Divisor-based problems

This chapter systematically explores fundamental concepts and techniques for analyzing divisors of integers, a cornerstone of number theory. A robust understanding of these methods is critical for solving complex problems encountered in the CMI BS Hons examination.

Chapter Contents

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| Topic |

|---|-------| | 1 | Number of divisors | | 2 | Sum of divisors | | 3 | Even and odd divisors | | 4 | Ordered factor pairs |

We begin with Number of divisors.

Part 1: Number of divisors

Number of Divisors

Overview

Many divisor-counting problems look difficult only until the prime factorization is written down. Once a number is expressed in the form $\qquad n = p_1^{a_1} p_2^{a_2}\cdots p_k^{a_k}$ the number of positive divisors becomes completely structured. In CMI-style questions, this topic is rarely about listing divisors one by one. Instead, it tests whether you can:

• use prime factorization correctly

• apply the divisor-count formula

• classify numbers with a fixed number of divisors

• count such numbers under a bound like $n\le 60$

• move between factorization patterns and divisor counts quickly

This topic is a standard bridge between factorization and counting. ---

Learning Objectives

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Prime Factorization First

All divisor-count questions begin from this form. ---

Divisor Count Formula

Why This Works

A divisor of $n$ has the form $\qquad p_1^{b_1} p_2^{b_2}\cdots p_k^{b_k}$ where each exponent satisfies $\qquad 0\le b_i\le a_i$ So:

• $b_1$ has $a_1+1$ choices

• $b_2$ has $a_2+1$ choices

• ...

• $b_k$ has $a_k+1$ choices

By the multiplication principle, $\qquad D(n)=(a_1+1)(a_2+1)\cdots(a_k+1)$ ---

Quick Examples

Example 1 Find $D(72)$. Prime factorization: $\qquad 72 = 2^3 \cdot 3^2$ Hence $\qquad D(72) = (3+1)(2+1)=4\cdot 3=12$ So $\boxed{D(72)=12}$. --- Example 2 Find $D(360)$. Prime factorization: $\qquad 360 = 2^3\cdot 3^2\cdot 5$ Hence $\qquad D(360)=(3+1)(2+1)(1+1)=4\cdot3\cdot2=24$ So $\boxed{D(360)=24}$. ---

Standard Patterns

These classifications come directly from factorizing the divisor count. ---

Why the $D(n)=6$ Pattern Matters

This is the main pattern behind the PYQ type. ---

Counting Under a Bound

This is the safest approach. ---

Example of Bounded Counting

Example 3 Count positive integers $n\le 50$ such that $D(n)=6$. From the classification, $\qquad n=p^5 \quad \text{or} \quad n=p^2q$

Case 1: $n=p^5$

We need $\qquad p^5\le 50$ Only $\qquad 2^5=32$ works. So this gives $1$ number.

Case 2: $n=p^2q$

We need distinct primes $p,q$ with $\qquad p^2q\le 50$

• If $p=2$, then $4q\le 50$, so $q\le 12.5$.

Valid primes $q\ne 2$ are $\qquad 3,5,7,11$ giving $4$ numbers.

• If $p=3$, then $9q\le 50$, so $q\le 5.5$.

Valid primes $q\ne 3$ are $\qquad 2,5$ giving $2$ numbers.

• If $p=5$, then $25q\le 50$, so $q\le 2$.

Valid prime is $\qquad 2$ giving $1$ number.

• If $p\ge 7$, then $p^2>50$, impossible.

So total from Case 2 is $\qquad 4+2+1=7$ Hence total count is $\qquad 1+7=8$ Therefore there are $\boxed{8}$ such integers. ---

Odd Number of Divisors

Why?

