Question
Product of two consecutive positive even numbers is
120. Find the sum of the digits of the two numbers.Solution
Let the numbers be ( 2n ) and ( (2n + 2) ) respectively. ATQ, (2n * (2n + 2) = 120 ) Or, ( n * (2n + 2) = 60 ) Or, ( 2n² + 2n = 60 ) Or, ( n² + n = 30 ) Or, ( n² + n - 30 = 0 ) Or, ( n² + 6n - 5n - 30 = 0 ) Or, ( n(n + 6) - 5(n + 6) = 0 ) Or, ( (n + 6)(n - 5) = 0 ) So, ( n = -6 ) or ( 5 ) So, (n = 5) So, numbers are ( 10 ) and ( 12 ). Therefore, required sum = ( 1 + 0 + 1 + 2 = 4 )
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