answer:1. We start with the expression (underbrace{2 times 2 times ldots times 2}_{20 text{ times}}). This expression is the same as (2^{20}). 2. We are asked to find the units digit of (2^{20} - 1). 3. To do this, let's first look at the pattern in the units digits of powers of 2: - (2^1 = 2), units digit is (2) - (2^2 = 4), units digit is (4) - (2^3 = 8), units digit is (8) - (2^4 = 16), units digit is (6) - (2^5 = 32), units digit is (2) 4. We observe that the units digits repeat every 4 powers: - (2, 4, 8, 6) - For (2^6), the units digit is again (2), and this pattern continues. 5. Finding the units digit of (2^{20}): - Note the repeating pattern length: 4 - Compute the effective exponent modulo 4: [ 20 mod 4 = 0 ] - This means that (2^{20}) has the same units digit as (2^4). - From our earlier pattern: (2^4), the units digit is (6). 6. Since we now know that the units digit of (2^{20}) is (6), we need to subtract 1 from it to get the units digit of (2^{20} - 1): [ 6 - 1 = 5 ] # Conclusion: The units digit of (2^{20} - 1) is (boxed{5}).

answer:1. Start by squaring the given equation: [ left(n + frac{1}{n}right)^2 = 10^2 ] This simplifies to: [ n^2 + 2left(n cdot frac{1}{n}right) + frac{1}{n^2} = 100 ] Simplifying further: [ n^2 + 2 + frac{1}{n^2} = 100 ] 2. Isolate ( n^2 + frac{1}{n^2} ) by subtracting 2 from both sides: [ n^2 + frac{1}{n^2} = 100 - 2 = 98 ] 3. Add 6 to the expression: [ n^2 + frac{1}{n^2} + 6 = 98 + 6 = 104 ] Therefore, the final answer is: [ boxed{104} ]

answer:Let's denote the speed of the car during the forward journey as ( v ) km/hr. The time taken for the forward journey is 7 hours, so the distance ( d ) traveled can be expressed as: [ d = v times 7 ] During the return journey, the car increases its speed by 12 km/hr, so its speed is ( v + 12 ) km/hr. The time taken for the return journey is 5 hours, so the distance traveled can also be expressed as: [ d = (v + 12) times 5 ] Since the distance ( d ) is the same for both the forward and return journeys, we can set the two expressions for ( d ) equal to each other: [ v times 7 = (v + 12) times 5 ] Now, we can solve for ( v ): [ 7v = 5v + 60 ] [ 7v - 5v = 60 ] [ 2v = 60 ] [ v = 30 ] km/hr Now that we have the speed during the forward journey, we can find the distance ( d ): [ d = v times 7 ] [ d = 30 times 7 ] [ d = 210 ] km So, the distance traveled is boxed{210} km.

answer:1. **Assumption of the cost of each poster**: Let the regular price per poster be 4. This makes the mathematics straightforward and doesn't affect the general outcome since we only need to know the number of posters. 2. **Calculate total money Jeremy has**: If Jeremy can buy 24 posters at 4 each, then he has: [ 24 times 4 = 96 text{ dollars} ] 3. **Calculate the cost of posters under the sale conditions**: Under the sale, the first poster costs 4 and the second costs half of that, which is: [ frac{1}{2} times 4 = 2 text{ dollars} ] Therefore, every pair of posters (one at full price and one at the discounted price) costs: [ 4 + 2 = 6 text{ dollars} ] 4. **Determine how many pairs of posters Jeremy can buy**: With 96 dollars, the number of pairs of posters Jeremy can buy is: [ frac{96}{6} = 16 text{ pairs} ] 5. **Calculate the total number of posters**: Each pair consists of 2 posters, so the total number Jeremy can buy is: [ 16 times 2 = 32 text{ posters} ] The greatest number of posters Jeremy could buy is 32. The final answer is boxed{textbf {(C) } 32}