Problem:
How many nonnegative integers can be written in the form
where for ?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
Let be the set of integers, both negative and nonnegative, having the given form. Increasing the value of by for creates a one-to-one correspondence between and the ternary (base ) representation of the integers from through , so contains elements. One of those is , and by symmetry, half of the others are positive, so contains elements.
First note that if an integer can be written in this form, then can also be written in this form as long as not all the in the representation of are equal to . A procedure to alter the representation of so that it will represent instead is to find the least value of such that , reduce the value of that by , and set for all lower values of . By the formula for the sum of a finite geometric series, the greatest integer that can be written in the given form is
Therefore, nonnegative integers can be written in this form, namely all the integers from through , inclusive. (The negative integers from through can also be written in this way.)
Think of the indicated sum as an expansion in base using "digits" , and . Note that the leftmost digit of any positive integer that can be written in this form cannot be negative and therefore must be . Then there are choices for each of the remaining digits to the right of , resulting in positive integers that can be written in the indicated form. Thus there are
positive numbers of the indicated form. Because can also be written in this form, the number of nonnegative integers that can be written in the indicated form is .
The problems on this page are the property of the MAA's American Mathematics Competitions