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 unit digit of \(7^{4}\)?
Answer: D
We look at the pattern of the unit digits of powers of 7. \(7^1=7\), \(7^2=49\) (unit digit 9), \(7^3=343\) (unit digit 3), \(7^4=2401\) (unit digit 1). The unit digit of \(7^4\) is 1.
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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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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 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 highest power of 45 that divides 100! ?
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.
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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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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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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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