Solutions
... Suppose the middle number is 2k, where k > 3, so the numbers are 2k – 1, 2k, 2k + 1. As one of the three numbers must be divisible by 3, either 2k – 1 or 2k + 1 would have to be a power of 3. We consider these two cases separately. It is now helpful to use modular arithmetic (the arithmetic of remai ...
... Suppose the middle number is 2k, where k > 3, so the numbers are 2k – 1, 2k, 2k + 1. As one of the three numbers must be divisible by 3, either 2k – 1 or 2k + 1 would have to be a power of 3. We consider these two cases separately. It is now helpful to use modular arithmetic (the arithmetic of remai ...
n = n//*,
... Ore studied numbers whose harmonic means are integers, and showed that perfect numbers have this property [9]. W. H. Mills demonstrated that any odd number with an integral harmonic mean must have a prime power factor greater than IO7. This bound in Mills' (unpublished) calculation arose from the li ...
... Ore studied numbers whose harmonic means are integers, and showed that perfect numbers have this property [9]. W. H. Mills demonstrated that any odd number with an integral harmonic mean must have a prime power factor greater than IO7. This bound in Mills' (unpublished) calculation arose from the li ...
Rational Numbers, Divisibility and the Quotient Remainder Theorem
... • Given integer n and positive integer d, there exist unique integers q and r such that n = dq + r, 0 ≤ r < n • q is called the quotient and r the remainder • q = n div d (n\d) ← Integer Division! • r = n mod d (n%d) • n%d = n – d(n\d) ...
... • Given integer n and positive integer d, there exist unique integers q and r such that n = dq + r, 0 ≤ r < n • q is called the quotient and r the remainder • q = n div d (n\d) ← Integer Division! • r = n mod d (n%d) • n%d = n – d(n\d) ...
