Factorials & Power Cycles Questions and Answers

Answer: C

First, evaluate the factorials. \(4! = 24\) and \(2! = 2\). The expression becomes \(24^2\). To find the unit digit, we only need to look at the unit digit of the base, which is 4. The problem is to find the unit digit of \(4^2\), which is 16. The unit digit is 6.

Answer: A

Let N1 = 100! and N2 = 200!. When adding numbers, the number of trailing zeros in the sum is determined by the number with the fewer trailing zeros. We can factor out the smaller number of zeros. For example, \(100 + 1000 = 100(1+10) = 1100\). The number of zeros is 2, same as in 100. So we need to find the number of zeros in 100!. Number of zeros in 100! = \(\lfloor 100/5 \rfloor + \lfloor 100/25 \rfloor = 20 + 4 = 24\). Number of zeros in 200! is \(40+8+1=49\). The smaller number is 24, so the sum will have 24 trailing zeros.

Answer: C

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 \(105\) when divided by 4. \(105 \div 4\) gives a remainder of 1. The required unit digit is the 1st in the cycle, which is 7.

Answer: B

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=30 and p=2. The power is \(\lfloor 30/2 \rfloor + \lfloor 30/4 \rfloor + \lfloor 30/8 \rfloor + \lfloor 30/16 \rfloor = 15 + 7 + 3 + 1 = 26\). The highest power of 2 is 26.

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.