Distinct Possibilities for Renting Apartments

This document discusses the number of distinct possibilities for Jack and Jill to rent separate apartments in a new building by the river. Jack does not care about the view, while Jill only wants an apartment with a river view. The document provides a solution to calculate the number of possibilities.

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1 –
Jack and Jill want to rent separate apartments on the third floor of a new building by the river.
The building has nine apartments available, numbered 301, 302,. . . , 309. The odd-numbered
apartments have a river view, and the even-numbered apartments do not. Jill will only rent an
apartment with a river view, and Jack does not care about the view. How many distinct
possibilities exist for the pair of apartments they end up renting?
Solution here >
$Let$
$$A= set of apartments of third floor$$
$$B= Set of odd-numbered apartments $$
$$C= Set of even-numbered apartments$$
$$A \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$$
$$B \{1, 3, 5, 7, 9\}$$
$$C \{2, 4, 6, 8\}$$
$$=n(A)=9 n(B)= 5 n(C)= 4$$
$$= (A ∩ B)x(C)$$
$$= 5 x 4$$
$$= 20$$
2 –
$n ≥ 2$ distinguishable Hogwarts students participate in Professor Snape’s
experiment. Each student is given potion $A$, or philter $B$, or neither, or
both. We know that Harry and Hermione are the only students among the n that
are given both $A$ and $B$. In how many distinct ways could Snape have
distributed his experimental liquids?
Solution here >
$$n ≥ 2 $$
$$A U B U {∅} U AB$$
$$n(A)$$
$$n(B) $$
$$n({∅})$$
$$n(AB)$$
= $4 distinct ways$
3 –
Prove that for all integers $x$, if $x$ is odd and $x ≥ 3$, then $x^2 − 1$ is
divisible by 8.
Solution here >
If $x$ is odd, then there exists $k∈Z$, such that $x = 2k+1$. Therefore:
$$x^2−1=(2k+1)^2−1=4k^2+4k+1−1=4k^2+4k$$
$$=8×\frac{k(k+1)}{2}$$
$k(k+1)$ is even, since either $k$ or $k+1$ is even. Thus we can write it as
$2n$ where $n∈Z$, so now we obtain:
$$x^2−1=8\frac{2n}{2}=8n$$
Therefore $x^2−1$ is divisible by 8.
4 –

