Question

ththe sum of digits of two digit number is three these digits are reversed the number is nine less than the number what is the number

Answer 1

let the digits of the number be a(in the ten's place) and b(in the units's place).

so the number will be 10a+b.

=>reversed number= 10b+a.

given:- when the digits are reversed,the NEW number is 9 LESS THAN the ORIGINAL number.....that is..

(10a+b)-9=10b+a

(10a+b)-(10b+a)=9          (transposing)

10a+b-10b-a=9

9a-9b=9

9(a-b)=9

[a-b=1]       -equation 1

given:-sum of the digits = 3

that is....[a+b=3]           -equation 2

(adding equation 1 and equation 2)

a+b+a-b=4

2a=4

a=2

(substituting value of a in equation 2)

2+b=3

b=1

so a=2 and b=1

hence the number is 10(2)+1=20+1=21

Answer=>21