Factorials & Power Cycles Questions and Answers

Answer: D

Correct answer: 2

Only the unit digit of each base matters, and unit digits repeat in cycles of at most four.

\(13^{24}\): the cycle for 3 is 3, 9, 7, 1. As \(24\) is a multiple of 4, the unit digit is 1.
\(68^{57}\): the cycle for 8 is 8, 4, 2, 6. As \(57 \equiv 1 \pmod 4\), the unit digit is 8.
So the first product ends in \(1\times 8 = 8\).

\(24^{13}\): the cycle for 4 is 4, 6; an odd exponent gives 4.
\(57^{68}\): the cycle for 7 is 7, 9, 3, 1. As \(68\) is a multiple of 4, the unit digit is 1.
So the second product ends in \(4\times 1 = 4\).

Adding, \(8 + 4 = 12\), so the unit digit of the whole expression is 2.

Answer: D

We find the unit digit of each term separately.

Part 1: Unit digit of \(13^{25} \times 68^{57}\). This is the unit digit of \(3^{25} \times 8^{57}\). For \(3^{25}\), the cycle is (3,9,7,1) of length 4. \(25 \div 4\) gives rem 1, so unit digit is 3. For \(8^{57}\), the cycle is (8,4,2,6) of length 4. \(57 \div 4\) gives rem 1, so unit digit is 8. The unit digit of Part 1 is the unit digit of \(3 \times 8 = 24\), which is 4.

Part 2: Unit digit of \(24^{13} \times 57^{68}\). This is the unit digit of \(4^{13} \times 7^{68}\). For \(4^{13}\) (odd power), the unit digit is 4. For \(7^{68}\), the cycle is (7,9,3,1) of length 4. \(68\) is a multiple of 4, so the unit digit is the last in the cycle, which is 1. The unit digit of Part 2 is the unit digit of \(4 \times 1 = 4\).

The unit digit of the entire expression is the unit digit of the sum of the unit digits of the two parts: \(4 + 4 = 8\).

Answer: B

By definition \( 0! = 1 \).

It is the value that makes \( n! = n \times (n-1)! \) work at \( n = 1 \), and it counts the single way to arrange nothing at all - which is why the combination formula needs it.

Answer: C

Every positive power of 5 ends in 5:

\[ 5,\ 25,\ 125,\ 625,\ \ldots \]

So \( 5^{2025} \) ends in 5.

Answer: B

Correct answer: 6

Each trailing zero comes from a factor of 10, that is a pair of 2 and 5, so count both primes across the product.

\[20 = 2^2\times 5,\quad 25 = 5^2,\quad 30 = 2\times 3\times 5,\quad 35 = 5\times 7,\quad 40 = 2^3\times 5\]

Fives: \(1 + 2 + 1 + 1 + 1 = 6\). Twos: \(2 + 0 + 1 + 0 + 3 = 6\).

The number of zeros is the smaller count, so 6. Multiplying out confirms it: the product is 21000000.

Answer: C

Correct answer: 6

Count the factors of 5, since factors of 2 are always more plentiful.

\[\left\lfloor \frac{25}{5} \right\rfloor + \left\lfloor \frac{25}{25} \right\rfloor = 5 + 1 = 6\]

The second term is needed because 25 contributes two fives rather than one.

Answer: C

To find the unit digit of the sum, we look at the unit digits of the individual factorials. \(1! = 1\), \(2! = 2\), \(3! = 6\), \(4! = 24\) (unit digit 4). For any \(n \ge 5\), \(n!\) will have a unit digit of 0, as it contains factors of 2 and 5. Therefore, we only need to sum the unit digits of the first four factorials: \(1 + 2 + 6 + 4 = 13\). The unit digit of the sum is the unit digit of 13, which is 3.

Answer: A

To find the highest power of 6, we need to find the powers of its prime factors, 2 and 3, in 60!. The power will be limited by the factor that occurs fewer times. The factor 3 is less frequent than 2. So we just need to count the power of 3 in 60!. Power of 3 = \(\lfloor 60/3 \rfloor + \lfloor 60/9 \rfloor + \lfloor 60/27 \rfloor = 20 + 6 + 2 = 28\). The highest power of 6 is 28.

Answer: A

We find the unit digit of each term. For \(37^{37}\), we need the unit digit of \(7^{37}\). Cycle for 7 is (7,9,3,1), length 4. \(37\div 4\) rem 1. So unit digit is 7. For \(23^{23}\), we need the unit digit of \(3^{23}\). Cycle for 3 is (3,9,7,1), length 4. \(23\div 4\) rem 3. So unit digit is 7. The unit digit of the difference is the unit digit of \(7-7=0\). The result is 0.

Answer: B

Take out the common factor \( 10! \).

\[ 11! - 10! = 11 \times 10! - 10! = 10!(11 - 1) \]

\[ = 10 \times 10! \]