Problem:
For a positive integer n and nonzero digits a,b, and c, let An​ be the n-digit integer each of whose digits is equal to a; let Bn​ be the n-digit integer each of whose digits is equal to b; and let Cn​ be the 2n-digit (not n-digit) integer each of whose digits is equal to c. What is the greatest possible value of a+b+c for which there are at least two values of n such that Cn​−Bn​=An2​?
Answer Choices:
A. 12
B. 14
C. 16
D. 18
E. 20
Solution:
The equation Cn​−Bn​=An2​ is equivalent to
c⋅9102n−1​−b⋅910n−1​=a2(910n−1​)2.
Dividing by 10n−1 and clearing fractions yields
(9c−a2)⋅10n=9b−9c−a2
As this must hold for two different values n1​ and n2​, there are two such equations, and subtracting them gives
(9c−a2)(10n1​−10n2​)=0
The second factor is non-zero, so 9c−a2=0 and thus 9b−9c−a2=0. From this it follows that c=(3a​)2 and b=2c. Hence digit a must be 3, 6, or 9, with corresponding values 1, 4, or 9 for c, and 2,8, or 18 for b. The case b=18 is invalid, so there are just two triples of possible values for a,b, and c, namely (3,2,1) and (6,8,4). In fact, in these cases, Cn​−Bn​=An2​ for all positive integers n; for example, 4444−88=4356=662. The second triple has the greater coordinate sum, 6+8+4=(D)18​.
The problems on this page are the property of the MAA's American Mathematics Competitions