the first house.
The man who smokes Blend lives next door to the one who keeps cats.
The man who keeps horses lives next door to the man who smokes Dunhill.
The owner who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
The man who smokes Blend has a neighbour who drinks water.
This puzzle is usually attributed to Einstein, who may or may not have written it. The German owns the fish and the table below details the full answer:
Nationality: Norweg Dane Brit German Swede
Colour : Yellow Blue Red Green White
Beverage : water tea milk coffee beer
Smokes : Dunhill Blend Pall Mall Prince Blue Master
Pet : cats horses birds fish dogs
You have 12 coins, one of which is fake. The fake coin is indistinguishable from the rest except that it is either heavier or lighter, but you don't know which. Can you determine which is the fake coin and whether it is lighter or heavier using a balance scale and only 3 weighings?
One solution is to label the coins with the letters FAKE MIND CLOT and weigh the coins in the following three combinations:
MA DO -- LIKE
ME TO -- FIND
FAKE -- COIN
Logic will now allow you to find the fake coin based on the three results. For instance, if the results were left down, balanced, left down, we could work out which coin is fake in the following way:
From the middle weighing we know that the coins METOFIND are all normal. So one of the coins ACKL is fake. Therefore looking at these coins one at a time in the other two weighings, we can see that:
A - appears on the left twice and could be fake.
C - appears only once, therefore can't be fake (otherwise the first weighing would be balanced).
K - appears on opposite sides, so it can't make the left side go down both times.
L - appears only once, therefore can't be fake (otherwise the third weighing would be balanced).
Therefore the only possibility is A, which must be heavier. Any other combination of ups and downs will allow you to use the same logic to find the fake coin.
====è didn’t understand.
Assuming you have enough coins of 1, 5, 10, 25 and 50 cents, how many ways are there to make change for a dollar?
Sol: Let N(a, c) be the number of ways
to represent an amount a using coins no larger than
value c. Then N(a, 1) = 1 (there is only one
way to represent any amount using only pennies), and N(0, c) = 1 (there
is only one way to represent an amount of zero, no matter what coins are available).
N(5n, 5) = n+1 (you can use anywhere from 0 to n nickels,
and then have to make the rest up in pennies).
N(10n, 10) = (n+1)^2 (the sum of N(0, 5), N(10,5), N(20, 5), ...,
N(10n, 5), which is the sum of the first n+1 odd numbers - it's
an old fact that this sum yields the perfect squares).
N(10n+5, 5) = (n+1)*(n+2).
Now go top down. We want N(100, 50). N(100, 50) = N(0, 25) + N(50, 25) + N(100, 25),
representing the use of 2, 1, and 0 fifty-cent pieces, respectively.
N(50, 25) = N(0, 10) + N(25, 10) + N(50, 10).
N(100, 25) = N(0, 10) + N(25, 10) + N(50, 10) + N(75, 10) + N(100, 10).
Using our previous facts about N(a, 10), we find that
N(100, 25) = 1 + 12 + 36 + 72 + 121 = 242; N(50, 25) = 1 + 12 + 36 = 49;
and finally, N(100, 50) = 1 + 49 + 242 = 292.
==è solution not understandable, very lengthy.
Mr. Simkin, the new math teacher at school, was impressed by his students' ability to solve logic puzzles. He pulled aside three more students, and handed them each a sealed envelope with a number written inside. He told them that they each have a positive integer, and the sum of their numbers was 14.
Manny, Moe, and Jack each opened their envelopes. Mr. Simkin asks Manny if he knows anything about the numbers the other two are holding, and Manny says, "I know that Moe and Jack are holding different numbers."
Moe answers, "IN THAT CASE, I know that all three of our numbers are different."
Jack thinks for a bit, and then says, "Now I know all of our numbers."
Mr. Simkin turns to the class and asks if anyone in the class knows the numbers. Gretchen's hand shoots up into the air, and after waiting for a while to see if anyone else will get the answer, Mr. Simkin calls on Gretchen.
What numbers does she say they each are holding?
Manny has a 3, Moe has a 2, and Jack has a 9.
From Manny's statement, we can deduce that his number is odd. Since Moe did not know that they were all different until Manny said that, we know that Moe is not holding a 7, 9, or 11. (Otherwise, he would have already known they were all different.) If he were holding a 1, 3, or 5, he would not be able to be sure his number was different than Manny's. If he were holding a 4, 8, or 12, he could not know that Manny and Jack didn't have the same number, since there are odd pairs that would bring the total to 14. Therefore, Moe must be holding a 2, 6, or 10.
There are 10 triples for (Manny, Moe, Jack) that satisfy all three statements. They are:
(1, 2, 11)
(1, 6, 7)
(1, 10, 3)
(3, 2, 9)
(3, 6, 5)
(3, 10, 1)
(5, 2, 7)
(5, 6, 3)
(7, 2, 5)
(7, 6, 1)
Since Jack can reason flawlessly, he knows these are the possibilities. In order to make his statement, his number has to be a unique solution. Therefore, he must be holding a 9 or an 11. If he were holding 11, though, he would have known from Manny's statement that Manny had a 1, Moe had a 2, and he had an 11. Since he didn't know them all until after Moe spoke, he must have a 9, leaving Manny a 3 and Moe a 2.
Mr. Simkin suggests to Gretchen that she may have a career in law enforcement if she can further hone these impressive deductive reasoning skills.
A local grocer delivers lemonade using two full 10 pint jugs. Recently, the grocer delivered to a pair of ladies who had a five pint jug and a four pint jug between them. Each wanted two pints of lemonade though. How did the grocer measure two pints into each jug only using the two 10 pint jugs and the ladies jugs? The jugs are not marked in any way and there was no other container which could be used.
Ans : If we call the two 10 pint jugs jugA and jugB, the four pint jug4 and the five pint jug5 we can do the following.
Fill jug5 from jugA.
Fill jug4 from jug5.
Pour jug4 back into jugA.
Empty jug5 into jug 4.
Fill jug5 from jugA.
Fill jug4 from jug5.
Empty jug4 into jugA.
Fill jug4 from jugB.
Fill jugA from jug4.
Both jugs now contain 2 pints.
A ship is twice as old as the ship's boiler was when the ship was as old as the boiler is. What is the ratio of the boiler's age to the ship's age?
If we take the S to be the ship's age and B to be the boiler's age, and T to be the difference we get:
S - T = B
S = 2 x (B - T)
Eliminate T to get:
B / S = 3 / 4.
Some months have 30 days, some months have 31 days. How many months have 28 days?
ü All months have at least 28 days.
Ø If a doctor gives you 3 pills and tells you to take one pill every half-hour, how long would it be before all the pills had been taken?
ü You take the first pill now, the next pill in the next half hour, and the last in the next half hour... So the correct answer is 1 hour!
Ø I went to bed at eight o'clock in the evening and wound up my clock and set the alarm to sound at nine o'clock in the morning. If I went to sleep immediately, how many hours sleep would I get before being awoken by the alarm?
ü Its a WOUND UP clock. So the correct answer is 1 hour! Wound up clocks cannot tell pm from am. If you got this one, you might be showing your age
Ø A Farmer had 17 sheep. All but 9 died. How many live sheep were left?
ü If All but 9 died ... then there should be 9 sheep alive!
Ø A man builds a house with four sides, of rectangular construction, each side having a southern exposure. A big bear comes along. What colour is the bear?
ü The only place where your house can have four southern exposures in on the North Pole. So the bear will most likely be a white furred Polar Bear. (do they have blue tongues too?)