1.
What is the value of 5! ?
Answer: B
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
24 is \( 4! \) and 720 is \( 6! \), the neighbours on either side.
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Factorials & Power Cycles Questions and Answers
Important Instructions
2.
What is the unit digit of \(7^{4}\)?
Answer: D
The unit digits of powers of 7 cycle with period 4:
\[ 7,\ 9,\ 3,\ 1 \]
The fourth power lands on the last of the cycle, so the unit digit is 1. Indeed \( 7^{4} = 2401 \).
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3.
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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4.
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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5.
How many trailing zeros are in 25! ?
Answer: C
The number of trailing zeros in a factorial is determined by the number of times 5 is a factor in its prime factorization. We can find this using the formula: \(\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + ...\). For 25!, this is \(\lfloor 25/5 \rfloor + \lfloor 25/25 \rfloor = 5 + 1 = 6\). There are 6 trailing zeros.
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6.
What is the unit digit of \((1! + 2! + 3! + 4!)\)?
Answer: B
\[ 1 + 2 + 6 + 24 = 33 \]
The unit digit is 3. Only the unit digits of the terms matter: \( 1 + 2 + 6 + 4 = 13 \), ending in 3.
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7.
Find the unit digit of \(3^{57}\).
Answer: B
The power cycle of the unit digit for 3 is (3, 9, 7, 1), with a length of 4. We find the remainder of the power \(57\) when divided by 4. \(57 \div 4 = 14\) with a remainder of 1. The required unit digit is the 1st in the cycle, which is 3.
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8.
What is the highest power of 7 that divides 100! ?
Answer: C
To find the highest power of a prime p in n!, we use Legendre's formula: \(\lfloor n/p \rfloor + \lfloor n/p^2 \rfloor + ...\). Here, n=100 and p=7. The power is \(\lfloor 100/7 \rfloor + \lfloor 100/49 \rfloor = 14 + 2 = 16\). The highest power of 7 is 16.
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9.
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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10.
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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