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Math 347 C
HOUR EXAM II
27 July 2011
SOLUTIONS
- Let A and B be disjoint sets, with |A| = m and |B| = n. Let m, n, k be natural numbers satisfying k ≤ n and k ≤ m. Give combinatorial interpretations of
(m+n k
and of
(m i
)( (^) n k−i
in terms of the sets A and B. Use your interpretations to prove
∑^ k
i=
m i
n k − i
m + n k
SOLUTION
(m+n k
is the number of ways to choose a set of size k from a set of size m( + n, which is the size of A ∪ B, since A and B are disjoint. m i
)( (^) n k−i
is the number of ways to choose i elements from A and k − i elements from B. The sum
∑k i=
(m i
)( (^) n k−i
is the number of ways of choosing k elements from A ∪ B in all possible ways from A and B separately. But that is the same as choosing a set of size k from a set of size m + n, namely from A ∪ B. Hence the equality.
- Find all integer solutions to 1 60
x 5
y 12
SOLUTION Multiplying by 60 we get 1 = 12x + 5y. Using either the Euclidean Algorithm or by inspection we get the solution x = − 2 and y = 5. Now suppose that u, v is another solution, i.e., 1 = 12u + 5v. Subtracting, 1 = 12(−2) + 5(5) from 1 = 12u + 5v, we get 0 = 12(u + 2) + 5(v − 5). Therefore 12|5(v − 5) and since 12 and 5 are relatively prime, 12|(v − 5). Therefore v − 5 = 12k, for some integer k. Hence v = 5 + 12k. Similarly u + 2 = 5k′^ and since 0 = 12(u + 2) + 5(v − 5), k′^ = −k. Therefore every integral solution of 1 = 12x + 5y is
u = − 2 − 5 k, v = 5 + 12k,
for all integers k.
- Let p be a prime. The ”order” of the congruence class ¯a is the smallest natural number n such that (a)n^ = 1.
a) Show that the condition (a)n^ = 1 is equivalent to an^ ≡ 1(mod p). b) Let p = 5 and find the orders of 2, 3 and 4. c) Show that if a is a congruence class mod p, a 6 = 0 and d is the order of a, then d|(p − 1). (Hint: Use the Division Algorithm to divide p − 1 by d and then use Fermat’s Little Theorem.) SOLUTION a) By the definition of multiplication of congruence classes, (a)n^ = (an). Thus if (a)n^ = 1, then an^ ≡ 1 (mod p). b) Since 2^2 = 4 and 2^3 = 8 ≡ 3 (mod 5) and 2^4 = 16 ≡ 1 (mod 5), then the order of 2 is 4. Since 3^2 = 9 ≡ 4 (mod 5) and 3^3 = 27 ≡ 2 (mod 5) and 3^4 = 81 ≡ 1 (mod 5), then the order of 3 is 4. Since 4^2 = 16 ≡ 1 (mod 5), then the order of 4 is 2. c) Let d be the order of a, mod p. Then d is the smallest positive integer satisfying ad^ ≡ 1 (mod p). Now divide p − 1 by d and we get
p − 1 = dq + r, 0 ≤ r < d.
Since a 6 = 0, Fermat’s Little Theorem says that ap−^1 ≡ 1 (mod p) and since p − 1 = dq + r, then
1 ≡ ap−^1 = adq+r^ = adq^ ar^ = (ad)q^ ar^ ≡ 1 q^ ar^ = ar^ (mod p).
Now r < d and ar^ ≡ 1 (mod p). But d is the smallest positive integer satisfying (a)d^ = 1. Hence r = 0 and so p − 1 = dq, whence, d|(p − 1).
- Consider the polynomial f (x) = x^6 + x^5 + cx^4 + 1, where c is an integer. For which values of c does f (x) have rational roots? SOLUTION By the Rational Roots Theorem, the only possible rational roots are ±1. 1 is a root of f (x) if and only if f (1) = 1 + 1 + c + 1 = 3 + c = 0 if and only if c = − 3. -1 is a root of f (x) if and only if f (−1) = 1 − 1 + c + 1 = c + 1 = 0 if and only if c = − 1.
- Consider the set of natural numbers [10] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 ,. 9 , 10 }.
a) Show that if a, b ∈ [10] and a 6 = b then a 6 ≡ b (mod 10). b) Partition [10] into the smallest possible number of subsets T 1 , T 2 ,... , Tr , so that for each value of i, if a, b ∈ Ti, and a 6 = b, then a + b = 10. What is your value of r? c) Show that if R ⊂ [10], with |R| = 7, then there are c, d ∈ R, c 6 = d, so that c + d = 10.
