Question

A two digits number is seven times the sum of its digits . The number formed by reversing the digits is 18 less than the original number. Find the number .​

Answer 1

Hey there!

10x+y=7(x+y)=7x+7y

or 10x−7x=7y−y

or 3x=6y

or x=2y ........(1)

Number formed by reversing the digits is yx, i.e.,

10y+x=10x+y−18

Substituting x=2y, we get

10y+2y=10×2y+y18

or 2y=21y−18

or 21y−12y=18

or 9y=18

or y=2

∴x=2y=2×2=4

Therefore, the number is 42.

Answer 2

`▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪`

Let the two digit number be ' xy '

where, x is the ten's place digit, and y is one's place digit, now let's proceed further ~

The origin number can be written as :

`\qquad \sf  \dashrightarrow \: 10x + y`

[ because x is at ten's place ]

of the digits are swapped, the new number will be :

`\qquad \sf  \dashrightarrow \: 10y + x`

According to conditions provided in the question : -

`\qquad \sf  \dashrightarrow \: 10x + y = 7 * (x + y)`

`\qquad \sf  \dashrightarrow \: 10x + y = 7x + 7y`

`\qquad \sf  \dashrightarrow \: 10x - 7x + y - 7y= 0`

`\qquad \sf  \dashrightarrow \: 3x - 6y = 0 \: \: \: \: \: \:`

`\qquad \sf  \dashrightarrow \: x - 2y = 0 \: \: \: \: \: \: - (1)`

here we got our first equation, let's find the other one ~

`\qquad \sf  \dashrightarrow \: (10x + y) - 18= 10y + x`

`\qquad \sf  \dashrightarrow \: 10x - x + y - 10y = 18`

`\qquad \sf  \dashrightarrow \: 9x - 9y = 18`

`\qquad \sf  \dashrightarrow \: x - y = 2 \: \: \: \: \: \: - (2)`

we got our second equation ~

Now it's time to subtract equation (1) from equation (2) :

`\qquad \sf  \dashrightarrow \:( x - y ) - (x - 2y)= 2 - 0`

`\qquad \sf  \dashrightarrow \: \cancel{x }- y - \cancel{x} + 2y= 2 - 0`

`\qquad \sf  \dashrightarrow \: y= 2`

hence, value of y is 2, use this value in equation (2) to find x ~

`\qquad \sf  \dashrightarrow \: x - y= 2`

`\qquad \sf  \dashrightarrow \: x - 2= 2`

`\qquad \sf  \dashrightarrow \: x = 4`

Therefore, the required number is xy = 42