Hard math riddles for kids

logicmathsimplecleanclever

There are 100 ants on a board that is 1 meter long, each facing either left or right and walking at a pace of 1 meter per minute. The board is so narrow that the ants cannot pass each other; when two ants walk into each other, they each instantly turn around and continue walking in the opposite direction. When an ant reaches the end of the board, it falls off the edge. From the moment the ants start walking, what is the longest amount of time that could pass before all the ants have fallen off the plank? You can assume that each ant has infinitely small length.
The longest amount of time that could pass would be 1 minute. If you were looking at the board from the side and could only see the silhouettes of the board and the ants, then when two ants walked into each other and turned around, it would look to you as if the ants had walked right by each other. In fact, the effect of two ants walking into each other and then turning around is essentially the same as two ants walking past one another: we just have two ants at that point walking in opposite directions. So we can treat the board as if the ants are walking past each other. In this case, the longest any ant can be on the board is 1 minute (since the board is 1 meter long and the ants walk at 1 meter per minute). Thus, after 1 minute, all the ants will be off the board.
74.89 %
100 votes
logicmath

A farmer lived in a small village. He had three sons. One day he gave $100 dollars to his sons and told them to go to market. The three sons should buy 100 animals for $100 dollars. In the market there were chickens, hens and goats. Cost of a goat is $10, cost of a hen is $5 and cost of a chicken is $0.50. There should be at least one animal from each group. The farmer’s sons should spend all the money on buying animals. There should be 100 animals, not a single animal more or less! What do the sons buy?
They purchased 100 animals for 100 dollars. $10 spent to purchase 1 goat. $45 spent to purchase 9 hens. $45 spent to purchase 90 chickens.
73.64 %
95 votes
logicmathsimple

Two trains are traveling toward each other on the same track, each at 60 miles per hour. When they are exactly 120 miles apart, a fly takes off from the front of one of the trains, flying toward the other train at a constant rate of 100 miles per hour. When the fly reaches the other train, it instantly changes directions and starts flying toward the other train, still at 100 miles per hour. It keeps doing this back and forth until the trains finally collide. If you add up all the distances back and forth that the fly has travelled, how much total distance has the fly travelled when the trains finally collide?
The fly has travelled exactly 100 miles. We can figure this out using some simple math. Becuase the trains are 120 miles apart when the fly takes off, and are travelling at 60 mph each, they will collide in exactly 1 hour. This gives the fly exactly 1 hour of flying time, going at a speed of 100 miles per hour. Thus, the fly will travel 100 miles in this hour.
73.58 %
77 votes
logicmathclean

You are visiting NYC when a man approaches you. "Not counting bald people, I bet a hundred bucks that there are two people living in New York City with the same number of hairs on their heads," he tells you. "I'll take that bet!" you say. You talk to the man for a minute, after which you realize you have lost the bet. What did the man say to prove his case?
This is a classic example of the pigeonhole principle. The argument goes as follows: assume that every non-bald person in New York City has a different number of hairs on their head. Since there are about 9 million people living in NYC, let's say 8 million of them aren't bald. So 8 million people need to have different numbers of hairs on their head. But on average, people only have about 100,000 hairs. So even if there was someone with 1 hair, someone with 2 hairs, someone with 3 hairs, and so on, all the way up to someone with 100,000 hairs, there are still 7,900,000 other people who all need different numbers of hairs on their heads, and furthermore, who all need MORE than 100,000 hairs on their head. You can see that additionally, at least one person would need to have at least 8,000,000 hairs on their head, because there's no way to have 8,000,000 people all have different numbers of hairs between 1 and 7,999,999. But someone having 8,000,000 is an essential impossibility (as is even having 1,000,000 hairs), So there's no way this situation could be the case, where everyone has a different number of hairs. Which means that at least two people have the same number of hairs.
73.22 %
67 votes
cleanlogicmath

How do you make the number 7 an even number without addition, subtraction, multiplication, or division?
Drop the "S".
72.70 %
70 votes
logicmath

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.
72.18 %
60 votes
logicmathsimpleclean

I know a number which when multiplied by multiple of 9 i.e 9 18 27 36 45 ... The output consist of number containing only one digit. Can you identify the number?
12345679 12345679 × 9 = 111111111 (only 1s) 12345679 × 18 = 222222222 (only 2s) 12345679 × 27 = 333333333 (only 3s) 12345679 × 36 = 444444444 (only 4s) 12345679 × 45 = 555555555 (only 5s)
71.29 %
58 votes
logicmathstorylove

A man who lives in Middletown has two girlfriends, one in Northtown and one in Southtown. Trains from the Middletown train station leave for Northtown once every hour. Separate trains from the station also leave for Southtown once every hour. No trains go to both Northtown and Southtown. Each day he gets to the Middletown train station at a completely random time and gets onto the first train that is going to either Northtown or Southtown, whichever comes first. After a few months, he realizes that he spends 80% of his days with his girlfriend from Northtown, and only 20% of his days with his girlfriend from Southtown. How could this be?
The train to Northtown leaves every hour, on the hour (9:00AM, 10:00AM, etc...). The train to Southtown leaves at 12 after the hour (9:12AM, 10:12AM, etc...). So there is only a 12/60 (1/5) chance that he will end up on the train to Southtown each day, since he will usually get to the station during the 48 minutes of each hour when the train to Northtown will be the next to come.
70.90 %
53 votes
logicmathclever

Dean Sam and Castiel are three brothers. Interestingly their current age is prime. What's more interesting that difference between their ages is also prime. How old are they?
Sam : 2 Dean : 5 Castiel : 7 Age diff 7 - 2 = '5' is prime 7 - 5 = '2' is prime 5 - 2 = '3' is prime
68.87 %
109 votes
12
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