Question
A = 1200 – X where X is a real number. A > 0 and A is
divisible by 3, 5, 7 and 11. The sum of the two largest possible value of X is:Solution
Two largest Possible value of X = 45 and (-1110) [ Both are real numbers ] For X = 45, A = 1155 [Divisible by 3,5,7,11] For X = (-1110), A = 2310 [ Divisible by 3,5,7,11 ] Both of the above values of X satisfy the given conditions. Therefore, the sum of two largest possible values of X for which A is divisible by the given numbers (A > 0) will be 45 + (-1110) = (-1065) ∴ The largest possible value of X is -1065 for which the given condition is satisfied.
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