answer:**Part (a):** 1. We need to evaluate ( v_{p}(mathrm{N}!) ), which represents the exponent of the prime ( p ) in the factorization of ( mathrm{N}! ). 2. We represent this factorization by: [ v_{p}(mathrm{N}!) = sum_{i=1}^{infty} leftlfloor frac{N}{p^i} rightrfloor ] This sum counts the number of times ( p ) appears as a factor in the numbers from 1 to ( mathrm{N} ). 3. To derive ( sum_{i=1}^{infty} leftlfloor frac{N}{p^i} rightrfloor ), we use the following steps: - Write down the numbers from 1 to ( N ) in a list. - For every ( i ) from 1 to ( v_{p}(mathrm{N}) ), place a mark (point) at all multiples of ( p^i ). 4. The total number of points corresponds to the term ( sum_{i=1}^{infty} leftlfloor frac{N}{p^i} rightrfloor ), as every number ( k in [1, N] ) receives ( v_{p}(k) ) points. 5. Hence, the total number of points counts the sum and therefore satisfies the equality: [ sum_{i=1}^{infty} leftlfloor frac{N}{p^i} rightrfloor = v_{p}(N!) ] **Part (b):** 1. Using the property ( binom{N}{k} = frac{N!}{k!(N-k)!} ), we aim to find ( v_{p}left(binom{N}{k}right) ), which is the exponent of ( p ) in ( binom{N}{k} ). 2. By the property of ( v_{p} ), we recall the additivity of ( v_{p} ) over multiplication: [ v_{p}left(binom{N}{k}right) = v_{p}left( frac{N!}{k!(N-k)!} right) = v_{p}(N!) - v_{p}(k!) - v_{p}((N-k)!) ] 3. Given ( N = p^n ) and ( N geq k ), we use the previously established formula for ( v_{p}(N!) ): 4. For all ( i in [1, v_{p}(k)] ): [ leftlfloor frac{N}{p^i} rightrfloor = leftlfloor frac{k}{p^i} rightrfloor + leftlfloor frac{N-k}{p^i} rightrfloor ] 5. For ( i in [v_{p}(k) + 1, v_{p}(N)] ): [ leftlfloor frac{N}{p^i} rightrfloor = leftlfloor frac{k}{p^i} rightrfloor + leftlfloor frac{N-k}{p^i} rightrfloor ] Essentially, the contribution from these terms is 1. 6. Combining these results, we get: [ v_{p}left( binom{N}{k} right) = v_{p}(N) - v_{p}(k) - v_{p}(N-k) = v_{p}(N) - v_{p}(k) - 0 ] Since ( v_{p}(N-k) = 0 ) when ( N = p^n ). 7. Ultimately, we conclude: [ v_{p}left( binom{N}{k} right) = v_{p}(N) - v_{p}(k) ] Thus, we have derived the desired results in both parts. [ boxed{ v_{p}(N!) = sum_{i=1}^{infty} leftlfloor frac{N}{p^i} rightrfloor } ] [ boxed{ v_{p} left( binom{N}{k} right) = v_{p}(N) - v_{p}(k) } ]

answer:Let's first calculate the total number of legs before any chairs were damaged and disposed of. Kenzo has 80 chairs with 5 legs each, so that's a total of: 80 chairs * 5 legs/chair = 400 legs from chairs He also has 20 tables with 3 legs each, so that's a total of: 20 tables * 3 legs/table = 60 legs from tables The total number of legs before any damage is: 400 legs from chairs + 60 legs from tables = 460 legs Now, we know that after the damage, Kenzo has 300 legs remaining. This means that the number of legs that were disposed of is: 460 original legs - 300 remaining legs = 160 legs disposed of Since each chair has 5 legs, we can find out how many chairs were disposed of by dividing the number of disposed legs by the number of legs per chair: 160 disposed legs / 5 legs/chair = 32 chairs disposed of Now, to find the percentage of chairs that were damaged and disposed of, we divide the number of disposed chairs by the original number of chairs and multiply by 100 to get the percentage: (32 disposed chairs / 80 original chairs) * 100 = 40% So, boxed{40%} of the chairs were damaged and disposed of.

answer:Since ∠C = 90° in right triangle ABC, the lengths of the sides a, b, and c opposite angles A, B, and C must satisfy the triangle inequality. Thus, we have a+b > c, which implies that a+b=cx > c. Therefore, x > 1. Given that by the Pythagorean theorem, a^2+b^2 = c^2, we substitute c^2 for a^2 + b^2 in the equation a+b=cx to get (a+b)^2 = c^2x^2. Expanding (a+b) and substituting c^2 with a^2 + b^2, we have: [ x^2= frac{a^2 + b^2 + 2ab}{a^2 + b^2} leq frac{2(a^2 + b^2)}{a^2 + b^2} = 2. ] Thus, x leq sqrt{2}. Combining both conditions, we find that the range of x is x in (1, sqrt{2}]. Therefore, the final range of x is: [ boxed{x in (1, sqrt{2}]}. ]

answer:To find the sale in the fifth month, we first need to calculate the total sales for the six months to achieve an average sale of Rs. 6500. The formula for the average sale is: Average Sale = Total Sales / Number of Months We know the average sale is Rs. 6500 and the number of months is 6, so we can write: 6500 = Total Sales / 6 Now, we solve for Total Sales: Total Sales = 6500 * 6 Total Sales = Rs. 39000 We have the total sales for the first four months and the required sale for the sixth month. We add those up to find out how much is already accounted for: Total Sales (First 4 Months) = Rs. 6535 + Rs. 6927 + Rs. 6855 + Rs. 7230 Total Sales (First 4 Months) = Rs. 27547 We also know the sale for the sixth month: Sale (Sixth Month) = Rs. 4891 Now, we add the sales of the first four months and the sixth month: Total Sales (First 4 Months + Sixth Month) = Rs. 27547 + Rs. 4891 Total Sales (First 4 Months + Sixth Month) = Rs. 32438 To find the sale in the fifth month, we subtract the total sales of the first four months and the sixth month from the total sales required for six months: Sale (Fifth Month) = Total Sales - Total Sales (First 4 Months + Sixth Month) Sale (Fifth Month) = Rs. 39000 - Rs. 32438 Sale (Fifth Month) = Rs. 6562 Therefore, the sale in the fifth month was Rs. boxed{6562} .