Problem:
Call a positive integer N a 7β10 double if the digits of the base-7 representation of N form a base-10 number that is twice N. For example, 51 is a 7β10 double because its base-7 representation is 102. What is the largest 7β10 double?
Solution:
Suppose that akβ7k+akβ1β7kβ1+β―+a2β72+a1β7+a0β is a 7β10 double, with akβξ =0. In other words, akβ10k+akβ1β10kβ1+β―+a2β102+a1β7+a0β is twice as large, so that
akβ(10kβ2β
7k)+akβ1β(10kβ1β2β
7kβ1)+β―+a2β(102β2β
72)+a1β(10β2β
7)+a0β(1β2)=0.
Since the coefficient of aiβ in this equation is negative only when i=0 and i=1, and no aiβ is negative, it follows that k is at least 2. Because the coefficient of aiβ is at least 314 when i>2, and because no aiβ exceeds 6 it follows that k=2 and 2a2β=4a1β+a0β. To obtain the largest possible 7β10 double, first try a2β=6. Then the equation 12=4a1β+a0β has a1β=3 and a0β=0 as the solution with the greatest possible value of a1β. The largest 7β10 double is therefore 6β
49+3β
7=315β.
The problems on this page are the property of the MAA's American Mathematics Competitions