Puzzle Sum 17
What American city does this sum spell?
What American state does this sum spell?
SOLUTION
CAN + TON = CANTON
INDIAN + ARAB + BIT – RABBIT = INDIANA
Category: Puzzles
Samuel Loyd (1841 – 1911), born in Philadelphia and raised in New York City, was an American chess player, chess composer, puzzle author, and recreational mathematician. As a chess composer, he authored a number of chess problems, often with interesting themes. At his peak, Loyd was one of the best chess players in the US and was ranked 15th in the world, according to chessmetrics.com.
Space Traffic
Help the police hunt down the red robot on the run. On the way, try to find:
- one red-domed roof; two yellow-capped chimneys;
- three red-rimmed chimneys;
- a
skyscraper with four triangle-topped roofs; - a road that branches into five paths.
SOLUTION
The solution is shown below.
Move Move Chess
Consider a chess game in which every player is allowed to move twice per turn. Show that Black does not have a winning strategy.
SOLUTION
Assume Black has a winning strategy. Then if White plays Kb1-Kc3 and Kc3-b1 on his first turn, the game basically will start all over again, but with Black moving first. Therefore White will have a winning strategy, which is a contradiction.
Codex Enigmatum
These are a few enigmas from the puzzle book CODEX ENIGMATUM. What is the answer to puzzle #9?
SOLUTION
- Puzzle #2 After turning the first wheel 22 times to the right, then 19 times to the left, then 15 times to the right, and finally 11 times to the left, the final wheel will spell the word EXIT.
- Puzzle #3 The total number of spots on the hidden sides of the die on the left is 6, which corresponds to the sixth letter in the alphabet – F. Therefore, the four dice on the right correspond to the letters K, I, N, G.
- Puzzle #4 In the mosaic on the right, you can find a little star which contains pieces with letters H, I, D, E.
- Puzzle #6 If you trace the signature on the paper, starting from the large C, you will pass through the letters C, O, N, T, I, N, U, O, U, S.
- Puzzle #8 The picture on the left and the answer to puzzle #6 (“continuous”) suggest that we have to consider the images on the right which can be drawn continuously, without taking off the pencil from the paper or passing through any segment twice. These images are labeled with the letters N, O, S, E.
- Puzzle #9 The first 2 letters from the word NOSE spell NO. The last letter from the word EXIT is T. The first letter from the word HIDE is H. The last three letters from the word KING spell ING. When you combine all of them, you get the word NOTHING.
Source:
Moms’ Talk
Two moms, Sarah and Courtney, are talking to each other.
Sarah: I have two children
What is the probability that both of Sarah’s children are boys?
Courtney: Me too! Do
What is the probability that both of Courtney’s children are boys?
Sarah: Yes, I do! What is your younger child?
What is the probability that both of Sarah’s children are boys?
Courtney: It is a boy. He is so mischievous!
What is the probability that both of Courtney’s children are boys?
Sarah: Is he Sagittarius? Sagittarius boys are known to drive their mothers crazy. I can testify from personal experience.
What is the probability that both of Sarah’s children are boys?
Courtney: No, but actually I have the opposite personal experience to yours.
What is the probability that both of Courtney’s children are boys?
Sarah: Well, I guess astrology does not always get it right.
Courtney: I assume it does about half of the time.
SOLUTION
The answers are: ~1/4, ~1/4, ~1/3, ~1/2, ~23/47, 1.
Explanation:
Initially, we do not have any information about the children and therefore the chance that both of them boys is 1/2 × 1/2. This applies to the first and the second question.
After Sarah says that she has at least one boy, there are equal possibilities that she has Boy + Boy, Boy + Girl, or Girl + Boy. Therefore, the chance that both children are boys is 1/3.
After Courtney says that her younger child is a boy, the only remaining question is what is the gender of her older child, and therefore the chance is 1/2.
The fifth exchange implies that Sarah has a Sagittarius boy. There are 23 combinations such that both children are boys and at least one of them is Sagittarius. There are 47 combinations such that at least one of the children is a Sagittarius boy. Therefore, the chance that both children are boys is 23/47.
Finally, Courtney says that her younger child, which is a boy, is not Sagittarius, but her personal experience with Sagittarius boys is positive. Therefore, her older child is a Sagittarius boy and the chance is 1.
10 Pounds of Flour
You have a big bag of flour, two 5lbs weights, and an inaccurate balance scale. How can you measure exactly 10lbs of flour?
SOLUTION
Put both weights on one side, then fill the other side with flour so that the scale balances out. Then remove the weights and replace them with flour so that the balance scale is balanced again. The amount of flour you put there is exactly 10lbs.
35 Moves
Design a game that takes less than 35 moves to get to the position below.
SOLUTION
One possible solution is:
1.d3 h6 2.Bxh6 f5 3.Qd2 f4 4.Qxf4 a5 5.Qxc7 Kf7 6.g3 Kg6 7.Bg2 Kh5 8.Bxb7 Kg4 9.Nf3 Kh3 10.Bxc8 e5 11.Bxg7 e4 12.Kd2 e3+ 13.Kxe3 Kg2 14.Ng1 Kf1 15.Kf3 Ke1 16.Qxa5+ Bb4 17.Nc3+ Kd2 18.Rf1 Rh3 19.Bxd7 Nh6 20.Nd1 Kc1 21.Bxh6+ Kb1 22.Bc1 Na6 23.Kg2 Rc8 24.Bxh3 Rc3 25.Nxc3+ Ka1 26.Nb1 Nc5 27.Rd1 Be1 28.Qxe1 Ne4 29.Kf1 Nd2+ 30.Rxd2 Qd5 31.Qd1 Qg2+ 32.Ke1 Qf1+ 33.Bxf1
Vowels and Even Numbers
“If there is a vowel written on one side of a card, then there is an even number written on the other side.”
How many of these four cards do you need to flip in order to check the validity of this sentence?
What would the answer be if you know that each card contains one letter and one number?
SOLUTION
You need to flip all cards except for the second one. If each card contains one letter and one number, then you need to flip only A and 7.
Lakes but No Water
I have forests but no trees.
I have lakes but no water.
I have roads but no cars.
What Am I?
SOLUTION
The answer is MAP.
Progression Banned
I give you a pen and paper and ask you to write the numbers from 1 to 100 in succession so that there are no three numbers such that twice the second one is equal to the sum of the first and the third one. The three numbers do not need to be successive in the sequence.
Remark: The sequence 3, 1, 2, 5, 4 works, but the sequence 1, 4, 2, 5, 3 does not because of the numbers 1, 2, and 3.
SOLUTION
Start with the following sequences:
1 → 1, 2 → 2, 4, 1, 3 → 4, 8, 2, 6, 3, 7, 1, 5 → 8, 16, 4, 12, 6, 14, 2, 10, 7, 15, 3, 11, 5, 13, 1, 9
and keep iterating until you get a sequence with all numbers from 1 to 128. On each step you take the previous sequence, multiply all elements by 2, and then add the same result but with all elements decreased by 1. This will ensure that the first half contains only even numbers and the second half contains only odd numbers. Since the sum of an odd and an even number is not divisible by 2, if some sequence violates the property, then the previous sequence would have violated it as well.
Once you construct a sequence with 128 numbers, simply remove the numbers from 101 to 128 and you are done. To speed up the process, you can reduce the sequence 8, 16, 4, 12, 6, 14, 2, 10, 7, 15, 3, 11, 5, 13, 1, 9 to 8, 4, 12, 6, 2, 10, 7, 3, 11, 5, 13, 1, 9 and then continue the process.