Table of Contents
- 1 How many pairs p q of integers satisfy the equation 4 p q 2 15?
- 2 How many ordered pairs M N satisfy the equation MN 2m 2n 2020 if?
- 3 How many ordered pairs M N satisfy the equation MN 2m 2n 2020 if M n belongs to natural numbers?
- 4 How many ordered pairs of integers (m) are there?
- 5 What is 1/m + 1/n = 1/4?
- 6 How many integer solutions are there by reversing m and N?
How many pairs p q of integers satisfy the equation 4 p q 2 15?
There are 3 pairs of positive integers that satisfy the given conditions.
How many ordered pairs M N satisfy the equation MN 2m 2n 2020 if?
Answer: Thus we have obtained six pairs of that satisfy the given equation.
How many ordered pairs M N satisfy the equation MN 2m 2n 2020 if M n belongs to natural numbers?
Hence, there are 8 ordered pairs of integers that satisfy .
How many pairs of positive integers Mn satisfy 1 4 1 mn 12 where n an odd integer less than 60 a 4 b7 C 5 D 3?
How many pairs of positive integers m,n satisfy 1/m + 4/n = 1/12, where n is an odd integer less than 60? Thus n take only three values– 49, 51 and 57. Hence there are three pairs of (m,n) which satisfy the given conditions.
How many pairs(m) of positive integer satisfy m^2+105=N^2?
Q. How many pairs (m, n) of positive integer satisfy the equation m^2+105=n^2? So, only 4 cases possible to get values of n, m as positive. Thus number of solution = 4
How many ordered pairs of integers (m) are there?
So, there are 16 ordered pairs of integers (if possibility of m = n is also included). How many ordered pairs of positive integers (m,n) satisfy GCF (m,n) = 3 and LCM (m,n) = 108?
What is 1/m + 1/n = 1/4?
We are looking for positive integers m and n, and since 1/m + 1/n = 1/4, both m and n have to be more than 4. If one of them is less than 4 then the other number would be negative and if one of them is 4 then the other number would be infinity.
How many integer solutions are there by reversing m and N?
There are 2 more solutions by reversing m and n and they are 12, 6 and 20, 5. There are no more integer solutions, which is not very difficult to prove and is being left as an exercise. (HINT: When one number is greater than 20 then the other number has to be smaller than 5!)