Estefano R. answered • 07/19/24
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(LaTex) Notice that \( b \) can only take integer values, which means \( 2b \) is always an even number. Let's rewrite the equation in terms of an auxiliary variable \( k \):
\[
k = 2b
\]
where \( k \) is an even integer. Now, the equation becomes:
\[
a + k + c = 128
\]
with the additional constraint that \( k \) is even and non-negative.
\( k \) can be any even number from 0 to 128. Thus, \( k = 2j \) where \( j \) ranges from 0 to 64 (since \( 2 \times 64 = 128 \)).
For each value of \( k \), we need to solve:
\[
a + c = 128 - k
\]
where \( k = 2j \). The number of non-negative integer solutions to \( a + c = 128 - k \) is \( 128 - k + 1 \).
We sum over all valid \( k \):
\[
\sum_{j=0}^{64} (128 - 2j + 1)
\]
\[
\sum_{j=0}^{64} (129 - 2j)
\]
This can be split into two separate sums:
\[
\sum_{j=0}^{64} 129 - \sum_{j=0}^{64} 2j
\]
\[
\sum_{j=0}^{64} 129 = 129 \times 65 = 8385
\]
\[
\sum_{j=0}^{64} 2j = 2 \sum_{j=0}^{64} j = 2 \times \frac{64 \times 65}{2} = 64 \times 65 = 4160
\]
Hence,
\[
8385 - 4160 = 4225
\]
So the number of non-negative integer solutions to the equation \( a + 2b + c = 128 \) is \( 4225 \).
