+8 For distinct positive integers a, b, and c, N = 5a + 3b + 5c = 4a + 5b + 4c. If 131 < N < 150, then
what is the maximum possible value of a + 2b + 3c?
+964 We can simplify the given equation: 5a + 3b + 5c = 4a + 5b + 4c a - 2b - c = 0 a = 2b + c
Now we can substitute a in the expression for N: N = 5(2b + c) + 3b + 5c N = 13b + 10c
We are given that 131 < N < 150. Let's try some values for b and c:
If b = 10 and c = 1, then N = 140, which is within the given range.
If b = 11 and c = 1, then N = 153, which is outside the given range.
Therefore, the maximum possible value of b is 10.
Now we can find the maximum possible value of a + 2b + 3c: a + 2b + 3c = (2b + c) + 2b + 3c = 5b + 4c
Substituting b = 10 and c = 1, we get: 5b + 4c = 5(10) + 4(1) = 54
Therefore, the maximum possible value of a + 2b + 3c is 54.
