On the first day they cover one quarter of the total distance.
The next day they cover one quarter of what is left.
The following day they cover two fifths of the remainder and on the fourth day half of the remaining distance.
The group now have 14 miles left, how many miles have they walked?
You have been given the task of transporting 3,000 apples 1,000 miles from Appleland to Bananaville. Your truck can carry 1,000 apples at a time. Every time you travel a mile towards Bananaville you must pay a tax of 1 apple but you pay nothing when going in the other direction (towards Appleland). What is highest number of apples you can get to Bananaville?
833 apples.
Step one: First you want to make 3 trips of 1,000 apples 333 miles. You will be left with 2,001 apples and 667 miles to go.
Step two: Next you want to take 2 trips of 1,000 apples 500 miles. You will be left with 1,000 apples and 167 miles to go (you have to leave an apple behind).
Step three: Finally, you travel the last 167 miles with one load of 1,000 apples and are left with 833 apples in Bananaville.
Using only and all the numbers 3, 3, 7, 7, along with the arithmetic operations +,-,*, and /, can you come up with a calculation that gives the number 24? No decimal points allowed.
[For example, to get the number 14, we could do 3 * (7 - (7 / 3))]
Consider the following explanation for why 1=2:
1. Start out Let y = x
2. Multiply through by x xy = x2
3. Subtract y2 from each side xy - y2 = x2 - y2
4. Factor each side y(x-y) = (x+y)(x-y)
5. Divide both sides by (x-y) y = x+y
6. Divide both sides by y y/y = x/y + y/y
7. And so... 1 = x/y + 1
8. Since x=y, x/y = 1 1 = 1 + 1
8. And so... 1 = 2
How is this possible?
Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not followinng basic mathematical rules, we are able to get strange results like these.
You are on a gameshow and the host shows you three doors. Behind one door is a suitcase with $1 million in it, and behind the other two doors are sacks of coal. The host tells you to choose a door, and that the prize behind that door will be yours to keep.
You point to one of the three doors. The host says, "Before we open the door you pointed to, I am going to open one of the other doors." He points to one of the other doors, and it swings open, revealing a sack of coal behind it.
"Now I will give you a choice," the host tells you. "You can either stick with the door you originally chose, or you can choose to switch to the other unopened door."
Should you switch doors, stick with your original choice, or does it not matter?
You should switch doors.
There are 3 possibilities for the first door you picked:
You picked the first wrong door - so if you switch, you win
You picked the other wrong door - again, if you switch, you win
You picked the correct door - if you switch, you lose
Each of these cases are equally likely. So if you switch, there is a 2/3 chance that you will win (because there is a 2/3 chance that you are in one of the first two cases listed above), and a 1/3 chance you'll lose. So switching is a good idea.
Another way to look at this is to imagine that you're on a similar game show, except with 100 doors. 99 of those doors have coal behind them, 1 has the money. The host tells you to pick a door, and you point to one, knowing almost certainly that you did not pick the correct one (there's only a 1 in 100 chance). Then the host opens 98 other doors, leave only the door you picked and one other door closed. We know that the host was forced to leave the door with money behind it closed, so it is almost definitely the door we did not pick initially, and we would be wise to switch.
Search: Monty Hall problem
You are standing in a pitch-dark room. A friend walks up and hands you a normal deck of 52 cards. He tells you that 13 of the 52 cards are face-up, the rest are face-down. These face-up cards are distributed randomly throughout the deck.
Your task is to split up the deck into two piles, using all the cards, such that each pile has the same number of face-up cards. The room is pitch-dark, so you can't see the deck as you do this.
How can you accomplish this seemingly impossible task?
Take the first 13 cards off the top of the deck and flip them over. This is the first pile. The second pile is just the remaining 39 cards as they started.
This works because if there are N face-up cards in within the first 13 cards, then there will be (13 - N) face up cards in the remaining 39 cards. When you flip those first 13 cards, N of which are face-up, there will now be N cards face-down, and therefore (13 - N) cards face-up, which, as stated, is the same number of face-up cards in the second pile.
How can you divide a pizza into 8 equal slices using only 3 straight cuts?
Cut 1: Cut the pizza straight down the middle into two halves.
Cut 2: Keeping the two halves in the place, cut the pizza straight down the middle at right angles to the first cut (you will be left with 4 equal quarters)
Cut 3: Pile the 4 quarters on top of each other and cut through the middle of the pile. You will be left with 8 equal slices.
Your friend shows you two jars, one with 100 red marbles in it, the other with 100 blue marbles in it.
He proposes a game. He'll put the two jars behind his back and tell you to pick one of them at random. You'll then close your eyes, he'll hand you the jar you picked, and you'll pick a random marble from that jar.
You win if the marble you pick is blue, and you lose otherwise.
To give you the best shot at winning, your friend gives you the two jars before the game starts and says you can move the marbles around however you'd like, as long as all 200 marbles are in the 2 jars (that is, you can't throw any marbles away).
How should you move the marbles around to give yourself the best chance of picking a blue marble?
Put one blue marble in one jar, and put the rest of the marbles in the other jar. This will give you just about a 75% chance of picking a blue marble.
Two words are anagrams if and only if they contain the exact same letters with the exact same frequency (for example, "name" and "mean" are anagrams, but "red" and "deer" are not).
Given two strings S1 and S2, which each only contain the lowercase letters a through z, write a program to determine if S1 and S2 are anagrams. The program must have a running time of O(n + m), where n and m are the lengths of S1 and S2, respectively, and it must have O(1) (constant) space usage.
First create an array A of length 26, representing the counts of each letter of the alphabet, with each value initialized to 0. Iterate through each character in S1 and add 1 to the corresponding entry in A. Once this iteration is complete, A will contain the counts for the letters in S1. Then, iterate through each character in S2, and subtract 1 from each corresponding entry in A. Now, if the each entry in A is 0, then S1 and S2 are anagrams; otherwise, S1 and S2 aren't anagrams.
Here is pseudocode for the procedure that was described:
def areAnagrams(S1, S2)
A = new Array(26)
A.initializeValues(0)
for each character in S1
arrayIndex = mapCharacterToNumber(character) //maps "a" to 0, "b" to 1, "c" to 2, etc...
A[arrayIndex] += 1
end
for each character in S2
arrayIndex = mapCharacterToNumber(character)
A[arrayIndex] -= 1
end
for (i = 0; i < 26; i++)
if A[i] != 0
return false
end
end
return true
end
