Riddle #730

math

Number of Fs

Count the number of times the letter "F" appears in the following paragraph: FAY FRIED FIFTY POUNDS OF SALTED FISH AND THREE POUNDS OF DRY FENNEL FOR DINNER FOR FORTY MEMBERS OF HER FATHER'S FAMILY.
It appears 14 times. Make sure to count the "F"s in the word "OF", which people commonly miss.
89.07 %
38 votes

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cleandirtylogicmathshort

Which number?

What number should appear next in this sequence? 1 5 12 34 92 252 ?
688. Add the two previous numbers and multiply by 2.
84.17 %
35 votes

logicmath

Tiling Without Corners

You can easily "tile" an 8x8 chessboard with 32 2x1 tiles, meaning that you can place these 32 tiles on the board and cover every square. But if you take away two opposite corners from the chessboard, it becomes impossible to tile this new 62-square board. Can you explain why tiling this board isn't possible?
Color in the chessboard, alternating with red and blue tiles. Then color all of your tiles half red and half blue. Whenever you place a tile down, you can always make it so that the red part of the tile is on a red square and the blue part of the tile is on the blue square. Since you'll need to place 31 tiles on the board (to cover the 62 squares), you would have to be able to cover 31 red squares and 31 blue squares. But when you took away the two corners, you can see that you are taking away two red spaces, leaving 30 red squares and 32 blue squares. There is no way to cover 30 red squares and 32 blue squares with the 31 tiles, since these tiles can only cover 31 red squares and 31 blue squares, and thus, tiling this board is not possible.
90.26 %
43 votes

logicmath

Camel and Banana

The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels. What is the most bananas you can bring over to your destination?
First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A. So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again. Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between. When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are <= 1000, then he should not return and only move forward. In the first part, P1, to shift the bananas by 1Km, the Camel will have to Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana Note: After point 5 the Camel does not need to return to point A again. So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas. After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas. Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km. Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up this 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Note: After point 3 the Camel does not need to return to the starting point of P2. So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas. After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas. The Camel will actually be able to cover 333.33 km, I have ignored the decimal part because it will not make a difference in this example. Hence the length of part P2 is 333 Km. Now, for the last part, P3, the Camel only has to move forward. He has already covered 533 (200+333) out of 1000 km in Parts P1 & P2. Now he has to cover only 467 km and he has 1000 bananas. He will eat up 467 bananas on the way forward, and at point B the Camel will be left with only 533 Bananas.
85.76 %
59 votes

logicmathshort

7+2+5 = ?

5+3+2 = 151022 9+2+4 = 183652 8+6+3 = 482466 5+4+5 = 202541 Then; 7+2+5 = ?
143547.
89.33 %
39 votes

logicmath

The Witch

A witch owns a field containing many gold mines. She hires one man at a time to mine this gold for her. She promises 10% of what a man mines in a day, and he gives her the rest. Because she is blind, she has three magic bags who can talk. They report how much gold they held each day, and this is how she finds out if men are cheating her. Upon getting the job, each man agrees that if he isn't honest, then he will be turned into stone. So around the witch's mines, many statues lay! Now comes an honest man named Garry. He accepts the job gladly. The witch, who didn't trust him said, "If I wrongly accuse you of cheating me, then I'll be turned into stone." That night, Garry, having honestly done his first day's job, overheard the bags talking to the witch. He then formulated a plan... The next night, he submitted his gold, and kept 1.6 pounds of gold. Later, the witch talked with her bags. The first bag said it held 16 pounds that day. The second one said it held 5 pounds. The third one said it held 2 pounds. Beaming, the witch confronted Garry. "You scoundrel, you think you could fool me. Now you shall turn into stone!" the witch cried. One second later, the witch was hard as a rock, and very grey-looking. How did Garry brilliantly deceive the witch?
Garry put 2 lbs. in bag #1. 3 lbs. were put in bag #2. 11 lb. were put into bag #3. He then put bag #2 into bag #3, and bag #1 into bag #2. The bags only felt the weight of the gold above it. Thus they inadvertently gave the message that 23 lbs. were taken.
88.20 %
48 votes

