Factorials & Power Cycles Questions and Answers

Answer: C

The unit digits of powers of 4 alternate:

\[ 4,\ 6,\ 4,\ 6,\ \ldots \]

Even exponents end in 6 and odd ones in 4. Since 1024 is even, the unit digit is 6.

Answer: A

\( 8! = 1 \times 2 \times \cdots \times 8 \) includes 7 as a factor, so it is a multiple of 7.

\[ \text{remainder} = 0 \]

Every factorial from \( 7! \) onward is divisible by 7 for the same reason.

Answer: C

To find the unit digit of \(32^{32}\), we only need to consider the unit digit of the base, which is 2. So the problem is to find the unit digit of \(2^{32}\). The power cycle for 2 is (2, 4, 8, 6), with a length of 4. We need to find the remainder of the power \(32\) when divided by 4. Since 32 is a multiple of 4, the remainder is 0. A remainder of 0 corresponds to the last digit in the cycle, which is 6.

Answer: A

To find the highest power of 12, we need to find the highest powers of its prime factors, which are \(2^2\) and 3.

Power of 3 in 50! = \(\lfloor 50/3 \rfloor + \lfloor 50/9 \rfloor + \lfloor 50/27 \rfloor = 16+5+1 = 22\).

Power of 2 in 50! = \(\lfloor 50/2 \rfloor + \lfloor 50/4 \rfloor + \lfloor 50/8 \rfloor + \lfloor 50/16 \rfloor + \lfloor 50/32 \rfloor = 25+12+6+3+1 = 47\).

Since \(12=2^2 \times 3\), we need pairs of (one 3) and (two 2s). We have 22 threes. For the twos, we have 47, so we can make \(\lfloor 47/2 \rfloor = 23\) pairs of 2s. The power of 12 is limited by the smaller of these counts (22 for threes, 23 for pairs of twos). The limiting factor is the number of 3s. So the highest power of 12 is 22.

Answer: B

The power cycle of the unit digit for 7 is (7, 9, 3, 1), with a length of 4. We find the remainder of the power \(71\) when divided by 4. \(71 = 17 \times 4 + 3\). The remainder is 3. The required unit digit is the 3rd in the cycle, which is 3.

Answer: B

The factorial 25! is the product \(25 \times 24 \times 23 \times ... \times 1\). Since this product contains the number 23 as a factor, 25! is a multiple of 23.

Therefore, the remainder when 25! is divided by 23 is 0.

The problem then becomes finding the remainder of \((0 + 1)\) when divided by 23, which is simply 1.

Answer: A

To find the unit digit of the expression, we only need to consider the unit digits of the bases. The problem is equivalent to finding the unit digit of the product \(1^{111} \times 2^{222} \times 3^{333}\).

Unit digit of \(1^{111}\) is always 1.

For \(2^{222}\), the power cycle of 2 is (2, 4, 8, 6), which repeats every 4 powers. We find the remainder of \(222 \div 4\), which is 2. So the unit digit is the second in the cycle, which is 4.

For \(3^{333}\), the power cycle of 3 is (3, 9, 7, 1), which repeats every 4 powers. We find the remainder of \(333 \div 4\), which is 1. So the unit digit is the first in the cycle, which is 3.

The unit digit of the entire product is the unit digit of \(1 \times 4 \times 3 = 12\), which is 2.

Answer: C

The 8! cancels.

\[ \frac{10!}{8!} = 10 \times 9 = 90 \]

Answer: D

The unit digits of powers of 9 alternate:

\[ 9,\ 1,\ 9,\ 1,\ \ldots \]

Odd exponents end in 9. Since 99 is odd, the unit digit is 9.

Answer: B

We need to find the number of factors of 5 in 1000!. Using Legendre's formula: \(\lfloor 1000/5 \rfloor + \lfloor 1000/25 \rfloor + \lfloor 1000/125 \rfloor + \lfloor 1000/625 \rfloor = 200 + 40 + 8 + 1 = 249\). There are 249 trailing zeros.