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 5 that divides the product \(1 \times 3 \times 5 \times ... \times 99\)?
Answer: B
We need to count the number of factors of 5 in the product of all odd numbers from 1 to 99. The multiples of 5 in this product are 5, 15, 25, 35, 45, 55, 65, 75, 85, 95.
Let's count the factors of 5 from each term:
5 = \(1 \times 5\) (one 5)
15 = \(3 \times 5\) (one 5)
25 = \(5 \times 5\) (two 5s)
35 = \(7 \times 5\) (one 5)
45 = \(9 \times 5\) (one 5)
55 = \(11 \times 5\) (one 5)
65 = \(13 \times 5\) (one 5)
75 = \(3 \times 25 = 3 \times 5 \times 5\) (two 5s)
85 = \(17 \times 5\) (one 5)
95 = \(19 \times 5\) (one 5)
Total number of factors of 5 is \(1+1+2+1+1+1+1+2+1+1 = 12\).
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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 remainder when \(1! + 2! + 3! + ... + 100!\) is divided by 8?
Answer: A
We need to find the remainder of the sum when divided by 8. Let's look at the remainders of the first few factorials. \(1! = 1\) (rem 1). \(2! = 2\) (rem 2). \(3! = 6\) (rem 6). \(4! = 24\). Since 24 is a multiple of 8, the remainder is 0. For any \(n \ge 4\), \(n!\) contains the factors 4 and 2, so it is a multiple of 8. The remainder for all subsequent factorials will be 0. We only need to sum the remainders of the first three factorials: \(1 + 2 + 6 = 9\). The remainder of \(9 \div 8\) is 1.
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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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Find the number of trailing zeros in the expression \(100! + 200!\)
Answer: A
Let N1 = 100! and N2 = 200!. When adding numbers, the number of trailing zeros in the sum is determined by the number with the fewer trailing zeros. We can factor out the smaller number of zeros. For example, \(100 + 1000 = 100(1+10) = 1100\). The number of zeros is 2, same as in 100. So we need to find the number of zeros in 100!. Number of zeros in 100! = \(\lfloor 100/5 \rfloor + \lfloor 100/25 \rfloor = 20 + 4 = 24\). Number of zeros in 200! is \(40+8+1=49\). The smaller number is 24, so the sum will have 24 trailing zeros.
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Find the unit digit of \(7^{105}\).
Answer: C
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 \(105\) when divided by 4. \(105 \div 4\) gives a remainder of 1. The required unit digit is the 1st in the cycle, which is 7.
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