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There is a 2 digit number such that the sum of its digits is 6 while the product of the digits is one-third the original number Find the number?
Wiki User
∙ 14y ago
if the digits are A and B, then we are given:
A + B = 6
A * B = (10A + B) / 3
∴ B = 6 - A
∴ A * (6 - A) = (10A + [6 - A]) / 3
∴ 6A - A2 = (9A + 6) / 3
∴ 6A - A2 = 3A + 2
∴ 6A - A2 - 3A - 2 = 0
∴ A2 - 3A + 2 = 0
∴(A - 2)(A - 1) = 0
∴A ∈ {1, 2}
A + B = 6
∴ B ∈ {5, 4}
∴ The numbers 15 or 24 both meet these conditions.
Wiki User
∙ 14y ago
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