the sum of the digits of a two -digit number is 9.if 9 is added to the number formed by reversing the digits then the result is thrice the original number. find the original number
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let the tens place digit no. be x
and ones place digit no. be y
So, two digit no.=10x+y
By reversing order,
two digit number=10y+x
According to question
x+y=9
x=9-y. 1
10y+x+9=3(10x+y)
10y +9-y+9=3[10(9-y)+y]
9y+18=3[90-10y+y]
9y+18=270-27y
9y+27y=270-18
36y=252
y=7
from 1
x=2
and ones place digit no. be y
So, two digit no.=10x+y
By reversing order,
two digit number=10y+x
According to question
x+y=9
x=9-y. 1
10y+x+9=3(10x+y)
10y +9-y+9=3[10(9-y)+y]
9y+18=3[90-10y+y]
9y+18=270-27y
9y+27y=270-18
36y=252
y=7
from 1
x=2
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