Here you can find all of the past weekly math problems and their answers! For the most recent math problems, check the home page.
There are chickens and rabbits in the same cage. There are a total of 35 heads and 94 legs. How many chickens and rabbits are there? (this problem came from ancient Chinese math book)
Assume x chickens and y rabbits in the cage, then we can have
x + y = 35 (1)
2x + 4y = 94 (2)
Solve the above equation by (2) - (1)*2, we got
2y = 24 => y = 12
Therefore, x = 23. So we have 23 chickens and 12 rabbits in the cage!
Jack and his three friends went trick-or-treating together. If Jack got 11 pieces of candy, and each of his friends got three times as much candy as he did, how many pieces of candy did Jack and his friends get in all?
If Jack’s friends got three times as much candy as Jack, and Jack got 11 pieces of candy, that means that all three of Jack’s friends got 33 pieces of candy. So in order to find the sum, we can do this calculation.
33*3 + 11 = 110
Since 3 of Jack’s friends each got 33 pieces of candy, we multiplied 33 by 3. Therefore, Jack and his friends got 110 pieces of candy in total.
More than 200 students signed up to participate in the Science Olympiad competition. If teams of 3 could be made with no students left out, and teams of 4 could also be made with no students left out, what is the fewest number of students who could have signed up for the Geography Bowl?
We need an integer divisible by both 3 and 4, and greater than 200. If an integer is divisible by both 3 and 4, it is divisible by 12. 200 divided by 12 is 16 remainder 4. Therefore, the next multiple of 12 up from 200 is 204. Therefore, the fewest amount of students who could have signed up is 204.
Goldbach’s conjecture states that every even number greater than two can be expressed as the sum of two prime numbers. For example, 2022 = 191 + 1831. How many ordered pairs of prime numbers have a sum of 60?
The ordered pairs who's components are prime numbers with sum of 60 are: (7,53), (13,47), (17,43), (19,41), (23,37), and (29,31). We see 6 ordered pairs, but we must also account for the reverse of each of these ordered pairs. Therefore, there are 6 x 2 = 12 ordered pairs.
Fyodor and his three sons, Ivan, Dmitri and Alyosha, are standing exactly on the corners of a rectangular room. Fyodor is 3 meters from Dmitri and 5 meters from Ivan. What is the minimum possible distance that Fyodor could be from Alyosha?
With a rectangle, two vertices may be the endpoints of a short side, of a long side, or of a diagonal. Given two such lengths, the shorter must be a side, while the longer may be a side or a diagonal. If the 5 meters is a diagonal, then we have a 3-4-5 right triangle and the unknown length is the second side, which must be 4 meters. If the 5 meter is the longer side, the unknown length is the diagonal, even longer than the 5 meters. Thus, the minimum possible distance in question is 4 meters.
A ship anchored in a port has a ladder which hangs over the side. The length of the ladder is 200cm, the distance between each rung in 20cm and the bottom rung touches the water. The tide rises at a rate of 10cm an hour. When will the water reach the fifth rung?
Never. The tide raises both the water and the boat so the water will never reach the fifth rung.
While driving, Carl notices that his odometer reads 25,952 miles, which happens to be a palindrome. He thought this was pretty rare, but 2.5 hours later, his odometer reads as the next palindrome number of miles. What was Carl’s average speed during those 2.5 hours?
The first palindrome is at 25952. Any 5-digit palindrome starting with 25 must end in 52 and we can have an arbitrary digit in the middle, but getting the next palindrome in the 25 thousands would require the next digit after 9 for the middle digit, but there is no such digit. Thus, we must go to the 26 thousands, and the palindrome must end in 62; the least middle digit is 0. The next palindrome after 25952 is 26062, with a separation of 110. Therefore, the travel is 11- miles in 2.5 hours, for an average speed of (110mi/2.5h) = 44 mi/h
A machine randomly generates one of the nine numbers 1, 2, 3, ... , 9 with equal likelihood. What is the probability that when Tsuni uses this machine to generate four numbers their product is divisible by 14? Express your answer as a common fraction.
To have the product of the 4 digits be divisible by 14, the digit 7 must be chosen at least once, and one or more of the digits 2,4,6, and 8 must be chosen at least once. This is the complement of never choosing 7, which has probability (8/9)^4 = 4096/6561 or never choosing an even digit, which has the probability (5/9)^4 = 625/6561. however, these two cases overlap (choosing 1, 3, 5 or 9 every time), which has probability (4/9)^4 = 256/6561. Thus, the answer is 1 – ( 4096+625-256/6561) = 2096/6561
A positive integer divisor of 11! Is chosen at random. What is the probability that this divisor is prime? Express your answer as a common fraction?
11! = 11 x (2 x 5) x 3^2 x 2^3 x 7 x (2 x 3) x 5 x 2^2 x 3 x 2 = 2^8 x 3^4 x 5^2 x 7 x 11, which has (8 + 1)(4 + 1)(2 + 1) (1 + 1)(1 + 1) = 9 x 5 x 3 x 2 x 2 = 27 x 20 = 540 divisors. The only prime divisors are 2, 3, 5, 7, and 11 – thus, 5 of them. Therefore, the probability of a randomly chosen divisor of 11! Being prime is 5/540 or 1/108.
The sum of the integer n and eighteen is equal to the product of four and five. What is the value of n?
N + 18 = 4 x 5 = 20, so n = 20 - 18 = 2
There are 49 dogs signed up for a dog show. There are 36 more small dogs than large dogs. How many small dogs have signed up to compete
To figure out how many small dogs are competing, you have to subtract 36 from 49 and then divide that answer, 13 by 2, to get 6.5 dogs, or the number of big dogs competing. But you’re not done yet! You then have to add 6.5 to 36 to get the number of small dogs competing, which is 42.5. Of course, it’s not actually possible for half a dog to compete in a dog show, but for the sake of this math problem let’s assume that it is.