Question

a two digit number is 3 more than 4 times the sum of its digits . if 18 is added the number its digits are reversed. find the number

Answer 1

Hello,

Let's assume a and the digit of the unknowed number

$\overline{ab}=3+4(a+b)\\ \Rightarrow\ 10a+b=3+4a+4b\\ \Rightarrow\ 6a-3b=3\\ \Rightarrow\ 2a-b=1\\ 18+\overline{ab}=\overline{ba}\\ \Rightarrow\ 18+10a+b=10b+a\\ \Rightarrow\ 9a-9b=-18\\ \Rightarrow\ -a+b=2\\ \left \{ {{2a-b=1} \atop {-a+b=2}} \right.\\\\ \left \{ {{a=3} \atop {b=5}} \right.$

Your number is 35.

Answer 2

solutions :-

let the digit at once digit be x

and tens place be y

then

10y+x

when digit reversed

it becomes 

10x+y

case 1


10y+x=4(x+y)+3

10y+x=4x+4y+3

10y+x-4x-4y=3

6y-3x=3....... ( both side divided by 3)

2y-x=1________(1)


case 2


(10y+x)+18=10x+y

10y+x-10x-y=18

9x-9y=18......... ( both side divided by 9)

x-y=2________(2)



x=2+y

put the value of x in equation (1)


2y-X=1

2y-(2+y)=1

2y-2-y=1

y=3

put value of y in equation 2


x-y=2

x-3=2

x=5

then the number is


10y+x= 10*3+5

=35

Thanks