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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Find the unit digit of \(4^{1024}\).
Answer: C
The power cycle for the unit digit of 4 is (4, 6), with a length of 2. For even powers, the unit digit is 6. For odd powers, the unit digit is 4. Since the power 1024 is even, the unit digit is 6.
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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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How many trailing zeros are there in the product \(20 \times 25 \times 30 \times 35 \times 40\)?
Answer: B
To find the number of trailing zeros, we need to count the pairs of 2 and 5 in the prime factorization of the product.
\(20 = 2^2 \times 5^1\)
\(25 = 5^2\)
\(30 = 2 \times 3 \times 5^1\)
\(35 = 5 \times 7\)
\(40 = 2^3 \times 5^1\)
Total number of 2s = \(2+1+3 = 6\). Total number of 5s = \(1+2+1+1+1 = 6\). Wait, 35 is 5x7. Yes. 40 is 8x5. Yes. Total 5s is 6. Total 2s is 6. So there are 6 pairs. The answer should be 6. Let me re-count. 20 (one 5), 25 (two 5s), 30 (one 5), 35 (one 5), 40 (one 5). Total 5s = 1+2+1+1+1=6. Total 2s. 20 (two 2s), 30 (one 2), 40 (three 2s). Total 2s = 2+1+3=6. So number of zeros is 6. The answer key says 5. Why? Let me check again. 20x25 = 500. 30x35=1050. 40. So 500x1050x40. 5x105x4 x 10000. 20x105x10000. 2100 x 10000. 21 x 10^6. So 6 zeros. The answer key is wrong. I will correct it to C.
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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 N = 1! + 2! + 3! + ... + 1000!. What is the unit digit of N?
Answer: C
To find the unit digit of the sum, we observe the unit digits of the individual factorials. \(1!=1\), \(2!=2\), \(3!=6\), \(4!=24\) (unit digit 4). For any factorial \(n!\) where \(n \ge 5\), the unit digit is 0. So all terms from 5! onwards have a unit digit of 0. We only need to sum the unit digits of the first four terms: \(1 + 2 + 6 + 4 = 13\). The unit digit of the sum is 3.
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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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What is the unit digit of \(2^{10}\)?
Answer: B
The power cycle of the unit digit of 2 is (2, 4, 8, 6), which has a length of 4. To find the unit digit of \(2^{10}\), we find the remainder of the power when divided by the cycle length: \(10 \div 4\) gives a remainder of 2. The required unit digit is the 2nd in the cycle, which is 4.
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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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