(2003 / 30) (2003 / 33) (2003 / 36) (2003 / 44) (2003 / 45) (2003 / 55) (2003 / 60) (2003 / 66) (2003 / 90) (2003 / 99) (2003 / 110) (2003 / 132) (2003 / 165) (2003 / 180) (2003 / 198) (2003 / 220) (2003 / 330) (2003 / 396) (2003 / 495) (2003 / 660) (2003 / 990) (2003 / 1980) So I could find 22 of such numbers giving a remainder of 23 with the number 2003.
A positive integer N leaves the same remainder of 35 when divided by both 2009 and 2010.
What is the remainder when N is divided by 42?
If 1^3 + 2^3 + 3^3 = m^2 where m is also an integer. What are the next three consecutive positive integers such that the sum of their individual cubes is equal to a perfect square?
