Finding Divisors of 132

To solve this problem, we first need to find all the positive divisors of the number 132. A divisor is a number that divides another number completely, leaving no remainder.

One way to find divisors is by using prime factorization. Let's find the prime factorization of 132:

$$132 = 2 \times 66$$

$$132 = 2 \times 2 \times 33$$

$$132 = 2^2 \times 3^1 \times 11^1$$

The positive divisors of 132 can be formed by taking combinations of these prime factors: $2^a \times 3^b \times 11^c$, where $a$ can be 0, 1, or 2; $b$ can be 0 or 1; and $c$ can be 0 or 1.

Let's list all the possible combinations and the resulting divisors:

• $2^0 \times 3^0 \times 11^0 = 1$
• $2^1 \times 3^0 \times 11^0 = 2$
• $2^2 \times 3^0 \times 11^0 = 4$
• $2^0 \times 3^1 \times 11^0 = 3$
• $2^1 \times 3^1 \times 11^0 = 6$
• $2^2 \times 3^1 \times 11^0 = 12$
• $2^0 \times 3^0 \times 11^1 = 11$
• $2^1 \times 3^0 \times 11^1 = 22$
• $2^2 \times 3^0 \times 11^1 = 44$
• $2^0 \times 3^1 \times 11^1 = 33$
• $2^1 \times 3^1 \times 11^1 = 66$
• $2^2 \times 3^1 \times 11^1 = 132$

So, the positive divisors of 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, and 132.

Arranging Divisors in Descending Order

The question asks for the divisors to be arranged in descending order. This means listing them from the largest divisor to the smallest divisor.

Arranging the divisors (1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132) in descending order, we get:

132, 66, 44, 33, 22, 12, 11, 6, 4, 3, 2, 1.

Identifying the First Divisor and its Unit Digit

The first divisor in this descending order list is the largest divisor, which is 132.

The question asks for the digit at the unit place of this first divisor. The unit place is the rightmost digit of a number.

In the number 132, the digits are 1 (hundreds place), 3 (tens place), and 2 (unit place).

The digit at the unit place of 132 is 2.

Therefore, when all positive divisors of 132 are arranged in descending order, the first divisor is 132, and the digit at its unit place is 2.