1. Consider the seven congruence classes of the naturals, modulo
7. Then exactly six of these classes contain a prime number.
2. Let a, b, and c be any three naturals. If ab and ac are
congruent modulo 26, then we may conclude that b and c are also
congruent modulo 26.
3. Let n be any positive integer. Then the inverse of n,
modulo n+1, is itself, and the inverse of n+1, modulo n, is
-(n-1).
4. Let x, y, and z be positive integers such that x and y are
relatively prime to one another, and x and z are relatively prime
to one another. Then x is also relatively prime to the
product yz.
5. Given any five consecutive naturals, at least one must have an
inverse modulo 70.
6. Let a and b be two naturals that are relatively prime. Then the
naturals 5a + 3b and 2a + b are also relatively prime.
7. If n is a prime number, it is possible that n! + 1 is not
prime.
8. Duncan's food comes in cans that each contain six servings, and
he gets four servings of food each day. If today we opened a
new can for his first serving, and we continue to feed him on
schedule, we will never open a new can for his fourth serving on
any future day.
9. Since the number 250 satisfies the congruences x = 16 (mod 39)
and x = 55 (mod 65), the next largest integer to satisfy these two
congruences is 250 + 39(65) = 2785.
10. Any system of congruences of the form "x ≡ a (mod 7) and x ≡ b
(mod 11) and x ≡ c (mod 13)" has a solution in the range {2000,...,
2999}.