Seven times a two number is equal to four times the number obtained by reversing the order of its digits. If the difference of the digits is $3$ , determine the number.

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Solution

Let;-
$y =$ once place
$x =$ thence place
digit $= 10 x + y$
$\begin{matrix} 7 (10 x + y) = 4 (10 y + x) 70 x + 7 y = 40 y + 4 x 70 x - 4 x = 40 y - 7 y 66 x = 33 y 2 x = y . . . . . . . . . . . . . . e q u a t i o n 1 x - y = 3 x - 2 x = 3 - x = 3 x = - 3 \end{matrix}$
putting the value of $x$ in equation $1$
we have,
$\begin{matrix} x - y = 3 - 3 - y = 3 y = - 6 \end{matrix}$
hence
there are $36$ digits

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Seven times a two number is equal to four times the number obtained by reversing the order of its digits. If the difference of the digits is 3, determine the number.

Let;-y=once placex=thence placedigit=10x+y 7(10x+y)=4(10y+x)70x+7y=40y+4x70x−4x=40y−7y66x=33y2x=y..............equation1x−y=3x−2x=3−x=3x=−3putting the value of x in equation1we have,x−y=3−3−y=3y=−6hence there are 36 digits

Seven times a two number is equal to four times the number obtained by reversing the order of its digits. If the difference of the digits is $3$ , determine the number.