Mathematical-Reasoning · Unit 20 · Video 2 · Interactive Practice

Functions Without Formulas: Birthdays and Sums of Divisors

IKey Formulas

Statement	Meaning
$b(\text{Andrew Wiles}) = \text{April } 11$	Birthday function: the input (a person) goes inside the parentheses, the output (a month and day, an element of the codomain) on the right
$s(n) =$ the sum of the distinct natural-number divisors of $n$ , e.g. $s(6) = 1 + 2 + 3 + 6 = 12$	Sum of the divisors function, with domain and codomain $\mathbb{N}$
$p$ prime $\Rightarrow s(p) = 1 + p$ ; $n > 1 \Rightarrow s(n) \ge 1 + n$	A prime has exactly the divisors $1$ and $p$ ; any $n > 1$ has at least the two divisors $1$ and $n$
For each $y$ in the codomain there exists $x$ with $f(x) = y$ ; if $x_1 \ne x_2$ , then $f(x_1) \ne f(x_2)$	Two questions to ask of any function; neither is part of the definition

Key Insight: A function needs only a domain, a codomain and a rule giving each input exactly one output, and the rule can be a sentence. Two inputs may share an output, and an element of the codomain may be the output of no input at all.

IIThe Birthday Function: 366 Days, 367 People

Spread birthdays out as far as possible: 366 people can avoid sharing a day, but 367 cannot.

People 367

IIIComputing $s(n)$ from Its Divisors

Each $n$ has one list of divisors and so one sum; a prime sits exactly on the bound $1 + n$ .

n 6

IVIs $m$ an Output of $s$ ?

Since $s(n) \ge 1 + n$ for $n > 1$ , deciding whether $s(n) = m$ has a solution takes only finitely many checks.

Target m 5

VQuiz Questions

Problem 1 · Writing a Fact in Function Notation

Given: the birthday function $b$ , whose domain is the set of all people and whose codomain is the set of the $366$ days of a leap year. The mathematician Emmy Noether was born on March 23, 1882. Which line writes this fact in function notation?

✅ Correct! The input is the person, and the output is a month and day, which is an element of the codomain.

❌ Input and output are swapped. The inputs of

b

are people, so the person goes inside the parentheses. A day is not an element of the domain.

❌ The year is not a possible output. The codomain contains only months and days, so "March 23, 1882" is not one of its elements, even though the year is a true fact about her.

❌ A year is not an element of the codomain. The outputs of

b

are days of a leap year (a month and a day), so the output here is March 23.

Show solution

Function notation puts an input inside the parentheses and its output on the right. The input of $b$ is a person, so the input is Emmy Noether. The output is that person's birthday as a month and a day:

b(\text{Emmy Noether}) = \text{March } 23.

The year 1882 is true but does not appear: an output must be an element of the codomain, and the codomain contains only the $366$ months-and-days of a leap year.

Problem 2 · Computing a Value of $s$

Given: $s(n)$ is the sum of the distinct natural-number divisors of $n$ . Find $s(18)$ .

✅ Correct! The divisors of

18

are

1, 2, 3, 6, 9, 18

, and they add to

39

❌ That formula is for primes.

s(p) = 1 + p

holds only when

p

is prime;

18 = 2 \cdot 9

has more divisors than

1

and

18

❌ $18$ itself is missing.

1 + 2 + 3 + 6 + 9 = 21

leaves out the divisor

18

; every

n

divides itself, so

n

is always in the sum.

❌ A divisor is missing.

1 + 2 + 3 + 9 + 18 = 33

leaves out

6

, which divides

18

because

18 = 6 \cdot 3

Show solution

List the natural numbers that divide $18$ by pairing each divisor $d$ with $18/d$ :

18 = 1 \cdot 18 = 2 \cdot 9 = 3 \cdot 6,

so the divisors are $1, 2, 3, 6, 9, 18$ . Adding them,

s(18) = 1 + 2 + 3 + 6 + 9 + 18 = 39.

Since $18$ is not prime, $s(18)$ is more than $1 + 18 = 19$ , as the bound $s(n) \ge 1 + n$ predicts.

Problem 3 · Thirteen People, Twelve Months

Given: a club has $13$ members. The function $m$ assigns to each member the month in which that member was born; its domain is the set of the $13$ members and its codomain is the set of the $12$ months. Which statement must be true?

✅ Correct! There are

13

inputs and only

12

possible outputs, so two different members

x \ne y

must have

m(x) = m(y)

❌ Not forced. All

13

members could be born in January; then the other

11

months are the output of no input, and

m

is still a function.

❌ Shared outputs are allowed. The definition asks only that each input have exactly one output. Two members with the same birth month break nothing; one member with two birth months would.

❌ Not forced. Two members born in January, two in February and one in each of nine other months make

13

members with two shared months.

Show solution

(b) must hold. Each of the $13$ members has exactly one birth month, one of only $12$ months. If all $13$ outputs were different, the codomain would need at least $13$ elements. So some month is the output for two different members $x \ne y$ : $m(x) = m(y)$ .

(a) can fail: if all $13$ were born in January, no member has a birth month of February.

(c) is false: $m$ is a function, since each member has exactly one birth month. Two inputs sharing an output is allowed.

(d) can fail: with $2$ members born in January, $2$ in February, and $1$ in each of March through November ( $2 + 2 + 9 = 13$ ), two months are shared.

Problem 4 · Is 10 an Output of $s$ ?

Given: $s(n)$ is the sum of the distinct natural-number divisors of $n$ , for $n \in \mathbb{N}$ . Decide: is there a natural number $n$ with $s(n) = 10$ ?

✅ Correct! The bound

s(n) \ge 1 + n

turns an infinite question into nine checks, and the outputs

1, 3, 4, 7, 6, 12, 8, 15, 13

miss

10

❌ $9$ is not prime. Its divisors are

1, 3, 9

, so

s(9) = 13

; the rule

s(p) = 1 + p

applies only to primes.

❌ That adds every number up to $4$ .

3

does not divide

4

; the divisors of

4

are

1, 2, 4

, so

s(4) = 7

❌ Only finitely many $n$ need checking. For

n > 1

the two divisors

1

and

n

give

s(n) \ge 1 + n

, so

s(n) = 10

forces

n \le 9

Show solution

If $n = 1$ , then $s(1) = 1 \ne 10$ . If $n > 1$ , then $1$ and $n$ are two different divisors of $n$ , so $s(n) \ge 1 + n$ . Suppose $s(n) = 10$ with $n > 1$ ; then $1 + n \le 10$ , so $2 \le n \le 9$ .

Checking these:

s(2) = 3,\quad s(3) = 4,\quad s(4) = 7,\quad s(5) = 6,\quad s(6) = 12,

s(7) = 8,\quad s(8) = 15,\quad s(9) = 13.

None equals $10$ , so no natural number $n$ has $s(n) = 10$ : $10$ is an element of the codomain $\mathbb{N}$ that is the output of no input.

Solved: 0 / 4

Functions Without Formulas: Birthdays and Sums of Divisors

IKey Formulas

IIThe Birthday Function: 366 Days, 367 People

IIIComputing s(n)s(n)s(n) from Its Divisors

IVIs mmm an Output of sss?

VQuiz Questions

IIIComputing $s(n)$ from Its Divisors

IVIs $m$ an Output of $s$ ?