How many trailing zeros are in 125! ?
Answer: C
The number of trailing zeros in a factorial n! is determined by the number of times 5 is a factor. We use the formula: \(\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + ...\)
For 125!, this is \(\lfloor 125/5 \rfloor + \lfloor 125/25 \rfloor + \lfloor 125/125 \rfloor = 25 + 5 + 1 = 31\).
There are 31 trailing zeros in 125!.
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What is the value of 5! ?
Answer: B
The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. So, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
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What is the remainder when 8! is divided by 7?
Answer: A
\(8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\). Since 8! contains the factor 7, it is a multiple of 7. Therefore, when 8! is divided by 7, the remainder is 0.
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What is the unit digit of \(32^{32}\)?
Answer: C
To find the unit digit of \(32^{32}\), we only need to consider the unit digit of the base, which is 2. So the problem is to find the unit digit of \(2^{32}\). The power cycle for 2 is (2, 4, 8, 6), with a length of 4. We need to find the remainder of the power \(32\) when divided by 4. Since 32 is a multiple of 4, the remainder is 0. A remainder of 0 corresponds to the last digit in the cycle, which is 6.
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Find the highest power of 12 that divides 50!.
Answer: A
To find the highest power of 12, we need to find the highest powers of its prime factors, which are \(2^2\) and 3.
Power of 3 in 50! = \(\lfloor 50/3 \rfloor + \lfloor 50/9 \rfloor + \lfloor 50/27 \rfloor = 16+5+1 = 22\).
Power of 2 in 50! = \(\lfloor 50/2 \rfloor + \lfloor 50/4 \rfloor + \lfloor 50/8 \rfloor + \lfloor 50/16 \rfloor + \lfloor 50/32 \rfloor = 25+12+6+3+1 = 47\).
Since \(12=2^2 \times 3\), we need pairs of (one 3) and (two 2s). We have 22 threes. For the twos, we have 47, so we can make \(\lfloor 47/2 \rfloor = 23\) pairs of 2s. The power of 12 is limited by the smaller of these counts (22 for threes, 23 for pairs of twos). The limiting factor is the number of 3s. So the highest power of 12 is 22.
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What is the unit digit of the sum \(1^5 + 2^5 + 3^5 + ... + 20^5\)?
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.
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The unit digit of \((3547)^{153} \times (251)^{72}\) is:
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 \(7 \times 1 = 7\).
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Find the unit digit of \(111^{111} \times 222^{222} \times 333^{333}\)
Answer: A
To find the unit digit of the expression, we only need to consider the unit digits of the bases. The problem is equivalent to finding the unit digit of the product \(1^{111} \times 2^{222} \times 3^{333}\).
Unit digit of \(1^{111}\) is always 1.
For \(2^{222}\), the power cycle of 2 is (2, 4, 8, 6), which repeats every 4 powers. We find the remainder of \(222 \div 4\), which is 2. So the unit digit is the second in the cycle, which is 4.
For \(3^{333}\), the power cycle of 3 is (3, 9, 7, 1), which repeats every 4 powers. We find the remainder of \(333 \div 4\), which is 1. So the unit digit is the first in the cycle, which is 3.
The unit digit of the entire product is the unit digit of \(1 \times 4 \times 3 = 12\), which is 2.
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What is the value of \(\frac{10!}{8!}\)?
Answer: C
We can expand the factorials. \(10! = 10 \times 9 \times 8 \times 7 \times ... \times 1\) and \(8! = 8 \times 7 \times ... \times 1\). So, \(\frac{10!}{8!} = \frac{10 \times 9 \times 8!}{8!} = 10 \times 9 = 90\).
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What is the unit digit of \(9^{99}\)?
Answer: D
The power cycle for the unit digit of 9 is (9, 1), with a length of 2. For odd powers, the unit digit is 9. For even powers, the unit digit is 1. Since the power 99 is odd, the unit digit is 9.
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