Usually divisors come in pairs: $\qquad d \quad \text{and} \quad \dfrac{n}{d}$ These are distinct unless $\qquad d^2=n$ So an unpaired divisor occurs exactly when $n$ is a square. Equivalently, from the formula $\qquad D(n)=(a_1+1)\cdots(a_k+1)$, the product is odd iff each $(a_i+1)$ is odd, i.e. each $a_i$ is even, which means $n$ is a perfect square. ---

Useful Consequences

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $n=2^4\cdot 3^2$, then $D(n)$ equals" options=["$10$","$12$","$15$","$18$"] answer="C" hint="Use the divisor-count formula." solution="We have $\qquad D(n)=(4+1)(2+1)=5\cdot3=15$ So the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find $D(48)$." answer="10" hint="Factorize $48$ first." solution="Prime factorization: $\qquad 48=2^4\cdot 3$ Hence $\qquad D(48)=(4+1)(1+1)=5\cdot2=10$ Therefore the answer is $\boxed{10}$." ::: :::question type="MSQ" question="Which of the following forms can give exactly $6$ divisors?" options=["$p^5$","$p^2q$ with distinct primes $p,q$","$pq$ with distinct primes $p,q$","$p^3q$ with distinct primes $p,q$"] answer="A,B" hint="Factor the number $6$." solution="If $D(n)=6$, then the exponent-pattern product must equal $6$.

• $p^5$ gives $6$ divisors.

• $p^2q$ gives $(2+1)(1+1)=6$ divisors.

• $pq$ gives $(1+1)(1+1)=4$ divisors.

• $p^3q$ gives $(3+1)(1+1)=8$ divisors.

Hence the correct answer is $\boxed{A,B}$." ::: :::question type="SUB" question="Prove that a positive integer has an odd number of positive divisors if and only if it is a perfect square." answer="It is odd exactly for perfect squares." hint="Use divisor pairs or prime factorization." solution="Let $n$ be a positive integer. Usually divisors of $n$ come in pairs: $\qquad d \quad \text{and} \quad \dfrac{n}{d}$ These are distinct unless $\qquad d=\dfrac{n}{d}$, that is, $\qquad d^2=n$. So an unpaired divisor occurs exactly when $n$ is a perfect square. Hence the number of divisors is odd exactly for perfect squares. Equivalently, if $\qquad n=p_1^{a_1}\cdots p_k^{a_k}$, then $\qquad D(n)=(a_1+1)\cdots(a_k+1)$ is odd iff every factor $(a_i+1)$ is odd, i.e. every $a_i$ is even, which is exactly the condition that $n$ is a perfect square. Therefore, a positive integer has an odd number of positive divisors if and only if it is a perfect square." ::: ---

Summary

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Part 2: Sum of divisors

Sum of Divisors

Overview

The sum-of-divisors function is one of the most important arithmetic functions in elementary number theory. It converts the divisor structure of an integer into an additive quantity. In exam-level problems, it is usually used to compute divisor sums, compare numbers, classify perfect numbers, or solve equations involving divisors. ---

Learning Objectives

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Core Idea

Examples:

• $\qquad \sigma(1)=1$

• divisors of $6$ are $1,2,3,6$, so

$\qquad \sigma(6)=1+2+3+6=12$ ---

Prime Power Formula

This is the main computation formula. ---

Multiplicativity

So if $\qquad n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ then $\qquad \sigma(n)=\prod_{i=1}^{k}\dfrac{p_i^{a_i+1}-1}{p_i-1}$ ::: ---

Why Prime Factorisation Matters

This is much faster than listing divisors one by one for large $n$. ---

Examples

Example 1 Find $\sigma(12)$. Prime factorisation: $\qquad 12=2^2\cdot 3$ So $\qquad \sigma(12)=\sigma(2^2)\sigma(3)$ $\qquad =(1+2+4)(1+3)=7\cdot 4=28$ Hence $\qquad \sigma(12)=\boxed{28}$ --- Example 2 Find $\sigma(18)$. Prime factorisation: $\qquad 18=2\cdot 3^2$ So $\qquad \sigma(18)=\sigma(2)\sigma(3^2)$ $\qquad =(1+2)(1+3+9)=3\cdot 13=39$ Hence $\qquad \sigma(18)=\boxed{39}$ ---

Proper Divisors and Perfect Numbers

A positive integer $n$ is:

• perfect if $\qquad \sigma(n)=2n$

• deficient if $\qquad \sigma(n)<2n$

• abundant if $\qquad \sigma(n)>2n$

::: Example:

• $\qquad \sigma(6)=12=2\cdot 6$, so $6$ is perfect.