(a) A number $n ∈ N$ is called perfect if the sum of all of $n’s$ factors
other than $n$ itself is equal to $n$. For example, $6$ is perfect because its
factors are $1, 2, 3, and 6, and 1 + 2 + 3 = 6$. Prove that there are no
perfect prime numbers.
(b) Let $m, n ∈ Z +$, and suppose that $m$ is a factor of both $n$ and $n +
1$. Prove that $m = 1$.
Solution here >
a. Proof.
$Let N = l_{1} ^{e_{1}} l_{2}^{e_{2}} • • • l_{s}^{e_{s}}$ for some primes
$l_{1}, l_{2}, . . . , l_{s}$. Since $N$ is odd, all $l_{i}$ are odd. Finally,
$σ(N) = 2N$. Since $$σ(N) = σ(l_{1}^{e1} l_{2}^{e_{2}} • • • l_{s}^{e_{s}}) =
σ(l_{1}^{e_{1}})σ(l_{2}^{e_{2}})• • • σ(l_{s}^{e_{s}})$$, we take a look at $
$σ(l^{e} ) = 1 + l + l^{2} + . . . + l^{e}$$, a sum of $e + 1 $odd numbers.
This is odd only if $e$ is even. Since $$σ(l_{1}^{e1} l_{2}^{e_{2}} • •
•l_{s}^{e_{s}}) = σ(l_{1}^{e_{1}} )σ(l_{2}^{e_{2}})• • • σ(l_{s}^{e_{s}}) =
2l_{1}^{e_{1}}l_{2}^{e_{2}} • • • l_{s}^{e_{s}}$$ , we can only get one factor
of $2$. So the $e_{i}$ are even, all except one, say $e_{1}$. So $N =
l_{1}^{e_{1}} p_{1}^{2a_{1}}• • • p_{r}^{2a_{r}}$. We have $2 |
σ(l_{1}^{e_{1}} ) $ but $4 ∤ σ(l_{1}^{e_{1}} )$. Since $l_{1} is odd, e_{1}$
is odd. Now, $modulo 4$, we see that either $l_{1} ≡ 1 (mod 4)$ or $l_{1} ≡ −1
(mod 4)$. But if $l_{1} ≡ −1 (mod 4)$, then $σ(l_{1}^{e_{1}} ) = 1 + l_{1} +
(l_{1})^{2} + (l_{1})^{3} + . . . + (l_{1})^{e_{1}-1} + l_{1}^{e_{1}} ≡ 1 +
(−1) + 1 + (−1) + . . . + 1 + (−1) ≡ 0 (mod 4)$, which is clearly a
contradiction since $4 ∤ σ(l_{1}^{e_{1}} )$. Thus,$ l_{1} ≡ 1 (mod 4)$. Now,
$σ(l_{1}^{e_{1}} ) = 1 + l_{1} + (l_{1})^{2}+ (l_{1})^{3}+ . . . +
l_{1}^{e_{1}-1} + l_{1}^{e_{1}} ≡ 1 +(- 1) + 1 +(- 1 )+ . . . + 1 +(- 1) ≡
e_{1} + 1 (mod 4)$. Since$ e_{1}$ is odd, either $e_{1} + 1 ≡ 0 (mod 4)$ or
$e_{1} + 1 ≡ 2 (mod 4)$. If $e_{1} + 1 ≡ 0 (mod 4)$, then $4 |
σ(l_{1}^{e_{1}})$ which is again a contradiction. So $e_{1} + 1 ≡ 2 (mod 4) ⇔
e_{1} + 1 = 4e + 2$, that is, $e_{1} = 4e + 1$. Consequently, $N = q^{4e+1}
(p_{1})^{2a_{1}} • • • p_{r}^{2a_{r}}$ , for $q ≡ 1 (mod 4)$.
b.
Let $n$ be an integer. Suppose $a | n$ and $a | (n + 1)$. Then $as = n$ and
$at = n + 1$ for some $s, t ∈ Z$. But $1 = (n + 1) − n = at − as = a(t − s)$
therefore $a$ divides $1$. Therefore the only common divisors of $n$ and $n +
1$ are $1$ and $−1$.
5 –
(a) Give an example of three distinct (no two are the same), nonempty sets
$A, B, C$ such that • there are elements that are common to $A$ and $B$; •
every element of $A$ that is also in $B$ must also be in $C$; • there are
elements in $B$ that are not in $C$.
(b) Let $A$ be an arbitrary set such that ${∅} ∈ A$ and ${∅} ⊆ A$. List
four distinct subsets of $A$.
(c) Define the following set using set builder notation: $\{0, 1, 64\}$.
(d) Give examples of three sets $A, B, C ⊆ \{1, 2, 3, 4, 5, 6, 7\}$ such
that $A$ and $B$ are disjoint, $A \ C = \{1, 3, 7\}, B ∪ C = \{2, 4, 5, 6\},
|A| = 5$, and $B \ C 6= ∅$
Solution here >
a.

$$A \{1, 2, 3,\} B \{4, 3, 2\} C \{3, 2, 1\}$$
b.
If $A \{1, 2, 3, 4, 5, 6\}$, then the four subset would be $$\{1\}, \
{1, 2, 3\}, \{1, 2, 4\} and \{1, 2, 3, 4, 5\}$$
c.
$${x∈ Z|64≥x≥1}$$
d.
$A{1,3,5,7} B{2,4,5,6} C{1,2,3,4,5,6,7}$
6 –
(a) Let $S$ be the set ${a, b, c, d}$. What is its power set $2S$ ?
(b) Let $A = {2, 3}, B = {3, 4, 5}$, and $C = {3, 5, 7, 9}$. What is $A × (B ∩
C)$?
Solution here >
a. Power set= $2^{4}$
$$\{\{Null set\},\{a\},\{b\},\{c\},\{d\},\{a,a\},\{a,b\},\{a,c\}, \{a,d\},\
{b,b\},\{b, a\},\{b, c\},\{b, d\},\{c, c\},\{c, a\},\{c, b\},\{c, d\},\
{d ,d\},\{d,a\},\{d, b\},\{d, c\}\}$$
b.
$2 X (3, 5)$
= $(6,10)$
= $16$
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