Factorials & Power Cycles Questions and Answers

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.

Answer: A

We evaluate the exponents first. \(2!=2\) and \(5!=120\). The expression is \((2!)^2 \times (5!)^{120}\). We need to count factors of 2 and 5. \(2! = 2\). \(5! = 120 = 2^3 \times 3 \times 5\). So the expression is \(2^2 \times (2^3 \times 3 \times 5)^{120} = 2^2 \times 2^{360} \times 3^{120} \times 5^{120} = 2^{362} \times 3^{120} \times 5^{120}\). We have 362 factors of 2 and 120 factors of 5. The number of pairs is limited by the smaller number, which is 120. So there are 120 trailing zeros.

Answer: A

To find the unit digit of the product, we can look for numbers that produce a zero at the end. The term \(15^5\) has a unit digit of 5. The term \(16^6\) has a unit digit of 6 (an even number). The product of a number ending in 5 and an even number will result in a number ending in 0. Therefore, the unit digit of the entire expression is 0.

Answer: A

The prime factorization of 45 is \(3^2 \times 5\). We need to find the number of pairs of (two 3s) and (one 5) in 100!. Let's find the powers of 3 and 5. Power of 5 = \(\lfloor 100/5 \rfloor + \lfloor 100/25 \rfloor = 20+4=24\). Power of 3 = \(\lfloor 100/3 \rfloor + \lfloor 100/9 \rfloor + \lfloor 100/27 \rfloor + \lfloor 100/81 \rfloor = 33+11+3+1 = 48\). We need two 3s for each 45. We have 48 threes, so we can make \(\lfloor 48/2 \rfloor = 24\) pairs of 3s. We have 24 fives. The number of 45s is limited by the smaller count, which is 24 (from both the 3s and the 5s). So the highest power is 24.

Answer: C

The unit's digit is 6.

The unit's digit of \(47^{25}\) depends only on the unit's digit 7. Powers of 7 end in a repeating cycle of four: 7, 9, 3, 1. Since \(25 = 4 \times 6 + 1\), the 25th power is in position 1 of the cycle and ends in 7.

Subtracting 1 gives a number ending in \(7 - 1 = 6\).

7 is the unit's digit of \(47^{25}\) itself, before the 1 is subtracted, and 1 and 8 come from using the wrong position in the cycle.