MATH 115 TAKE HOME EXAM 1
Please work on the following problems independently. Please complete 1 − 3 and choose 3
of 4 − 7 to complete.
1. Let n ∈ Z. Show that n2 ≥ 0.
Proof. By trichotomy, n = 0 or n ∈ N or −n ∈ N. If n = 0, n2 = 0, so we’re done by a
lemma we prove later. If n ∈ N, then n2 ∈ N, so n2 > 0. Otherwise, −n ∈ N. We now
prove some lemmas.
First we show additive inverses are unique. Suppose x + y = 0 and x + z = 0. Then
(x + y) + z = 0 + z = z. Also (x + y) + z = x + (z + y) = (x + z) + y = 0 + y = y. So y = z.
Now we show 0 ∗ x = 0. We have 0 + 0 = 0. So 0 · x + 0 · x = 0 · x, so that (skipping steps) 0 · x = 0.
Now we show (−1) · x = −x. First we have 1 + −1 = 0. So that 1 · x + (−1) · x =
0 · x = x + (−1) · x = 0. So (−1) · x is an inverse for x so it must be equal to −x.
Now we’re supposing that −n ∈ N. So (−n)2 ∈ N. Let’s not worry about associativity and commutativity. If we can show (−n)2 = n2 , we’re done. We have (−n)2 =
(−n) · (−n) = (−1) · n · (−1) · n. So if we can show (−1) · (−1) = 1, we’re done. We have
(−1) · (−1) = −(−1). This is an additive inverse to −1, but so is 1, and since additive
inverses are unique, this must be 1.
2. Find integers x and y so that 4567x + 7842y = 1.
By doing Euclid’s algorithm and working backwards, we obtain x = 2701 and y = −1573.
3. Five bandits discover a chest containing N gold coins. They try dividing the coins
amongst themselves and find that there are 2 left over. The chief bandit responds by
shooting one of the bandits. Now they try to divide the coins amongst the remaining four
bandits. There are 3 left over. Before the chief bandit can draw his gun, he is bludgeoned
by one of the remaining bandits with a large shovel. The three remaining bandits are now
happy as they can divide the remaining coins amongst themselves evenly. What can N
be? How small can N be?
We have N ≡ 2 mod 5, N ≡ 3 mod 4, and N ≡ 0 mod 3. By CRT, there is a solution
that is unique up to a multiple of 60. We can construct N = 20x + 15y + 12z by reducing
1
2
MATH 115 TAKE HOME EXAM 1
modulo 3, 4, and 5. We see that z = 1, y = 1, and x = 0, so N = 27 + 60k, and 27 is the
smallest.
4. Prove that there are no integers n with φ(n) = 14.
First we find all p so that p − 1|14. Clearly only p = 2 and 3 satisfy this. Then such
an n is only divisible by 2 and 3. However, φ(2k 3j ) = 2k−1 (3 − 1)3j−1 is never equal to 14,
since for example this expression is never divisible by 7.
5. Let n > 0. Show that:
2n
4n
≤
2n
n
n
We first notice that 4 is the sum of the elements in the 2nth row of Pascal’s triangle. On
the other hand 2n
n is the largest element in this row which has 2n + 1 elements. To show
it is the largest, first notice that for 0 < i < n:
n
n!n!
=
n! ≥ 1
i!(n − i)!
i
2n
Thus 2n
n − i > 0. By symmetry, the other cases follow. We can take the first and last
1 and put them together and treat them as a 2, which gives 2n elements. Thus, 2n · 2n
n
is larger than the actual sum, which is 4n .
6. False, a = 3, n = 1.
Salvage: Let a, n > 1. Show that if an − 1 is prime, then a = 2 and n is prime.
Proof. To see that a = 2, notice that
an −1
a−1 = 1 + a
of an − 1. If
+ . . . an−1 . Since n > 1, if a 6= 2,
this would give a proper factorization
n = ab isn’t prime, then
1 + 2a + . . . 2a (b − 1) is another proper factorization.
2ab −1
2a −1
=
7. Let p be a prime different from 2 and 5. Show that p divides infinitely many of the
integers:
9, 99, 999, 9999, . . .
n
This is the same as showing 10 − 1 ≡ 0 mod p , or 10n ≡ 1 mod p for infinitely many n.
To show this, look at powers of 10 mod p. By assumption, (10, p) = 1, so 10 is invertible
modp. We want to show that there is an n > 0 so that 10n = 1 mod p. Then we can take
10kn for k > 0 and conclude. Since 10k can only take on finitely many values modp, by
the pigeonhole principle, 10m ≡ 10n mod p for some m > n. Since 10 is invertible, we have
10m−n ≡ 1 mod p and we are finished.