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

Answer 2

Step 1: Represent a in integer and fractional parts
Let \( a = n + f \), where \( n = [a] \) is an integer and \( f = \{a\} \) is the fractional part such that \( 0 \le f < 1 \).
Then the given equation becomes: \[ |2a - 1| = 3[a] + 2\{a\} \Rightarrow |2(n+f) - 1| = 3n + 2f. \] Simplify the left-hand side: \[ |2n + 2f - 1| = 3n + 2f. \]

Step 2: Consider the two possible cases for the modulus
Case 1: \( 2n + 2f - 1 \ge 0 \Rightarrow 2n + 2f - 1 = 3n + 2f \Rightarrow 2n - 1 = 3n \Rightarrow n = -1. \)
But for this case, \( 2n + 2f - 1 \ge 0 \Rightarrow 2(-1) + 2f - 1 \ge 0 \Rightarrow 2f - 3 \ge 0 \Rightarrow f \ge 1.5, \) which is impossible since \( 0 \le f < 1 \).
Hence, no solution in this case.

Case 2: \( 2n + 2f - 1 < 0 \Rightarrow -(2n + 2f - 1) = 3n + 2f \Rightarrow -2n - 2f + 1 = 3n + 2f. \)
Simplify: \[ 1 = 5n + 4f \Rightarrow f = \frac{1 - 5n}{4}. \] Since \( 0 \le f < 1 \), we have: \[ 0 \le \frac{1 - 5n}{4} < 1. \] Multiply through by 4: \[ 0 \le 1 - 5n < 4. \] From \( 1 - 5n \ge 0 \Rightarrow n \le 0 \), and \( 1 - 5n < 4 \Rightarrow -5n < 3 \Rightarrow n > -0.6. \) So the only integer \( n \) satisfying both is \( n = 0. \)
Thus \( f = \frac{1 - 0}{4} = \frac{1}{4}. \)

Step 3: Verify the sign condition
Check \( 2n + 2f - 1 < 0 \): \( 2(0) + 2(\frac{1}{4}) - 1 = \frac{1}{2} - 1 = -\frac{1}{2} < 0 \), which is valid.

Step 4: Write the solution set and compute the required value
Thus \( S = \{\frac{1}{4}\} \).
\[ \sum_{a \in S} a = \frac{1}{4}. \] Therefore, \[ 72 \sum_{a \in S} a = 72 \times \frac{1}{4} = 18. \]

Final answer

18