interviewlogicmath

King and Wind Bottles

A bad king has a cellar of 1000 bottles of delightful and very expensive wine. A neighboring queen plots to kill the bad king and sends a servant to poison the wine. Fortunately (or say unfortunately) the bad king’s guards catch the servant after he has only poisoned one bottle. Alas, the guards don’t know which bottle but know that the poison is so strong that even if diluted 100,000 times it would still kill the king. Furthermore, it takes one month to have an effect. The bad king decides he will get some of the prisoners in his vast dungeons to drink the wine. Being a clever bad king he knows he needs to murder no more than 10 prisoners – believing he can fob off such a low death rate – and will still be able to drink the rest of the wine (999 bottles) at his anniversary party in 5 weeks time. Explain what is in mind of the king, how will he be able to do so?
Think in terms of binary numbers. (now don’t read the solution, give a try). Number the bottles 1 to 1000 and write the number in binary format. bottle 1 = 0000000001 (10 digit binary) bottle 2 = 0000000010 bottle 500 = 0111110100 bottle 1000 = 1111101000 Now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc. prisoner = 10 9 8 7 6 5 4 3 2 1 bottle 924 = 1 1 1 0 0 1 1 1 0 0 For instance, bottle no. 924 would be sipped by 10,9,8,5,4 and 3. That way if bottle no. 924 was the poisoned one, only those prisoners would die. After four weeks, line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit. The number that you get is the bottle of wine that was poisoned. 1000 is less than 1024 (2^10). If there were 1024 or more bottles of wine it would take more than 10 prisoners.
90.26 %
43 votes

logicmathshort

Eight eights

Using eight eights and addition only, can you make 1000?
888 + 88 + 8 + 8 + 8 = 1000
89.33 %
39 votes

logicmathshort

Three positive whole numbers

Find three positive whole numbers that have the same answer added together or when multiplied together.
1,2, & 3. 1 x 2 x 3 = 6 and 1 + 2 + 3 = 6
83.45 %
42 votes

logicmathshort

Magic number

Ramanujan discovered 1729 as a magic number. Why 1729 is a magic number ?
It can be expressed as the sum of the cubes of two different sets of numbers. 10^3 + 9^3 = 1729 and 12^3 + 1^3 = 1729
88.20 %
48 votes

logicmath

2 Player and N Coin – Strategy Puzzle

There are n coins in a line. (Assume n is even). Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins. Would you rather go first or second? Does it matter? Assume that you go first, describe an algorithm to compute the maximum amount of money you can win. Note that the strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners. Example 18 20 15 30 10 14 First Player picks 18, now row of coins is 20 15 30 10 14 Second player picks 20, now row of coins is 15 30 10 14 First Player picks 15, now row of coins is 30 10 14 Second player picks 30, now row of coins is 10 14 First Player picks 14, now row of coins is 10 Second player picks 10, game over. The total value collected by second player is more (20 + 30 + 10) compared to first player (18 + 15 + 14). So the second player wins.
Going first will guarantee that you will not lose. By following the strategy below, you will always win the game (or get a possible tie). (1) Count the sum of all coins that are odd-numbered. (Call this X) (2) Count the sum of all coins that are even-numbered. (Call this Y) (3) If X > Y, take the left-most coin first. Choose all odd-numbered coins in subsequent moves. (4) If X < Y, take the right-most coin first. Choose all even-numbered coins in subsequent moves. (5) If X == Y, you will guarantee to get a tie if you stick with taking only even-numbered/odd-numbered coins. You might be wondering how you can always choose odd-numbered/even-numbered coins. Let me illustrate this using an example where you have 6 coins: Example 18 20 15 30 10 14 Sum of odd coins = 18 + 15 + 10 = 43 Sum of even coins = 20 + 30 + 14 = 64. Since the sum of even coins is more, the first player decides to collect all even coins. He first picks 14, now the other player can only pick a coin (10 or 18). Whichever is picked the other player, the first player again gets an opportunity to pick an even coin and block all even coins.
86.91 %
43 votes