Factorials & Power Cycles Questions and Answers

Answer: A

\[ 6! - 5! = 720 - 120 = 600 \]

\[ 600 = 100 \times 6 \quad\Longrightarrow\quad \text{remainder } 0 \]

Answer: B

To find the unit digit of \(7^{7^7}\), we need to find the remainder of the power \(7^7\) when divided by the cycle length of 7, which is 4. So we need to find \(7^7 \pmod 4\). We can write 7 as \(8-1\). So, \(7 \equiv -1 \pmod 4\). Then \(7^7 \equiv (-1)^7 = -1 \equiv 3 \pmod 4\). The remainder of the power is 3. We need the 3rd unit digit in the cycle of 7 (7, 9, 3, 1), which is 3.

Answer: C

To find the unit digit of the sum, we observe the unit digits of the individual factorials. \(1!=1\), \(2!=2\), \(3!=6\), \(4!=24\) (unit digit 4). For any factorial \(n!\) where \(n \ge 5\), the unit digit is 0. So all terms from 5! onwards have a unit digit of 0. We only need to sum the unit digits of the first four terms: \(1 + 2 + 6 + 4 = 13\). The unit digit of the sum is 3.

Answer: B

When multiplying numbers, the number of trailing zeros in the product is the sum of the number of trailing zeros in each number. Zeros in 50! = \(\lfloor 50/5 \rfloor + \lfloor 50/25 \rfloor = 10+2=12\). Zeros in 20! = \(\lfloor 20/5 \rfloor = 4\). Total zeros = \(12+4=16\).

Answer: A

To find the last digit of a product, we only need to look at the last digits of the numbers being multiplied. The product includes the number 85, which has a unit digit of 5. It also includes several even numbers, like 82, 84, 86, and 88. The product of a number ending in 5 and any even number will result in a number ending in 0. Therefore, the unit digit of the entire product will be 0.

Answer: A

The unit digit of \(n^5\) is the same as the unit digit of n. This is because the cycle length for all digits' powers divides 4, and \(5 \equiv 1 \pmod 4\). So, the unit digit of \(k^5\) is the same as the unit digit of k. We need to find the unit digit of the sum \(1+2+3+...+20\). The sum is \(\frac{20(21)}{2} = 210\). The unit digit of this sum is 0.

Answer: C

The highest power of 10 is determined by the number of trailing zeros, which is limited by the number of factors of 5. Power of 5 in 75! = \(\lfloor 75/5 \rfloor + \lfloor 75/25 \rfloor = 15 + 3 = 18\). The highest power of 10 is 18.

Answer: C

\[ \frac{6!}{2! \times 3!} = \frac{720}{2 \times 6} = \frac{720}{12} = 60 \]

Answer: B

To find the last non-zero digit, we divide the factorial by its powers of 10, which means removing pairs of 2 and 5. There are more 2s than 5s. Let's count them. 5s in 20! = \(\lfloor 20/5 \rfloor = 4\). 2s in 20! = \(10+5+2+1 = 18\). We remove four 5s and four 2s. We need the unit digit of \(20! / 10^4\). This is a complex calculation. A known formula is that the last non-zero digit of n! is \((2^{a_2-a_5} \times (\text{unit digits of other primes})) \pmod{10}\), where \(a_2\) and \(a_5\) are powers of 2 and 5. A simpler method involves finding the unit digit of the product of numbers after removing all 5s and an equal number of 2s. The calculation is complex but the result for 20! is known to be 4.

Answer: B

The unit digit of powers of 3 follows a cycle of length 4: (3, 9, 7, 1). To find the unit digit, we need the remainder of the power when divided by 4. The power is \(4n+1\). When \(4n+1\) is divided by 4, the term \(4n\) is perfectly divisible, leaving a remainder of 1. A remainder of 1 corresponds to the first digit in the cycle, which is 3.