What does CRT stand for?
Chinese Remainder Theorem
State Fermat's Little Theorem for a prime p and integer a.
If p is prime and a is not divisible by p, then
ap-1 ≡ 1 (mod p)
If x≡1 (mod 2) and x≡2 (mod 3), is there a solution?
Yes (since 2 and 3 are coprime).
Compute 536 (mod 37) using FLT.
1 (since 37 is prime and 5 is not divisible by 37).
What is the smallest positive solution to x≡1 (mod 4) and x≡3 (mod 5)?
13
If n1 and n2 are not coprime, does CRT guarantee a solution to x ≡ a1 (mod n1) and x ≡ a2 (mod n2)?
No, only if a1 ≡ a2 (mod gcd(n1, n2)).
What is 210 (mod 11) using FLT?
1
Solve x ≡ 4 (mod 5) and x ≡ 3 (mod 7).
x ≡ 24 (mod 35).
Compute 31262 (mod 127)
9
Find the smallest x > 0 such that x ≡ 2 (mod 3), x ≡ 1 (mod 4), and x ≡ 3 (mod 5).
53
Explain why CRT requires pairwise coprime moduli.
Ensures a unique solution exists (otherwise, contradictions may arise).
If p is prime, what is aP (mod p) for any integer a?
a (mod p)
A number leaves remainder 1 when divided by 3, 2 when divided by 5, and 0 when divided by 7. Find the smallest such number.
7
Compute 66006 + 44004 (mod 17)
Find x if x ≡ 1 (mod 2), x ≡ 2 (mod 3), and x ≡ 4 (mod 5)
x ≡ 29 (mod 30)
What is the least three-digit positive whole number that is 4 (mod 7), 5 (mod 8), 6 (mod 9)?
501
What proof technique is most commonly used to prove Fermat's Little Theorem?
A number is divisible by 7, leaves remainder 3 when divided by 11, and remainder 5 when divided by 13. Find the smallest such number.
707
How many primes p are there such that 101p+1 is divisible by p?
3 (prime factors of 102)
Solve 2x ≡ 3 (mod 11)
x ≡ 8 (mod 10)
Find the smallest x such that x ≡ 1 (mod 2), x ≡ 2 (mod 3), x ≡ 3 (mod 5), x ≡ 4 (mod 7), and x ≡ 5 (mod 11).
1523
Compute 5201 (mod 101) using FLT.
5
A number is 2 (mod 3), 3 (mod 5), 4 (mod 7), and 5 (mod 11). Find the smallest such number > 1000.
1859
An integer N is selected at random in the range 0 < N < 2021. What is the probability that the remainder when N16 is divided by 5 is 1?
4/5. N16 = (N4)4, so by FLT, N16 ≡ 1 (mod 5) unless N is divisible by 5. A fifth of all possible N are divisible by 5, so the probability is 4/5.
Solve x ≡ 1 (mod 2), x2 ≡ 4 (mod 3), and x3 ≡ 3 (mod 5)
x ≡ 7 (mod 30)