sum of digits in a two-digit equal 7, if the digits are switched the value of the two-digit number will increase by 9, what is the original two-digit number.

Question

The sum of the digits of a two-digit number equals 7, if the digits in the two-digit number are switched the value of the two-digit number will increase by 9...what is the original two-digit number...please show steps.

Answer 1

Original number is  10a + b    ten's digit is a units digit is b

 

facts:     a  +  b  =  7

           10a + b + 9 = 10b + a

 

a  +  b  =  7

9a - 9b + 9 = 0

 

9a + 9b = 63

9a - 9b = -9

 

adding    gives   18a = 54    thus  a =   3

substitution gives b =3

 

Thus original 2-digit number was 34.

Answer 2

using algebra

 

xy is the number

 

x+y=7

 

yx is the number with the digits switched

 

yx means 10y+x

 

10y+x=10x+y+9  (xy means 10x+y)

 

x+y=7------->y=7-x

 

substitute

 

10(7-x)+x=10x+7-x+9

 

70-10x+x=9x+16

 

70-9x=9x+16

 

70-16=9x+9x

 

54=18x

 

x=54/18

 

x=3

 

3+y=7

 

y=7-3

 

y=4

 

34 is the number because 43=34+9

 

 

using arithmetic

 

16, 25, 34, 43, 52, 61 are the possibilities (I left out 70 because 07 is not a two-digit number; you don't write 7 as 07, although it is not mathematically wrong, just not customary)

 

it's easy from here...

 

34 to 43 is an increase of 9