Factorials & Power Cycles Questions and Answers

Answer: B

The power cycle of the unit digit for 3 is (3, 9, 7, 1), with a length of 4. We find the remainder of the power \(57\) when divided by 4. \(57 \div 4 = 14\) with a remainder of 1. The required unit digit is the 1st in the cycle, which is 3.

Answer: C

To find the highest power of a prime p in n!, we use Legendre's formula: \(\lfloor n/p \rfloor + \lfloor n/p^2 \rfloor + ...\). Here, n=100 and p=7. The power is \(\lfloor 100/7 \rfloor + \lfloor 100/49 \rfloor = 14 + 2 = 16\). The highest power of 7 is 16.

Answer: B

We find the unit digit of each term. For \(17^{1999}\), we look at \(7^{1999}\). The cycle for 7 is (7, 9, 3, 1), length 4. \(1999 \div 4\) has a remainder of 3. So the unit digit is 3. For \(11^{1999}\), the unit digit is always 1. For \(7^{1999}\), the unit digit is 3. So, we have \(3 + 1 - 3 = 1\). The unit digit is 1.

Answer: C

The number of trailing zeros in a factorial n! is determined by the number of times 5 is a factor. We use the formula: \(\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + ...\)

For 125!, this is \(\lfloor 125/5 \rfloor + \lfloor 125/25 \rfloor + \lfloor 125/125 \rfloor = 25 + 5 + 1 = 31\).

There are 31 trailing zeros in 125!.

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.