Problem:
Find the sum of all positive integers b<1000 such that the base- b integer 36b is a perfect square and the base- b integer 27b is a perfect cube.
Solution:
The integer b must satisfy the equations 3b+6=j2 and 2b+7=k3 for positive integers j and k. Because 3b+6 is a perfect square and is divisible by 3 , it must be divisible by 9 , implying that b≡1(mod3). The perfect cubes modulo 9 are 0,1 , and 8 , so 2b+7≡0,1, or 8(mod9), from which b≡1, 6 , or 5(mod9). Taken together, the conditions imply that b≡1(mod9). Therefore k3=2b+7 is an odd multiple of 9 , so k is an odd multiple of 3 . Only if k=3 or 9 can b have a value less than 1000 . If k=3, then b=10, and in this case 3b+6=36, which is a perfect square. If k=9, then b=361, and 3b+6=1089, which is also a perfect square. Thus the requested sum is 10+361=371.
OR
If b<1000, then 27b<2007. The number 27b=2b+7 must be odd, and the odd cubes up to 2007 are 13=1,33=27,53=125,73=343,93=729, and 113=1331. The corresponding values of b are −3,10,59,168,361, and 662, making 3b+6 equal to −3,36,183,510,1089, and 1992 , respectively. Only 36 (when b=10 ) and 1089 (when b=361 ) are perfect squares. The requested sum is 10+361=371.
The problems on this page are the property of the MAA's American Mathematics Competitions