Download Solutions to Practice Midterm 1

Solutions to Practice Midterm 1
1. True or False:
a) If (x, y) = d then (ax + by, y) = d for all integers a, b > 0. FALSE: for example,
(2, 3) = 1 but (3 · 2 + 1 · 3, 3) = (9, 3) = 3.
b) The linear congruence 124x ≡ 10 (mod 1040) has a solution. FALSE: (124, 1040) = 4
which is not a divisor of 10.
c) If p1 , p2 , p3 are consecutive primes then 2p1 p2 p3 + 1 is a prime number. FALSE:
consider the consecutive primes 5, 7, 11: we have 2 · 5 · 7 · 11 + 1 is divisible by 3.
d) If a ≡ b (mod m) and c|a, c|b, then a/c ≡ b/c (mod m). FALSE: 8 ≡ 2 (mod 6) but
4 6≡ 1 (mod 6) (here a = 8, b = 2, m = 6, c = 2).
e) Any system
x≡a
mod m
x ≡ b mod n
4x ≡ c mod k
of linear congruences has a unique solution modulo mnk. FALSE: the last linear congruence may not even have a solution.
f) If 3|16a then 3|a. TRUE: 3 is prime.
g) 2x3 + x2 + 2 is not divisible by 3 for any integer x. TRUE: if x ≡ 0 (mod 3) then
2x3 + x2 + 2 ≡ 2 6= 0 (mod 3); if x ≡ 1 (mod 3) then 2x3 + x2 + 2 ≡ 2 6= 0 (mod 3); if
x ≡ 2 (mod 3) then 2x3 + x2 + 2 ≡ 1 6= 0 (mod 3).
2. Find gcd(587, 345) and lcm(587, 345) using the Euclidean algorithm.
We have
587 = 345 + 242,
345 = 242 + 103,
242 = 103 · 2 + 36,
103 = 36 · 2 + 31,
36 = 31 + 5,
1
31 = 5 · 6 + 1,
5 = 1 · 5 + 0,
so the gcd is 1.
3. Find gcd(n, n3 − 1) for any n ≥ 1.
There are a couple ways to solve this. We will use the Euclidean algorithm:
n3 − 1 = n · (n2 − 1) + (n − 1),
n = (n − 1) + 1,
n − 1 = 1 · (n − 1) + 0,
so the gcd is 1 if n ≥ 1.
4. Suppose n has prime factorization
n = p51 · p2 · p43
for some distinct primes p1 , p2 , p3 . How many positive divisors does n have?
Because every number has a unique prime factorization, the set of positive divisors of
n is
{pa1 pb2 pc3 | 0 ≤ a ≤ 5, 0 ≤ b ≤ 1, 0 ≤ c ≤ 4}.
Clearly, the number of elements in this set is 6 · 2 · 5 = 60.
5. Does there exist an integer n such that n, n + 2, and 5n + 2 are all primes numbers?
Yes: if n = 3 then these are 3, 5, 17. However, if n > 3, these will never all be simultaneously prime. If n > 3 and n ≡ 0 (mod 3), then n is not prime. If n ≡ 1 (mod 3), then
n + 2 ≡ 0 (mod 3) and so n + 2 is not prime (since n + 2 > 3). If n ≡ 2 (mod 3), then
5n + 2 ≡ 12 ≡ 0 (mod 3), so 5n + 2 is not prime (again, we are using that 5n + 2 > 3
too).
6. Suppose
lcm(a, b) >
ab
2
Find gcd(a, b).
We have lcm(a, b)gcd(a, b) = ab, and since lcm(a, b) >
In other words, it must be 1.
ab
2
7. Find the least non-negative residue of 32011 modulo 22.
2
this means that gcd(a, b) < 2.
This actually uses Euler’s theorem, which is more general than Fermat’s Little Theorem,
but the idea is the same. We know by Euler’s theorem that
3φ(22) = 310 ≡ 1
(mod 22).
Hence we have 32011 = (31 0)201 · 3 ≡ 1 · 3 = 3 (mod 22). So the answer is 3.
8. Determine, with proof, if the following system of congruences has a solution:
x≡1
mod 5
x≡3
mod 6
4x ≡ 2
mod 14
Note that the last congruence, 4x ≡ 2 mod 14, is equivalent to the congruence 2x ≡ 1
mod 7: we have 4x ≡ 2 mod 14 iff 14|4x − 2 iff 4x − 2 = 14k for some k ∈ Z, iff
2x − 1 = 7k iff 2x ≡ 1 mod 7. This last congruence holds true if x ≡ 4 mod 7. Now,
by the Chinese Remainder Theorem, we have that
x≡1
mod 5
x≡3
mod 6
x≡4
mod 7
does have a solution, since 5, 6, 7 are pairwise relatively prime. Alternately, you could
just cook up a solution, like x = 81, but that might take a while.
9. Let p ≥ 3 be a prime number. Find all solutions to
x3 ≡ x mod p
If x ≡ 0 (mod p), then x3 ≡ 0 ≡ x (mod p), so such an x is a solution.
If p 6 |x, then x has a multiplicative inverse y mod p (meaning xy ≡ 1 (mod p)), and so
x3 ≡ x (mod p) iff x3 y ≡ xy (mod p), which is iff x2 ≡ 1 (mod p). In class, we showed
that this happens iff x ≡ ±1 (mod p). So the solutions are any x which is 0, 1, or −1
(mod p).
10. Does the equation
4x + 16y = 22
have solutions with x and y integers?
No: (4, 16) = 4 which is not a divisor of 22, so there can be no solution.
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