dartboard arrangements
... must distinguish between even and odd values of n. For n = 2k, let (s1 , s2 , . . . , sk ) be any permutation of {1, 2, . . . , k}, which we refer to as small numbers, and (l1 , l2 , . . . , lk ) be any permutation of {k + 1, k + 2, . . . , 2k}, which we refer to as large numbers. The string s1 l1 s ...
... must distinguish between even and odd values of n. For n = 2k, let (s1 , s2 , . . . , sk ) be any permutation of {1, 2, . . . , k}, which we refer to as small numbers, and (l1 , l2 , . . . , lk ) be any permutation of {k + 1, k + 2, . . . , 2k}, which we refer to as large numbers. The string s1 l1 s ...
File
... Note : We can add or subtract two terms of a polynomial if and only if these two terms are similar , that is, when they are composed of the same variables respectively raised to the same exponents. Regardless of the value of their numerical coefficient. An irreducible ordered polynomial is a polynom ...
... Note : We can add or subtract two terms of a polynomial if and only if these two terms are similar , that is, when they are composed of the same variables respectively raised to the same exponents. Regardless of the value of their numerical coefficient. An irreducible ordered polynomial is a polynom ...
Leap Frog Solutions 2013
... Solution. We note that (n + 1)2 and (n + 2)2 are within 2013 of n2 exactly when (n + 2)2 ≤ n2 + 2013, or equivalently, n2 + 4n + 4 ≤ n2 + 2013. Subtracting n2 + 4 from both sides, the solution is the largest integer ...
... Solution. We note that (n + 1)2 and (n + 2)2 are within 2013 of n2 exactly when (n + 2)2 ≤ n2 + 2013, or equivalently, n2 + 4n + 4 ≤ n2 + 2013. Subtracting n2 + 4 from both sides, the solution is the largest integer ...