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Odd and Even Observations

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Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The value of $\sigma(12)$ is" options=["$24$","$26$","$28$","$30$"] answer="C" hint="Use prime factorisation." solution="We have $\qquad 12=2^2\cdot 3$ So $\qquad \sigma(12)=\sigma(2^2)\sigma(3)=(1+2+4)(1+3)=7\cdot 4=28$ Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find $\sigma(18)$." answer="39" hint="Use $18=2\cdot 3^2$." solution="Since $\qquad 18=2\cdot 3^2$ we get $\qquad \sigma(18)=\sigma(2)\sigma(3^2)$ $\qquad =(1+2)(1+3+9)=3\cdot 13=39$ Therefore the answer is $\boxed{39}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\sigma(p)=1+p$ for a prime $p$","If $\gcd(m,n)=1$, then $\sigma(mn)=\sigma(m)\sigma(n)$","The sum of proper divisors of $n$ is $\sigma(n)-n$","$\sigma(m+n)=\sigma(m)+\sigma(n)$ for all positive integers $m,n$"] answer="A,B,C" hint="Use the definition and the standard formula." solution="1. True. The divisors of a prime $p$ are $1$ and $p$.

True. This is the multiplicative property of $\sigma$ for coprime integers.

True. Proper divisors exclude $n$ itself.

False. The sum-of-divisors function is not additive in this way.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Show that if $n=p^2$ where $p$ is prime, then $\sigma(n)=1+p+p^2$." answer="$\sigma(p^2)=1+p+p^2$" hint="List the divisors of $p^2$." solution="If $n=p^2$ where $p$ is prime, then the positive divisors of $n$ are $\qquad 1,\ p,\ p^2$ So by definition, $\qquad \sigma(n)=1+p+p^2$ Hence $\qquad \boxed{\sigma(p^2)=1+p+p^2}$" ::: ---

Summary

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Part 3: Even and odd divisors

Even and Odd Divisors

Overview

Questions about even and odd divisors are really questions about prime factorisation, especially the role of the factor $2$. Once a number is written in the form $\qquad n = 2^a \cdot m$ where $m$ is odd, the entire problem becomes transparent: odd divisors come from the odd part only, while even divisors are the divisors that use at least one factor of $2$. ---

Learning Objectives

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Prime Factorisation Is the Main Tool

A positive divisor of $n$ must be of the form $\qquad d = p_1^{b_1} p_2^{b_2} \cdots p_r^{b_r}$ where each exponent satisfies $\qquad 0 \le b_i \le a_i$ ::: So divisor counting is really exponent-choice counting. ---

Total Number of Positive Divisors

Reason:

• exponent of $p_1$ can be chosen in $a_1+1$ ways

• exponent of $p_2$ can be chosen in $a_2+1$ ways

• and so on

Multiply the choices. ---

Separating the Even and Odd Parts

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Number of Odd Divisors

In words: ignore the factor of $2$ completely and count divisors of the odd part. ---

Number of Even Divisors

This is the main formula for the topic. ---

Why the Even-Divisor Formula Works

This direct counting argument is often cleaner than subtracting odd divisors from total divisors. ---

Powers of a Number

This is extremely common in exam questions. ---

Special Cases

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Minimal Worked Examples

Example 1 Find the number of even positive divisors of $\qquad 72 = 2^3 \cdot 3^2$ Total divisors: $\qquad (3+1)(2+1)=12$ Odd divisors: $\qquad 2^0 \cdot 3^j,\quad j=0,1,2$ So there are $\qquad 2+1=3$ odd divisors. Hence even divisors: $\qquad 12-3=9$ Or directly: $\qquad 3(2+1)=9$ --- Example 2 Find the number of odd positive divisors of $\qquad 180 = 2^2 \cdot 3^2 \cdot 5$ Ignore the factor $2^2$. So odd divisors are divisors of $\qquad 3^2 \cdot 5$ Hence count: $\qquad (2+1)(1+1)=6$ --- Example 3 Find the number of even positive divisors of $\qquad (2^3\cdot 3^2)^2 = 2^6 \cdot 3^4$ Even divisor count: $\qquad 6(4+1)=30$ ---

