Problem:
Call a positive integer N a 7−10 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 7−10 double because its base-7 representation is 102. What is the largest 7−10 double?
Solution:
Suppose that ak​7k+ak−1​7k−1+⋯+a2​72+a1​7+a0​ is a 7−10 double, with akâ€‹î€ =0. In other words, ak​10k+ak−1​10k−1+⋯+a2​102+a1​7+a0​ is twice as large, so that
ak​(10k−2⋅7k)+ak−1​(10k−1−2⋅7k−1)+⋯+a2​(102−2⋅72)+a1​(10−2⋅7)+a0​(1−2)=0.
Since the coefficient of ai​ in this equation is negative only when i=0 and i=1, and no ai​ is negative, it follows that k is at least 2. Because the coefficient of ai​ is at least 314 when i>2, and because no ai​ exceeds 6 it follows that k=2 and 2a2​=4a1​+a0​. To obtain the largest possible 7−10 double, first try a2​=6. Then the equation 12=4a1​+a0​ has a1​=3 and a0​=0 as the solution with the greatest possible value of a1​. The largest 7−10 double is therefore 6⋅49+3⋅7=315​.
The problems on this page are the property of the MAA's American Mathematics Competitions