SOLUTION: there is a two digit no. Such that sum of its digit is 6 while the product of digits is 1/3 the original number. find this number
Algebra.Com
Question 908607: there is a two digit no. Such that sum of its digit is 6 while the product of digits is 1/3 the original number. find this number
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
d = 6-u
(6-u)u = (1/3)[10(6-u) + u] = (1/3)(60 -9u)
6u - u^2 = 20-3u
u^2 -9u + 20 = 0
(u-5)(u-4) = 0, u= 4 , 24 the number
u = 5, 15 the number
And...
8 = 24/3 and 5 = 15/3
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