How many trailing zeros are in the result of \(50! \times 20!\)?
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 = \(12+4=16\).
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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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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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Find the unit digit of the expression \(17^{1999} + 11^{1999} - 7^{1999}\).
Answer: B
We find the unit digit of each term. For \(17^{1999}\), we look at \(7^{1999}\). The cycle for 7 is (7, 9, 3, 1), length 4. \(1999 \div 4\) has a remainder of 3. So the unit digit is 3. For \(11^{1999}\), the unit digit is always 1. For \(7^{1999}\), the unit digit is 3. So, we have \(3 + 1 - 3 = 1\). The unit digit is 1.
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Find the unit digit of \(6^{6^{6}}\)
Answer: D
The unit digit of any positive integer power of a number ending in 6 is always 6. Therefore, the unit digit of \(6^{6^{6}}\) is 6.
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What is the unit digit of \(4!^{2!}\)?
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.
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What is the value of 0! ?
Answer: B
By definition, the value of 0! (zero factorial) is 1. This is a convention used to make many mathematical formulas, like the binomial theorem, work correctly.
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Find the number of trailing zeros in \(2^5 \times 3^2 \times 4^8 \times 5^{10}\)
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.
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The number of zeros at the end of \((2!)^{2!} \times (5!)^{5!}\) is:
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.
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Find the unit digit of 13!
Answer: A
For any integer \(n \ge 5\), the factorial \(n!\) will contain the factors 2 and 5. The product of 2 and 5 is 10. Any number multiplied by 10 will have a unit digit of 0. Since 13 is greater than 5, \(13!\) will have a unit digit of 0.
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