14
n^3+8=(n+2)*(n^2-2n+4) -----> (1) n^2-4=(n+2)*(n-2) -----> (2) (1)/(2)= (n^2-2n+4)/(n-2)= n+ 4/(n-2) -----> (3) For (3) to be an integer 4/(n-2) must be integer, then n-2 = {-4,-2,-1,1,2,4} => n={-2,0,1,3,4,6} n=-2 is excluded as n^2-4 can't be zero due to divisibility rule Sum of all n will be 0+1+3+4+6=14
(n^3 + 2^3) = (n + 2)×(n^2 - 2n + 4) (n^2 - 2^2) = (n + 2)(n - 2) (n^3 + 2^3)/(n^2 - 2^2) = (n^2 - 2n + 4)/(n - 2) = n + (4/(n - 2)) This means for n = -2, 0, 1, 3, 4 & 6 we have (n^3 + 2^3)/(n^2 - 2^2) = -3, -2, -3, 7, 6 & 7 as for n > 6 & n < (-2), 4/(n - 2) will always yield a fraction making (n^2 - 2n + 4)/(n - 2) = n + (4/(n - 2)) always a non integer beyond n > 6 & n < (-2). Therefore the required sum = -2 + 0 + 1 + 3 + 4 + 6 = 12.
Consider a list of numbers where the integer n appears for n times, 1 ≤ n ≤ 200.
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, …, 200, …, 200
What is the median of the numbers in the list?
For some positive integer k both 4^k and 5^k start with the same digit x in base 10 i.e. 4^k = x..... 5^k = x..... What is the sum of all possible values of x.
The digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 can be arranged into an addition sum to add up to almost any total, except that nobody has yet found a way to add up to 1984. However 9 digits can equal 1984 by an addition sum. Which digit is omitted?