Question
Product of two consecutive positive even numbers is 440.
Find the sum of the digits of the two numbers.ÂSolution
Let the numbers be ( 2n ) and ( (2n + 2) ) respectively. ATQ, (2n * (2n + 2) = 440 ) Or, (n * (2n + 2) = 220 ) Or, ( 2n² + 2n = 220 ) Or, ( n² + n = 110 ) Or, ( n² + n - 110 = 0 ) Or, ( n² + 11n - 10n - 110 = 0 ) Or, ( n(n + 11) - 10(n + 11) = 0 ) Or, ( (n + 11)(n - 10) = 0 ) So, ( n = -11 ) or ( 10 ) So, ( n = 10 ) So, numbers are ( 20 ) and ( 22 ). Therefore, required sum = ( 2 + 0 + 2 + 2 = 6 )Â
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