Standard Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $n=2^4\cdot 3^2$, then the number of odd positive divisors of $n$ is" options=["$5$","$3$","$10$","$15$"] answer="B" hint="Ignore the factor of $2$ when counting odd divisors." solution="Odd divisors come only from the odd part $3^2$. So the number of odd positive divisors is $\qquad 2+1=3$ Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the number of even positive divisors of $72$." answer="9" hint="Write $72=2^3\cdot 3^2$." solution="We have $\qquad 72=2^3\cdot 3^2$ Even divisors require the exponent of $2$ to be chosen from $1,2,3$, so there are $3$ choices. For the prime $3$, the exponent can be chosen in $2+1=3$ ways. Hence the number of even positive divisors is $\qquad 3\cdot 3=9$ Therefore the answer is $\boxed{9}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $n$ is odd, then every positive divisor of $n$ is odd","If $n=2^a m$ with $m$ odd, then the number of odd divisors of $n$ equals the number of divisors of $m$","If $n=2^a m$ with $m$ odd, then the number of even divisors of $n$ is always equal to the number of odd divisors of $n$","If $n=2^a p_1^{b_1}\cdots p_r^{b_r}$, then the number of even divisors is $a(b_1+1)\cdots(b_r+1)$"] answer="A,B,D" hint="Compare the exponent choices for the prime $2$." solution="1. True.

True.

False. This happens only in special cases such as $a=1$.

True.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Let $n=2^a p_1^{b_1}p_2^{b_2}\cdots p_r^{b_r}$, where $p_1,\dots,p_r$ are odd primes. Show that the number of even positive divisors of $n$ is $a(b_1+1)(b_2+1)\cdots(b_r+1)$." answer="$a(b_1+1)(b_2+1)\cdots(b_r+1)$" hint="Count the exponent choices directly." solution="A positive divisor of $n$ has the form $\qquad d=2^e p_1^{c_1}p_2^{c_2}\cdots p_r^{c_r}$ where $\qquad 0\le e\le a$ and $\qquad 0\le c_i\le b_i$ For the divisor to be even, we need $\qquad e\ge 1$ So the exponent $e$ of $2$ can be chosen in exactly $\qquad a$ ways, namely $\qquad 1,2,\dots,a$ For each odd prime $p_i$, the exponent $c_i$ can be chosen in $\qquad b_i+1$ ways. By the multiplication principle, the number of even positive divisors is $\qquad a(b_1+1)(b_2+1)\cdots(b_r+1)$ Hence the required result is $\boxed{a(b_1+1)(b_2+1)\cdots(b_r+1)}$." ::: ---

Summary

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Part 4: Ordered factor pairs

Ordered Factor Pairs

Overview

This topic sits at the intersection of factorisation, parity, and counting divisors. In many exam problems, an equation such as $\qquad a^2-b^2 = N$ is not solved directly. Instead, it is rewritten as a product $\qquad (a-b)(a+b)=N$ and the problem becomes one of counting suitable factor pairs. The real skill is not just factoring, but understanding which factor pairs actually produce valid positive integers. ---

Learning Objectives

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Core Definitions

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The Key Identity

This identity is the main bridge between quadratic-looking equations and divisor-counting problems. ---

Standard Conversion

If we know $u$ and $v$, then $\qquad a=\dfrac{u+v}{2},\qquad b=\dfrac{v-u}{2}$ ::: So every valid solution comes from a factor pair $(u,v)$ satisfying extra conditions. ---

Validity Conditions

This third condition is crucial. ---

Parity Logic

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Important Solvability Criterion

So:

• odd numbers work

• multiples of $4$ work

• numbers of the form $4k+2$ do not work

::: ---

Counting Solutions: Odd Case

If $\tau(N)$ is the number of positive divisors of $N$, then:

• if $N$ is not a perfect square, the number of such pairs is

$\qquad \dfrac{\tau(N)}{2}$

• if $N$ is a perfect square, the number of such pairs is

$\qquad \dfrac{\tau(N)-1}{2}$ because the middle pair $u=v$ gives $b=0$, not a positive solution. ::: ---

