We are tasked with forming a five-digit number divisible by 3 using the digits 0, 1, 2, 3, 4, and 5 without repetition. A number is divisible by 3 if the sum of its digits is divisible by 3.
Step 1: Total sum of the digits The sum of the digits 0, 1, 2, 3, 4, and 5 is: \[ 0 + 1 + 2 + 3 + 4 + 5 = 15. \] Since the total sum of the digits is 15, which is divisible by 3, the sum of the digits of any five-digit number formed from these digits will also be divisible by 3, provided that we leave out one of the digits.
Step 2: Choosing one digit to leave out We can leave out any of the six digits (0, 1, 2, 3, 4, 5). If we leave out a digit, the sum of the remaining digits will still be divisible by 3. We need to calculate the number of five-digit numbers that can be formed with the remaining digits.
Step 3: Counting the number of five-digit numbers We must choose 5 digits from the 6 available digits. We have 6 choices for the digit to leave out. For each selection of 5 digits, the number of ways to arrange them is given by the number of permutations of 5 digits. However, the first digit cannot be 0, so we need to adjust the counting. - If 0 is not selected, all 5 digits can be arranged in \( 5! \) ways. - If 0 is selected, the first digit cannot be 0, so we must select the first digit from the remaining 4 digits (1, 2, 3, 4, or 5) and arrange the other 4 digits.
Step 4: Calculating the number of arrangements - If 0 is not selected, we have \( 5! = 120 \) ways. - If 0 is selected, the number of valid arrangements is \( 4 \times 4! = 4 \times 24 = 96 \). Thus, the total number of arrangements is: \[ 120 + 96 = 216. \] Thus, the total number of ways to form a five-digit number divisible by 3 is \( 216 \).
Let \(S=\left\{ z\in\mathbb{C}:\left|\frac{z-6i}{z-2i}\right|=1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right|=\frac{3}{5} \right\}.\)
Then $\sum_{z\in S}|z|^2$ is equal to
Which of the following are ambident nucleophiles?
[A.] CN$^{\,-}$
[B.] CH$_{3}$COO$^{\,-}$
[C.] NO$_{2}^{\,-}$
[D.] CH$_{3}$O$^{\,-}$
[E.] NH$_{3}$
Identify the anomers from the following.

The standard Gibbs free energy change \( \Delta G^\circ \) of a cell reaction is \(-301 { kJ/mol}\). What is \( E^\circ \) in volts?
(Given: \( F = 96500 { C/mol}\), \( n = 2 \))