Question

the sum of the digits of a two-digit number is 13.If 27 is added to the number, the digits get interchange . What is the number​

Answer 1

Answer:

ll say that the two digit number is AB where A is the tens digit and B is the ones digit (I'm not multiplying A and B).

AB can be rewritten as 10A + B.

The question tells me when I add the digits, I get 13.

A + B = 13

It also tells me that if I add 27 to the original number it reverses the digits:

AB + 27 = BA

Now by AB we actually mean 10A + B and by BA we actually mean 10B + A, so we rewrite this equation to be:

10A + B + 27 = 10B + A

Rewrite each side:

9A + (A + B) + 27 = 9B + (B + A)     (one way to do this is to replace any A + B s I see with 13 because of the first equation I have)

9A + 27 = 9B

A + 3 = B             (divide everything in the equation by 9)

Now I can substitute for B in that first equation:

A + B = 13

A + A + 3 = 13

A = 5

Then B has to add to this to give me 13, so B = 8.

My number is 58.

(Check: 5 and 8 add to 13, check. 58 + 27 = 85, check.)

Step-by-step explanation:

Answer 2

Answer:

58

Step-by-step explanation:

the two digit number is AB where A is the tens digit and B is the ones digit (I'm not multiplying A and B).

AB can be rewritten as 10A + B.

The question tells me when I add the digits, I get 13.

A + B = 13

It also tells me that if I add 27 to the original number it reverses the digits:

AB + 27 = BA

Now by AB we actually mean 10A + B and by BA we actually mean 10B + A, so we rewrite this equation to be:

10A + B + 27 = 10B + A

Rewrite each side:

9A + (A + B) + 27 = 9B + (B + A) (one way to do this is to replace any A + B s I see with 13 because of the first equation I have)

9A + 27 = 9B

A + 3 = B (divide everything in the equation by 9)

Now I can substitute for B in that first equation:

A + B = 13

A + A + 3 = 13

A = 5

Then B has to add to this to give me 13, so B = 8.

My number is 58.