A country has three denominations of coins, worth 7, 10, and 53 units of value. What is the maximum number of units of currency which one cannot have if they are only carrying these three kinds of coins?
I know the formula for 2 denominations, but I don't know how to do one with 3.
Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will 20! be the resulting product?
We have fraction m/n. I wrote the prime factorization of 20!, but I can't seem to find the largest possible value of m. Once I get the largest value of m, then I can just count the number of divisiors of m, but what do I do to get m.
Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms.
so apparently, we can have 2513 and stuff. So like we just need 251 in order and then can have extra digits. 251 /999 is interesting, because 251 is close to 250, a divisor of 1000, which is cllose to 999.
Anyways, idk about the 1st quesiton
EDIT: we don't need to have a repeated decimal
I found that these numbers satisfy Q2.
I sort of see what you are doing here, but then again, I don't quite understand. We try to make sure that no numbers in our set are the arithemtic mean of two others. I've been suggested to use base 3, but I don't know how that would work.
EDIT: Turn our set into base 3 integers. Selecting those expressed with only 0 and 1 (base 3) gives us 2^k integers. I think this works, but I have to prove that this satisfies the condition of one number not being the AM of 2 others.
Let b be an integer greater than 2, and let(the sum contains all valid base b numbers up to 100_b). Compute the number of values of b for which the sum of the squares of the base b digits are less than or equal to 512. I understand what N_b's value is, but i don't know about the values. Maybe trying smaller values of b would work.
[edited for clarity - bobbym]
Prove that from the setone can choose 2^k numbers so that none of them can be represented as the arithmetic mean of some pair of distinct chosen numbers.