68 71!x

Could you clarify what you want to do with "68 71!x"? There are a few common interpretations, and I can solve it once I know which one you mean. Here are the most likely: 1) If you mean 68 divides 71! × x (written as 68 | 71! x) for integer x: - Since 71! = 1·2·...·71 contains the factor 68 (because 68 ≤ 71), we can write 71! = 68 · k for some integer k. - Then 71! x = 68 · (k x), which is a multiple of 68. - Therefore 68 | 71! x for any integer x (provided x ∈ Z). 2) If you mean the product 68 · 71! · x: - This is just (68 · 71!) · x. If you want to simplify or evaluate, you’d leave it as is or note that the factor is 68 · 71! times x. 3) If you mean something about gcd, remainder, or a different relation: - For example, gcd(68, 71! x) = gcd(68, x) because 71! is already a multiple of 68. If you paste the exact problem (e.g., with a bar for divisibility: 68 | 71! x, or with an operator between 68 and 71!x), I’ll tailor the solution precisely.