SOLUTION: Find all 2-digit positive integers in which the difference between the integer and the product of its two digits is 12.

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Question 1090826: Find all 2-digit positive integers in which the difference between the integer and the product of its two digits is 12.
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!

If the tens digit is t and the units digit is u, then the integer itself is 10t+u, and the product of the digits is tu. If the difference between the integer itself and the product of its digits is 12, then we need to solve the equation
%2810t%2Bu%29-tu+=+12

So we have a single equation with two unknowns, both of which have to be single digit integers.

This is a Diophantine equation. A general procedure for solving this kind of equation is to solve it for one variable in terms of the other, and then use the requirement that the solutions have to be integers (in this example, single digit positive integers) to find the solutions.
%2810t%2Bu%29-tu+=+12
10t%2Bu-tu+=+12
%2810t-tu%29%2Bu+=+12
t%2810-u%29+=+12-u
t+=+%2812-u%29%2F%2810-u%29

We could use this equation for t to try to find values of u that give a valid value for t. But it is easier to do this if we "perform the division" on the right to get an integer plus a remainder:


t+=+1+%2B+2%2F%2810-u%29

Now we can see that the fraction on the right has to have an integer value; that means (10-u) has to be a factor of 2. So 10-u is either 1 or 2, which means u is either 9 or 8.

u = 9 --> t = 1+2/1 = 1+2 = 3 --> the integer is 39
u = 8 --> t = 1+2/2 = 1+1 = 2 --> the integer is 28

An easy check shows that these answers do indeed satisfy the requirements of the problem:
28-2*8 = 28-16 = 12
39-3*9 = 39-27 = 12