Question
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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?
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?
Answers
GPT-4o mini
Given the equations
\[
N = 5a + 3b + 5c
\]
and
\[
N = 4a + 5b + 4c,
\]
we can set them equal to each other as follows:
\[
5a + 3b + 5c = 4a + 5b + 4c.
\]
Rearranging this equation, we get:
\[
5a - 4a + 3b - 5b + 5c - 4c = 0,
\]
which simplifies to:
\[
a - 2b + c = 0.
\]
From this, we can express \( c \) in terms of \( a \) and \( b \):
\[
c = 2b - a.
\]
Substituting this back into \( N \), we find:
\[
N = 5a + 3b + 5(2b - a) = 5a + 3b + 10b - 5a = 13b.
\]
Thus we have:
\[
N = 13b.
\]
Given the condition \( 131 < N < 150 \), substituting for \( N \) yields:
\[
131 < 13b < 150.
\]
Dividing through by 13, we find:
\[
\frac{131}{13} < b < \frac{150}{13},
\]
which simplifies to:
\[
10.0769 < b < 11.5385.
\]
Since \( b \) must be a distinct positive integer, the only possible value for \( b \) is \( 11 \).
Now substituting \( b = 11 \) back into the expression for \( N \):
\[
N = 13 \times 11 = 143.
\]
Now substituting \( b = 11 \) into the equation \( c = 2b - a \):
\[
c = 2 \times 11 - a = 22 - a.
\]
Next, we need to ensure that \( a \), \( b \), and \( c \) are distinct positive integers. We have \( b = 11 \) and \( c = 22 - a \).
To keep \( c \) positive, we require:
\[
22 - a > 0 \implies a < 22.
\]
To maintain distinctness among \( a \), \( b \), and \( c \), we require:
- \( a \neq 11 \) (since \( b = 11 \)),
- \( a \neq 22 - a \) (i.e., \( 2a \neq 22 \implies a \neq 11 \)).
Thus, we can choose \( a \) as any integer from 1 to 21, excluding 11.
Let's evaluate \( a + 2b + 3c \):
\[
a + 2b + 3c = a + 2(11) + 3(22 - a) = a + 22 + 66 - 3a = -2a + 88.
\]
To maximize \( a + 2b + 3c \), we should minimize \( a \). The smallest value \( a \) can take, that is a distinct positive integer, is 1.
Substituting \( a = 1 \):
\[
a + 2b + 3c = -2(1) + 88 = 86.
\]
We can verify with larger values of \( a \):
- If \( a = 2 \): \( -2(2) + 88 = 84 \)
- If \( a = 3 \): \( -2(3) + 88 = 82 \)
- Continuing this way increases \( -2a \), thus lowering the total.
The maximum achievable is when \( a = 1\):
Therefore, the maximum possible value of \( a + 2b + 3c \) is
\[
\boxed{86}.
\]
\[
N = 5a + 3b + 5c
\]
and
\[
N = 4a + 5b + 4c,
\]
we can set them equal to each other as follows:
\[
5a + 3b + 5c = 4a + 5b + 4c.
\]
Rearranging this equation, we get:
\[
5a - 4a + 3b - 5b + 5c - 4c = 0,
\]
which simplifies to:
\[
a - 2b + c = 0.
\]
From this, we can express \( c \) in terms of \( a \) and \( b \):
\[
c = 2b - a.
\]
Substituting this back into \( N \), we find:
\[
N = 5a + 3b + 5(2b - a) = 5a + 3b + 10b - 5a = 13b.
\]
Thus we have:
\[
N = 13b.
\]
Given the condition \( 131 < N < 150 \), substituting for \( N \) yields:
\[
131 < 13b < 150.
\]
Dividing through by 13, we find:
\[
\frac{131}{13} < b < \frac{150}{13},
\]
which simplifies to:
\[
10.0769 < b < 11.5385.
\]
Since \( b \) must be a distinct positive integer, the only possible value for \( b \) is \( 11 \).
Now substituting \( b = 11 \) back into the expression for \( N \):
\[
N = 13 \times 11 = 143.
\]
Now substituting \( b = 11 \) into the equation \( c = 2b - a \):
\[
c = 2 \times 11 - a = 22 - a.
\]
Next, we need to ensure that \( a \), \( b \), and \( c \) are distinct positive integers. We have \( b = 11 \) and \( c = 22 - a \).
To keep \( c \) positive, we require:
\[
22 - a > 0 \implies a < 22.
\]
To maintain distinctness among \( a \), \( b \), and \( c \), we require:
- \( a \neq 11 \) (since \( b = 11 \)),
- \( a \neq 22 - a \) (i.e., \( 2a \neq 22 \implies a \neq 11 \)).
Thus, we can choose \( a \) as any integer from 1 to 21, excluding 11.
Let's evaluate \( a + 2b + 3c \):
\[
a + 2b + 3c = a + 2(11) + 3(22 - a) = a + 22 + 66 - 3a = -2a + 88.
\]
To maximize \( a + 2b + 3c \), we should minimize \( a \). The smallest value \( a \) can take, that is a distinct positive integer, is 1.
Substituting \( a = 1 \):
\[
a + 2b + 3c = -2(1) + 88 = 86.
\]
We can verify with larger values of \( a \):
- If \( a = 2 \): \( -2(2) + 88 = 84 \)
- If \( a = 3 \): \( -2(3) + 88 = 82 \)
- Continuing this way increases \( -2a \), thus lowering the total.
The maximum achievable is when \( a = 1\):
Therefore, the maximum possible value of \( a + 2b + 3c \) is
\[
\boxed{86}.
\]