proof that number of sum-product numbers in any base is finite
Let be the base of numeration.
Suppose that an integer has digits when expressed in base (not counting leading zeros, of course). Then .
Since each digit is at most , we have that the sum of the digits is at most and the product is at most , hence the sum of the digits of times the product of the digits of is at most .
If is a sum-product number, then equals the sum of its digits times the product of its digits. In light of the inequalities of the last two paragraphs, this implies that , so . Dividing both sides, we obtain . By the growth of exponential function, there can only be a finite number of values of for which this is true. Hence, there is a finite limit to the number of digits of , so there can only be a finite number of sum-product numbers to any given base .
|Title||proof that number of sum-product numbers in any base is finite|
|Date of creation||2013-03-22 15:47:06|
|Last modified on||2013-03-22 15:47:06|
|Last modified by||rspuzio (6075)|