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Find two-digit number the quotient of whose division by the product of its digits is equal to $\frac{8}{3}$ and the difference between the required number and the number consisting of the same digits written in the reverse order is 18.

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answer is C.

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Detailed Solution

Let n(a,b) be two digit number.
n(a,b) = 10a+b
Given,

\frac{n (a, b)}{ab} = \frac{8}{3}

Or,

\frac{10 a + b}{ab} = \frac{8}{3}

Given that n(a,b)−n(b,a)= 18
9(a−b)= 18
Or, a−b = 2
Or, a = b+2
30a+3(a−2) = 8a(a−2)
33a−6 = 8

a^{2}

−16a
8

a^{2}

−49a+6= 0
(8a−1)(a−6)= 0
Then, a = 6, a

\neq \frac{1}{8}

b = a−2 = 4
The original number = 10a + b
= 10(6) + 4
= 64
64 is the required number.

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