Find the unit digit of \(7^{71}\).
Answer: B
The power cycle of the unit digit for 7 is (7, 9, 3, 1), with a length of 4. We find the remainder of the power \(71\) when divided by 4. \(71 = 17 \times 4 + 3\). The remainder is 3. The required unit digit is the 3rd in the cycle, which is 3.
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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 \(111^{111} \times 222^{222} \times 333^{333}\)
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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How many trailing zeros are in 1000! ?
Answer: B
We need to find the number of factors of 5 in 1000!. Using Legendre's formula: \(\lfloor 1000/5 \rfloor + \lfloor 1000/25 \rfloor + \lfloor 1000/125 \rfloor + \lfloor 1000/625 \rfloor = 200 + 40 + 8 + 1 = 249\). There are 249 trailing zeros.
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Find the unit digit of \(25^{45} \times 56^{78} \times 89^{12}\).
Answer: A
To find the unit digit of the product, we can find the unit digit of each factor.
The unit digit of \(25^{45}\) is 5, as any power of a number ending in 5 will end in 5.
The unit digit of \(56^{78}\) is 6, as any power of a number ending in 6 will end in 6.
The unit digit of the product will be the unit digit of \(5 \times 6 = 30\), which is 0. We don't even need to calculate the third term's unit digit, as any number multiplied by a number ending in 0 will also end in 0.
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Find the unit digit of \((1!)!\).
Answer: A
First, we evaluate the expression inside the brackets. \(1! = 1\). So the expression becomes \((1)!\), which is just 1. The unit digit is 1.
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What is the unit digit of \(8^{88}\)?
Answer: D
The power cycle for the unit digit of 8 is (8, 4, 2, 6), which has a length of 4. We need to find the remainder of the power \(88\) when divided by 4. Since 88 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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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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