-3
7, 37, 67, 97, 127, and an infinite number of other numbers in this series (starting with 7 and increasing in increments of 30) have a remainder of 1 when divided by 6 and a remainder of 2 when divided by 5.
Well, isn't that a happy little math problem! When a number is divided by another number, we get a quotient and a remainder. In this case, we are looking for natural numbers that leave a remainder of 5 when divided into 47. So, we start by listing some numbers that fit this description: 5, 10, 15, 20, 25, 30, 35, 40, and 45. That's 9 different numbers in total that will leave a remainder of 5 when divided into 47.
In mathematics, numbers that can be divided without a remainder are referred to as divisible numbers. For example, any integer can be divided by its factors, such as 2, 3, or 5, if those numbers evenly divide it. Additionally, whole numbers can be divided by themselves and 1 without a remainder. Essentially, any number can be divided evenly by its divisors.
The first five numbers which when divided by 5 leave a remainder of 4 are: 4 = 4/5 = 0 remainder 4 9 = 9/5 = 1 remainder 4 14 = 14/5 = 2 remainder 4 19 = 19/5 = 3 remainder 4 24 = 24/5 = 4 remainder 4 The pattern continues in this way.
209
5.4
5.6
5.8
7.4
5.8333
72
27 divided by 5 is 5 with remainder 2