In a 2-digit number, the tens digit is 5 less than the units digit. If you reverse the number, the result is 7 greater than double the original number. Find the original number.

The original number is 38

A 2-digit number can be written as:

N = a*10 + b*1

Where a is the tens digit, and b is the units digit, these two are single-digit numbers.

We know that:

"the tens digit is 5 less than the units digit."

This means that:

a = b - 5

(notice that a must be larger than zero and smaller than 10, from this, we can conclude that b is a number in the range {6, 7, 8, 9})

"If you reverse the number, the result is 7 greater than double the original number"

The reverse number is:

b*10 + a

and this is 7 greater than 2 times the original number, then:

b*10 + a = 7 + 2*(a*10 + b)

Then we found two equations:

a = b - 5

b*10 + a = 7 + 2*(a*10 + b)

Replacing the first equation in the second, we get:

b*10 + (b - 5) = 7 + 2*((b - 5)*10 + b)

Now let's solve that:

b*10 + b - 5 = 7 + 2*(11*b - 50)

11*b - 5 = 7 + 22*b - 100

-5 - 7 + 100 = 22*b - 11*b

88 = 11*b

88/11 = b = 8

Now that we know that b = 8, we can use the equation:

a= b - 5

a = 8 -5 = 3

Then the original number is:

a*10 +  b = 3*10 + 8 = 38

The original number is 38

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