Download Exam II - U.I.U.C. Math

Math 347 C1
HOUR EXAM II
30 July 2013
SOLUTIONS
1. a) Show that 2561 ≡ 2(mod 17).
(Hint: Use the fact that 561 = 3 · 11 · 17 and apply Fermat’s Little Theorem and
the Division Algorithm.)
b) Suppose that you know additionally that
2561 ≡ 2(mod 3)
2561 ≡ 2(mod 11)
2561 ≡ 2(mod 17)
Prove that 2561 ≡ 2(mod 561).
SOLUTION
a) 2561 = 23·11·17 = (217 )3·11 ≡ 23·11 ≡ 233 (mod 17),
since 217 ≡ 2(mod 17) by Fermat’s Little Theorem.
By the Division Algorithm, 33 = 17 · 1 + 16 and so
233 = 217+16 = 217 · 216 ≡ 2 · 216 ≡ 217 ≡ 2(mod 17).
Hence 2561 ≡ 2(mod 561).
b) Consider the system of congruences:
x ≡ 2(mod 3)
x ≡ 2(mod 11)
x ≡ 2(mod 17)
Clearly x = 2 is a solution to the system. Moreover, we are given above that 2561 is
also a solution to the system. Since the moduli 3, 11 and 17 are pairwise relatively
prime, we can apply the Chinese Remainder Theorem which says that any two
solutions to the system are congruent modulo the product of the moduli.
Thus
2561 ≡ 2(mod 561).
OR
It follows from the 3 congruences above that 2561 − 2 is divisible by the primes 3,
11 and 17. Hence, by the Fundamental Theorem of Arithmetic, it must be divisible
by their product. Therefore 2561 − 2 ≡ 0(mod 561), i..e., 2561 ≡ 2(mod 561).
2. Find all rational solutions to the equation:
2x3 + x2 + x + 2 = 0.
Are there any other real solutions to that equation? Prove your answer.
SOLUTION
According to the Rational Zeros Theorem, if the rational number ab is a solution to
the equation above, then a|2 and b|2. Hence the possible solutions are 1, 2, 12 , -1,
-2 and − 12 .
Since all coefficients of the polynomial are positive, we can ignore the positive
numbers in this list. We must therefore test -1, -2, and − 21 . A simple calculation
shows that only x = −1 is a solution.
Therefore x + 1 is a factor of the polynomial and it is easy to see that
2x3 + x2 + x + 2 = (x + 1)(2x2 − x + 2).
To see if 2x3 + x2 + x + 2 = 0 has other real roots, we need to see if 2x2 − x + 2 = 0
has real roots.
Using the quadratic formula, we get
x = (1 ±
√
1 − 16)/4,
which are imaginary numbers.
Thus the only real solution to 2x3 + x2 + x + 2 = 0 is x = −1.
3. Consider the congruence equation 49x ≡ 17(mod 149).
a) Show that the congruence has a solution. (Hint: Don’t try to calculate a specific
solution.)
b) Replace one of 49, 17, 149 with another natural number so that the resulting
congruence doesn’t have a solution. Explain.
SOLUTION
a) The congruence has a solution if 49 and 149 are relatively prime. The only prime
divisor of 49 is 7 and clearly 7 doesn’t divide 149. Thus 49 and 149 are relatively
prime and so the congruence has a solution.
b) For example, if we change 149 to 140, i.e., 49x ≡ 17(mod 140), then
gcd(49, 140) = 7. Now convert the congruence to the equation: 49x = 17 + 140k.
It follows that if a solution exists then since 7 is a divisor of 49 and 140, 7 must
divide 17, which is false. Thus 49x ≡ 17(mod 140) has no solution.
4. Let S = [0, 1).
a. Find a sequence in S which converges to sup(S).
b. Find a sequence in S which converges to inf (S).
SOLUTION
a) Clearly sup(S) = 1. Let the sequence hai be defined by an = 1 − n1 , for n =
1, 2, . . . . Since 0 ≤ 1 − n1 < 1, each an ∈ S and an → 1 = sup(S).
b) Since inf (S) = 0 and 0 ∈ S, we can choose the sequence hbi with bn = 0, for all
n and bn → 0 = inf (S).