Counting Solutions: Multiple of 4 Case

Hence the number of positive integer solutions to $\qquad a^2-b^2=4M$ equals the number of positive factor pairs $(r,s)$ of $M$ with $\qquad r<s$ ::: So the problem reduces to counting strict factor pairs of $M$. ::: ---

Divisor Counting

This is extremely useful when counting factor pairs. ---

Minimal Worked Examples

Example 1 Count the positive integer solutions of $\qquad a^2-b^2=45$ Factor: $\qquad (a-b)(a+b)=45$ Since $45$ is odd, every positive factor pair is odd-odd. The positive factor pairs with smaller factor first are $\qquad (1,45),\ (3,15),\ (5,9)$ So there are $\boxed{3}$ solutions. --- Example 2 Count the positive integer solutions of $\qquad a^2-b^2=144$ Since $144=4\cdot 36$, write $\qquad a-b=2r,\qquad a+b=2s$ Then $\qquad rs=36$ Now count factor pairs of $36$ with $r<s$: $\qquad (1,36),\ (2,18),\ (3,12),\ (4,9)$ The pair $(6,6)$ is excluded because it would give $b=0$. So there are $\boxed{4}$ positive solutions. ---

PYQ-Type Insight

This topic is therefore much more about controlled counting than about raw algebra. ::: ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="To count positive integer solutions of $a^2-b^2=N$, the most useful first step is to write" options=["$a^2+b^2=N$","$(a-b)(a+b)=N$","$(a+b)^2=N$","$ab=N$"] answer="B" hint="Use the difference-of-squares identity." solution="The identity $\qquad a^2-b^2=(a-b)(a+b)$ converts the problem into a factor-pair problem. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="How many positive integer pairs $(a,b)$ satisfy $a^2-b^2=45$?" answer="3" hint="Count valid strict factor pairs of $45$." solution="We have $\qquad (a-b)(a+b)=45$ Since $45$ is odd, all positive factor pairs are valid for parity. The strict factor pairs are $\qquad (1,45),\ (3,15),\ (5,9)$ So there are $\boxed{3}$ positive integer pairs $(a,b)$." ::: :::question type="MSQ" question="Which of the following are true?" options=["If $N\equiv 2\pmod 4$, then $N$ cannot be written as $a^2-b^2$ with integers $a,b$","If $a^2-b^2=N$, then $a-b$ and $a+b$ have the same parity","If $N$ is odd, then every positive factor pair of $N$ automatically has the same parity","If $u=a-b$ and $v=a+b$, then $a=\dfrac{u-v}{2}$"] answer="A,B,C" hint="Use parity and the reconstruction formulas carefully." solution="1. True.

True.

True, because factors of an odd number are odd.

False. The correct formula is

$\qquad a=\dfrac{u+v}{2}$ Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Show that no integer of the form $4k+2$ can be written as $a^2-b^2$ for integers $a,b$." answer="Use parity of $a-b$ and $a+b$." hint="Write $a^2-b^2=(a-b)(a+b)$." solution="Suppose $\qquad a^2-b^2 = (a-b)(a+b)$ Now $a-b$ and $a+b$ always have the same parity. So there are only two possibilities:

both are odd, in which case their product is odd

both are even, in which case their product is divisible by $4$

Therefore $a^2-b^2$ can be either odd or divisible by $4$, but it can never be congruent to $2$ modulo $4$. Hence no integer of the form $\boxed{4k+2}$ can be expressed as $a^2-b^2$." ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="Let $n$ be an integer such that $\tau(n)=6$ and $\sigma(n)=28$. What is the value of $n$?" options=["8","12","18","24"] answer="12" hint="Consider $n$ to be of the form $p^5$ or $p_1^2 p_2^1$. Test the sum of divisors for each case." solution="Given $\tau(n)=6$, $n$ must be of the form $p^5$ or $p_1^2 p_2^1$.
Case 1: $n=p^5$.
$\sigma(n) = \frac{p^6-1}{p-1} = 28$.
If $p=2$, $\sigma(2^5) = \frac{2^6-1}{2-1} = 63 \ne 28$.
If $p=3$, $\sigma(3^5) = \frac{3^6-1}{3-1} = \frac{728}{2} = 364 \ne 28$. No prime $p$ satisfies this.

