Given that:
4m+n=20; and
|n|≤20

How many ordered pairs (m,n) exist in which m and n both are integers?

Question

Given that:
4m+n=20; and
|n|≤20

How many ordered pairs (m,n) exist in which m and n both are integers?

Vinod · Accepted Answer

First, let's understand what the question is asking:We need to find integer pairs (m,n) that satisfy the equation 4m+n=20The absolute value of n is less than or equal to 20 (|n|&le;20)
Let's rearrange the equation to isolate m:4m = 20 - nm = (20 - n) / 4
For m to be an integer, (20 - n) must be divisible by 4.
Given that |n|&le;20, n can be any integer from -20 to 20.
Let's list out the possible values of n that make (20 - n) divisible by 4:When n = 0, 20 - 0 = 20, which is divisible by 4When n = 4, 20 - 4 = 16, which is divisible by 4When n = 8, 20 - 8 = 12, which is divisible by 4When n = 12, 20 - 12 = 8, which is divisible by 4When n = 16, 20 - 16 = 4, which is divisible by 4When n = 20, 20 - 20 = 0, which is divisible by 4
And in the negative direction:When n = -4, 20 - (-4) = 24, which is divisible by 4When n = -8, 20 - (-8) = 28, which is divisible by 4When n = -12, 20 - (-12) = 32, which is divisible by 4When n = -16, 20 - (-16) = 36, which is divisible by 4When n = -20, 20 - (-20) = 40, which is divisible by 4
Now, let's calculate the corresponding m values:When n = 0, m = 20/4 = 5When n = 4, m = 16/4 = 4When n = 8, m = 12/4 = 3When n = 12, m = 8/4 = 2When n = 16, m = 4/4 = 1When n = 20, m = 0/4 = 0When n = -4, m = 24/4 = 6When n = -8, m = 28/4 = 7When n = -12, m = 32/4 = 8When n = -16, m = 36/4 = 9When n = -20, m = 40/4 = 10
Therefore, the ordered pairs (m,n) are:(5,0), (4,4), (3,8), (2,12), (1,16), (0,20),(6,-4), (7,-8), (8,-12), (9,-16), (10,-20)
Count the number of pairs: There are 11 ordered pairs in total.
Therefore, there are 11 ordered pairs (m,n) where m and n are both integers that satisfy the given conditions.

Given that: 4m+n=20; and |n|≤20 How many ordered pairs (m,n) exist in which m and n both are integers?