answer:To calculate the value of 1.003 raised to the power of 25, we can use a calculator or a computer program. 1.003 ^ 25 = 1.077782 Rounded to six decimal places, the value is boxed{1.077782} .

answer:Given that log_{8}5 = y, we have 8^y = 5. Also, log_{2}125 = my implies 2^{my} = 125. We need to relate these equations. Note that 125 = 5^3, so we can cube the first equation: (8^y)^3 = 5^3 = 125, giving 8^{3y} = 125. Since 8 = 2^3, substituting gives (2^3)^{3y} = 2^{9y} = 125 = 2^{my}. Thus, 9y = my. Therefore, we conclude m = 9 and boxed{m = 9}.

answer:To find the value of n, we need to determine the highest power of 5 that divides evenly into 20!. The factorial of 20 (20!) is the product of all positive integers from 1 to 20. To find out how many times 5 is a factor in this product, we count how many multiples of 5 are there between 1 and 20, including the multiples of higher powers of 5 (like 25, which is 5^2). Multiples of 5 between 1 and 20: 5, 10, 15, 20. There are 4 of these. Multiples of 25 (which is 5^2) between 1 and 20: 25. There is 1 of these. Each multiple of 5 contributes at least one factor of 5, and each multiple of 25 contributes an additional factor of 5 (since 25 is 5^2). So, the total number of 5s in 20! is 4 (from the multiples of 5) + 1 (from the multiples of 25) = 5. This means that 5^5 is a factor of 20!, but 5^(5+1) = 5^6 is not, because there are not enough factors of 5 in 20! to make up a 5^6. Therefore, the value of n is boxed{5} .

answer:To find the smallest possible value of N, the total number of 1-cm cubes in the solid rectangular block, we start by understanding that the 231 cubes which are not visible are covered by exactly one layer of visible cubes. This means these hidden cubes form a smaller rectangular solid, which is one unit shorter in each dimension compared to the original block. Let's denote the dimensions of the original block as l times m times n. Given the condition for the hidden cubes, we have: [ (l - 1) times (m - 1) times (n - 1) = 231 ] The prime factorization of 231 is: [ 231 = 3 cdot 7 cdot 11 ] This gives us several possibilities for the dimensions of the smaller block (the one formed by the hidden cubes). For example, one might initially consider dimensions like l - 1 = 1, m - 1 = 11, and n - 1 = 3 cdot 7. However, to minimize N, which is the product l cdot m cdot n, we should aim to make the dimensions l, m, and n as close to each other as possible. The optimal way to achieve this is by arranging the factors of 231 into dimensions that are as balanced as possible. This leads us to choose the smaller block to have dimensions: [ 3 times 7 times 11 ] Adding the extra layer to each dimension to get the original block's dimensions, we have: [ (l, m, n) = (3 + 1, 7 + 1, 11 + 1) = (4, 8, 12) ] Therefore, the total number of 1-cm cubes in the original block, N, is the product of these dimensions: [ N = 4 times 8 times 12 = 384 ] So, the smallest possible value of N is boxed{384}.