How many distinct integer values of N between 1 and 1000 are there, such that
N=4a+b+4c and 2N=7a+6b+7c for some positive integers a, b and c?
21
58
58
To solve this problem, we need to find the number of distinct integer values of N between 1 and 1000 that satisfy the given conditions.
Let's start by analyzing the given equations:
N = 4a + b + 4c ---(1)
2N = 7a + 6b + 7c ---(2)
We can simplify equation (2) by substituting N from equation (1):
2(4a + b + 4c) = 7a + 6b + 7c
8a + 2b + 8c = 7a + 6b + 7c
a - 4b - c = 0 ---(3)
Now, we have two equations (1) and (3) to work with.
Since a, b, and c are positive integers, we can find the range of possible values for each variable.
From equation (1), we know that:
1 ≤ N ≤ 1000
Now, let's analyze equation (3). We will solve it for each variable:
a - 4b - c = 0
For a positive integer solution, we can start with the lowest value for a and calculate the corresponding values for b and c.
Case 1: a = 1
1 - 4b - c = 0
4b + c = 1
The value of b in this case is 0, and the value of c is 1.
Case 2: a = 2
2 - 4b - c = 0
4b + c = 2
The value of b in this case is 0 or 1, and the value of c is 2 or 0 respectively.
We can continue this process for increasing values of a. However, we will reach a point where c becomes negative, which is not possible given the positive integer requirement. Hence, we can stop there.
Using this process, we can determine the distinct values of N that satisfy the given conditions by substituting the values of a, b, and c back into equation (1) to find N.
To summarize:
Case 1: N = 4(1) + 0 + 4(1) = 12
Case 2: N = 4(2) + 0 + 4(2) = 16
Case 3: N = 4(2) + 1 + 4(0) = 9
Therefore, there are three distinct integer values of N between 1 and 1000 that satisfy the given conditions: 9, 12, and 16.