BIGGEST DIVISOR SAME REMAINDER SOLUTION
... numbers has a COMMON FACTOR because one remainder is subtracted from the other. To solve the big problem we have to find the biggest common factor of the differences between the 5 numbers: 1705, 3069, 1364, 682 and 6820. To do this factorize the differences: 1705 = 5 × 341 = 5 x 11 x 31 ...
... numbers has a COMMON FACTOR because one remainder is subtracted from the other. To solve the big problem we have to find the biggest common factor of the differences between the 5 numbers: 1705, 3069, 1364, 682 and 6820. To do this factorize the differences: 1705 = 5 × 341 = 5 x 11 x 31 ...
1. 1/(1 − 1 ) = 2. Dick is 6 years older than Jane. Six years ago he
... of 9 up to 81 can be achieved as n − d(n). If n is a 3-digit number a2 a1 a0 , then (n − d(n))/9 = 11a2 + a1 . Since a1 can go from 0 to 9, (n − d(n))/9 cannot equal 11 − 1, 22 − 1, . . . , 99 − 1, but can equal all other numbers in this range. The largest (n − d(n))/9 can be for a 3-digit number is ...
... of 9 up to 81 can be achieved as n − d(n). If n is a 3-digit number a2 a1 a0 , then (n − d(n))/9 = 11a2 + a1 . Since a1 can go from 0 to 9, (n − d(n))/9 cannot equal 11 − 1, 22 − 1, . . . , 99 − 1, but can equal all other numbers in this range. The largest (n − d(n))/9 can be for a 3-digit number is ...
2015 Solutions
... We know the product of these two factors is 2000, i.e. k · b = 2000 where b = 2m + k − 1. Note that if k is even, then b is odd, and vice versa. We must now go through each possibility and find allowed possible value of m. It is easy to decompose 2000 into an odd and even factor as follows: (i) (k, ...
... We know the product of these two factors is 2000, i.e. k · b = 2000 where b = 2m + k − 1. Note that if k is even, then b is odd, and vice versa. We must now go through each possibility and find allowed possible value of m. It is easy to decompose 2000 into an odd and even factor as follows: (i) (k, ...
Weber problem
In geometry, the Weber problem, named after Alfred Weber, is one of the most famous problems in location theory. It requires finding a point in the plane that minimizes the sum of the transportation costs from this point to n destination points, where different destination points are associated with different costs per unit distance.The Weber problem generalizes the geometric median, which assumes transportation costs per unit distance are the same for all destination points, and the problem of computing the Fermat point, the geometric median of three points. For this reason it is sometimes called the Fermat–Weber problem, although the same name has also been used for the unweighted geometric median problem. The Weber problem is in turn generalized by the attraction–repulsion problem, which allows some of the costs to be negative, so that greater distance from some points is better.