Case 2: $n=p_1^2 p_2^1$.
$\sigma(n) = \left(\frac{p_1^3-1}{p_1-1}\right)\left(\frac{p_2^2-1}{p_2-1}\right) = (p_1^2+p_1+1)(p_2+1) = 28$.
Since 28 is a small number, $p_1$ and $p_2$ must be small primes.
Try $p_1=2$. Then $(2^2+2+1)(p_2+1) = 7(p_2+1) = 28$.
$p_2+1 = 4 \implies p_2=3$.
So $n=2^2 \cdot 3^1 = 4 \cdot 3 = 12$.
Let's verify: $\tau(12) = (2+1)(1+1) = 3 \cdot 2 = 6$.
$\sigma(12) = (2^2+2+1)(3+1) = 7 \cdot 4 = 28$.
Both conditions are met. The value of $n$ is 12."
:::

:::question type="NAT" question="How many positive integers less than 100 have exactly 3 odd divisors?" answer="8" hint="A number has exactly 3 odd divisors if its odd part is the square of an odd prime. Consider numbers of the form $2^k \cdot p^2$ where $p$ is an odd prime." solution="Let $n$ be a positive integer. We can write $n = 2^k \cdot m$, where $m$ is an odd integer. The divisors of $n$ are of the form $2^a \cdot d$, where $0 \le a \le k$ and $d$ is a divisor of $m$.
The odd divisors of $n$ are precisely the divisors of $m$.
We are given that $n$ has exactly 3 odd divisors, so $\tau(m)=3$.
For $\tau(m)=3$, $m$ must be the square of a prime number. Since $m$ is odd, $m=p^2$ for some odd prime $p$.

We need to find numbers $n < 100$ of the form $2^k \cdot p^2$.

Possible odd primes $p$:

If $p=3$, then $m=3^2=9$.

$k=0: n=9 < 100$
$k=1: n=2 \cdot 9 = 18 < 100$
$k=2: n=4 \cdot 9 = 36 < 100$
$k=3: n=8 \cdot 9 = 72 < 100$
$k=4: n=16 \cdot 9 = 144 \not< 100$
(4 numbers)

If $p=5$, then $m=5^2=25$.

$k=0: n=25 < 100$
$k=1: n=2 \cdot 25 = 50 < 100$
$k=2: n=4 \cdot 25 = 100 \not< 100$ (The question asks for numbers less than 100)
(2 numbers)

If $p=7$, then $m=7^2=49$.

$k=0: n=49 < 100$
$k=1: n=2 \cdot 49 = 98 < 100$
$k=2: n=4 \cdot 49 = 196 \not< 100$
(2 numbers)

If $p \ge 11$, then $p^2 \ge 121 \not< 100$. So no more cases.

Total numbers: $4 + 2 + 2 = 8$.
The numbers are: 9, 18, 36, 72, 25, 50, 49, 98."
:::

:::question type="MCQ" question="Which of the following numbers has an odd number of ordered factor pairs?" options=["10","12","18","25"] answer="25" hint="The number of ordered factor pairs of an integer $n$ is $\tau(n)$. Recall when $\tau(n)$ is odd." solution="The number of ordered factor pairs $(x,y)$ such that $xy=n$ is equal to $\tau(n)$.
We need to find which of the given numbers has an odd value for $\tau(n)$.
It is a known property that $\tau(n)$ is odd if and only if $n$ is a perfect square.

Let's check the options:
* $n=10 = 2^1 \cdot 5^1$. $\tau(10) = (1+1)(1+1) = 2 \cdot 2 = 4$ (even).
* $n=12 = 2^2 \cdot 3^1$. $\tau(12) = (2+1)(1+1) = 3 \cdot 2 = 6$ (even).
* $n=18 = 2^1 \cdot 3^2$. $\tau(18) = (1+1)(2+1) = 2 \cdot 3 = 6$ (even).
* $n=25 = 5^2$. $\tau(25) = (2+1) = 3$ (odd).

Since 25 is a perfect square, it has an odd number of ordered factor pairs (which are (1,25), (5,5), (25,1))."
:::

What's Next?