unable to form the equiation ...help ...

[indiadineshkarthik]

six times the sum of the digits of a two digit number is 1 less than the number and 8 more than the number obtained by reversing the digits of the two digits number. what is the product of the digits? .....how to find the answer...?

 

[Yogi]

Create variables for the things you don't know and convert the sentences into equations. What have you tried so far? Please show your work.

 

[indiadineshkarthik]

Create variables for the things you don't know and convert the sentences into equations. What have you tried so far? Please show your work.

let the unknown no be 10X+Y
six times the sum of the digits of a two digit no in 1 less that the number
so 10x+y=6(X+Y)+1
and 8 more thatn the number obtained by reversing the digits of the two digit number
so 10y+x+8=10x+y
from 1 and 2
6(x+y)+1=10y+x+8 ...
by solving this equ i am not geting the write answer......

 

[Yogi]

six times the sum of the digits of a two digit number is 1 less than the number and 8 more than the number obtained by reversing the digits of the two digits number. what is the product of the digits? .....how to find the answer...?

let the unknown no be 10X+Y
six times the sum of the digits of a two digit no in 1 less that the number
so 10x+y=6(X+Y)+1 (so far so good)
and 8 more thatn the number obtained by reversing the digits of the two digit number
so 10y+x+8=10x+y (the mistake is here)
from 1 and 2
6(x+y)+1=10y+x+8 ...
by solving this equ i am not geting the write answer......

The way the problem reads I could rewrite it as:
six times the sum of the digits of a two digit number is 1 less than the number
six times the sum of the digits of a two digit number is 8 more than the number obtained by reversing the digits of the two digits number

Your first equation is correct. The 2nd one should be 6(X+Y)=10Y+X+8
After you define the number using X and Y you don't need lower-case x and y as they could be confused as different from X and Y.

 

[soroban]

Hello, indiadineshkarthik!

Please check for typos.
The second condition is impossible.

Six times the sum of the digits of a two-digit number is 1 less than the number
and 8 more than the number is obtained by reversing the digits of the two-digit number.
What is the product of the digits?

Let \(\displaystyle 10x+y\) = the two-digit number.


\(\displaystyle \begin{array}{cccccc}\text{Six} & \text{times} & \text{sum of digits}& \text{is} & \text{1 less than the number} \\ 6 & \times & (x+y) & = & (10x+y) - 1 & [1] \end{array}\)


\(\displaystyle \begin{array}{cccccc}\text{8 more than the number} & \text{is} & \text{reverse the digits} \\ (10x+y) + 8 & = & 10y + x & [2]\end{array}\)


Equation [2] is: .\(\displaystyle 10x + y + 8 \:=\:10y + x\)

. . . . . . . . . . . . . . . \(\displaystyle 9y - 9x \:=\:8 \)

i . . . . . . . . . . . . . .\(\displaystyle 9(y-x) \:=\:8\)


The left side is a multiple of 9 . . . and the right side isn't.

 

[Yogi]

Please refer to my post, the necessary hints to get to the solution are done already. Soro, read the question carefully.