Question

If \( S = \{ a \in \mathbb{R} : |2a - 1| = 3[a] + 2\{a\} \} \), where \([t]\) denotes the greatest integer less than or equal to \(t\) and \(\{t\}\) represents the fractional part of \(t\), then \( 72 \sum_{a \in S} a \) is equal to _________.

Answer 1

Correct Answer: 18

Given:

\[ |2a - 1| = 3[a] + 2\{a\} \]

Rewrite \(|2a - 1|\) in two forms depending on the value of \(a\):

In this case:

\[ 2a - 1 = [a] + 2a \]

Since \([a] = -1\), we find that \(a \in [-1, 0)\), which is a contradiction because \(a > \frac{1}{2}\). Therefore, this case is rejected.

Case 2: \(a < \frac{1}{2}\) In this case:

\[ -2a + 1 = [a] + 2a \]

Let \(a = I + f\) where \(I\) is the integer part and \(f\) is the fractional part, so \([a] = 0\) and \(\{a\} = f\).

Then we have:

\[ -2(I + f) + 1 = I + 2f \]

Substituting \(I = 0\), we get:

\[ 1 = 2f \implies f = \frac{1}{4} \]

Thus, \(a = \frac{1}{4}\).

Now, calculating \(72 \sum_{a \in S} a\): \[ 72 \times \frac{1}{4} = 18 \]