SOLUTION: How many 2 digit numbers exist such that the product of the digits is greater than the actual number?

There are no such two-digit numbers

Suppose that:

      tu > 10t + u

  tu - u > 10t

u(t - 1) > 10t

The tens digit is never 0.

Case 1:
When the tens digit is 1, the number is 
10×1 + u or 10+u

and the product of digits is
1×u or u, 

and u is always less than 10 + u

So there are no two-digit numbers
beginning with 1 whose product of
digits is greater than the number.

Case 2:
When the tens digit is 2 or more then
t-1 is positive and we can divide the
inequality

u(t - 1) > 10t

by t-1 and this inequality will hold:


            10t
       u > ————— 
            t-1

Subtract then add 10 to the numerator:

            10t-10+10
       u > ——————————— 
               t-1

Factor 10 out of the first two terms on top:

            10(t-1)+10
       u > ——————————— 
               t-1
       
Break into two fractions:

            10(t-1)    10
        u > ——————— + ————— 
              t-1      t-1

Cancel the t-1's in the first fraction:

                  10
        u > 10 + ————— 
                  t-1

That is clearly false since the right side is greater than 10 and
no units digit can be greater than 9.

This proves that no 2 digit numbers exist such that the product of 
the digits is greater than the actual number. 

Edwin