Factorials & Power Cycles Questions and Answers

Answer: B

We need to find the number of pairs of 2 and 5. First, express all numbers in their prime factors. \(2^5 \times 3^2 \times (2^2)^8 \times 5^{10} = 2^5 \times 3^2 \times 2^{16} \times 5^{10} = 2^{21} \times 3^2 \times 5^{10}\). We have 21 factors of 2 and 10 factors of 5. The number of pairs of (2,5) is limited by the smaller power, which is 10. So there are 10 trailing zeros.

Answer: A

We look at the unit digit of each part of the product. The unit digit of 13! is 0, because 13 is greater than 5, so 13! contains factors of 2 and 5. Any number multiplied by a number ending in 0 will also end in 0. Therefore, the unit digit of the entire expression is 0, regardless of the unit digit of \(137^{13}\).

Answer: D

Only the unit digit of the base matters, so consider powers of 2:

\[ 2,\ 4,\ 8,\ 6 \quad (\text{period } 4) \]

\[ 23 = 4 \times 5 + 3 \quad\Longrightarrow\quad \text{third in the cycle} \]

So the unit digit is 8.

Answer: B

To find the highest power of 15 (which is \(3 \times 5\)), we need to find the highest powers of 3 and 5 in 80!. The power will be limited by the prime factor that occurs fewer times, which is 5. We count the power of 5 in 80!. Power of 5 = \(\lfloor 80/5 \rfloor + \lfloor 80/25 \rfloor = 16 + 3 = 19\). The highest power of 15 is 19.

Answer: C

We need the unit digit of \(7^{153} \times 1^{72}\). The unit digit of \(1^{72}\) is 1. For \(7^{153}\), the cycle is (7,9,3,1), length 4. We find the remainder of \(153 \div 4\), which is 1. So the unit digit is the 1st in the cycle, which is 7. The unit digit of the product is the unit digit of \(7 \times 1 = 7\).

Answer: D

The expression has 50 terms with base 4 and 50 terms with base 9. Let's group them: \((4^1 \times 4^3 \times ... \times 4^{99}) \times (9^2 \times 9^4 \times ... \times 9^{100})\). The unit digit of any odd power of 4 is 4. There are 50 such terms. The unit digit of any even power of 9 is 1. There are 50 such terms. So the unit digit is the unit digit of \((4 \times 4 \times ... (50 \text{ times})) \times (1 \times 1 \times ... (50 \text{ times}))\). This simplifies to the unit digit of \(4^{50} \times 1\). Since 50 is an even power, the unit digit of \(4^{50}\) is 6.

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 = \(12+4=16\).