Exercises — Factors and multiples — all factors of a number, factor pairs

WHAT we do: check whether the division leaves remainder $0$. $56÷7=8$ exactly (remainder $0$), because $7\times 8=56$. So ==$7$ is a factor of $56$==. $56÷9=6$ remainder $2$ (since $9\times 6=54$, and $56-54=2\ne 0$). So ==$9$ is not a factor of $56$==.

A pair $(a,b)$ is a factor pair of $48$ only if $a\times b=48$ exactly.

• $(6,8)$: $6\times 8=48$ ✅
• $(5,9)$: $5\times 9=45\ne 48$ ❌
• $(4,12)$: $4\times 12=48$ ✅
• $(3,15)$: $3\times 15=45\ne 48$ ❌ Factor pairs among these: $(6,8)$ and $(4,12)$.

Every factor pair multiplies to $n$, so the partner of $3$ is $60÷3$. $60÷3=20$. Answer: $(3,20)$, and indeed $3\times 20=60$.

Step 1 — find the stopping point. We need the biggest whole $d$ with $d^2\le 45$. Since $6^2=36\le 45$ and $7^2=49>45$, we get $\lfloor \sqrt{45}\rfloor =6$. Step 2 — scan $d=1$ to $6$, recording $(d,45/d)$ each time $d$ divides:

All factors of $45$: $1,3,5,9,15,45$ — six factors.

$\lfloor \sqrt{100}\rfloor =10$ (since $10^2=100$ exactly — so $100$ is a perfect square).

($3,6,7,8,9$ don't divide $100$.) The middle pair is $(10,10)$: we write $10$ once. Factors: $1,2,4,5,10,20,25,50,100$ — nine factors (odd ✅). This matches the rule from Perfect squares: the square root pairs with itself, so the total is odd.

$\lfloor \sqrt{37}\rfloor =6$ (since $6^2=36\le 37<49=7^2$). Test $d=1,\dots ,6$: only $1$ divides $37$ (it's odd, not divisible by $3,5$; $37/2,37/4,37/6$ all leave remainders). Factors: $1,37$ — exactly two, so $37$ is a prime (Prime numbers and composite numbers).

Reasoning: factors come in pairs $(d,n/d)$, so they'd total an even number — unless one pair has $d=n/d$, i.e. $d=\sqrt{n}$. That happens exactly when $n$ is a perfect square. That lone self-paired factor is counted once, breaking the evenness. So a perfect square always has an odd number of factors. Verify with $49$: $\lfloor \sqrt{49}\rfloor =7$. Factors: $(1,49),(7,7)$. That's $1,7,49$ — three factors, odd ✅.

WHAT the count tells us: an odd count ($3$) means $n$ is a perfect square (from L3.1). But a perfect square has many factors unless it's built as simply as possible. WHY exactly 3: the three factors must be $1$, $\sqrt{n}$, and $n$. For there to be no others, $\sqrt{n}$ itself must be prime — because if $\sqrt{n}$ had a factor $f$ with $1<f<\sqrt{n}$, then $f$ and $n/f$ would be extra factors. Conclusion: $n$ is the square of a prime, $n=p^2$. Smallest three: $2^2=4$ (factors $1,2,4$), $3^2=9$ (factors $1,3,9$), $5^2=25$ (factors $1,5,25$). (Note: $4^2=16$ is skipped because $4$ is not prime — $16$ has factors $1,2,4,8,16$, five of them.)

Apply the $\sqrt{n}$ rule to each ($\lfloor \sqrt{50}\rfloor =7$, $\lfloor \sqrt{51}\rfloor =7$, $\lfloor \sqrt{52}\rfloor =7$).

• $50$: pairs $(1,50),(2,25),(5,10)$ → factors $1,2,5,10,25,50$ → 6.
• $51$: $51=3\times 17$; pairs $(1,51),(3,17)$ → factors $1,3,17,51$ → 4.
• $52$: pairs $(1,52),(2,26),(4,13)$ → factors $1,2,4,13,26,52$ → 6. Most factors: $50$ and $52$ tie with 6 each; $51$ has fewest with $4$.

Factors of $18$ ($\lfloor \sqrt{18}\rfloor =4$): pairs $(1,18),(2,9),(3,6)$ → $\{1,2,3,6,9,18\}$. Factors of $24$ ($\lfloor \sqrt{24}\rfloor =4$): pairs $(1,24),(2,12),(3,8),(4,6)$ → $\{1,2,3,4,6,8,12,24\}$. Common factors (in both lists): $\{1,2,3,6\}$. Largest common factor (HCF): $6$. This is exactly the HCF and LCM idea — the biggest number dividing both.

Idea: dividing top and bottom by a shared factor doesn't change the fraction's value (Fractions in lowest terms). Common factors of $24$ and $36$: from their factor lists, the largest shared factor (HCF) is $12$. $\frac{24}{36}=\frac{24÷12}{36÷12}=\frac{2}{3}.$ Now $2$ and $3$ share only the factor $1$, so $\frac{2}{3}$ is in lowest terms.

Translate: each rectangle is a factor pair $(a,b)$ with $a\times b=72$. Distinct shapes = distinct unordered pairs = number of factor pairs. Scan $d=1$ to $\lfloor \sqrt{72}\rfloor =8$ (since $8^2=64\le 72<81=9^2$):

($5,7$ don't divide $72$.) That's 6 factor pairs → 6 distinct rectangles. (Check: $72$ has $12$ factors; since $72$ is not a perfect square, pairs = $12/2=6$ ✅.)

Strategy: test numbers upward, counting factors with the $\sqrt{n}$ rule, until we hit a count of $6$.

• $2$: factors $1,2$ → $2$.
• $3$: $1,3$ → $2$.
• $4$: $1,2,4$ → $3$.
• $5$: $2$. $6$: $1,2,3,6$ → $4$. $7$: $2$. $8$: $1,2,4,8$ → $4$. $9$: $1,3,9$ → $3$. $10$: $1,2,5,10$ → $4$. $11$: $2$.
• $12$: $\lfloor \sqrt{12}\rfloor =3$; pairs $(1,12),(2,6),(3,4)$ → $\{1,2,3,4,6,12\}$ → 6 factors ✅. Every number from $2$ to $11$ has fewer than $6$ factors (max was $4$), so nothing smaller qualifies. Answer: $\boxed{12}$.

Insight: the two factors are closest together right at the "middle" of the pair list — near $\sqrt{84}$. That's precisely where the $\sqrt{n}$ scan ends, so the last valid pair it records is the tightest one. $\lfloor \sqrt{84}\rfloor =9$ (since $9^2=81\le 84<100$). Scanning:

($5,8,9$ don't divide $84$.) The last pair, $(7,12)$, has the smallest difference $5$. Answer: $(7,12)$. The pairs march from "far apart" toward "close together" as $d$ climbs to $\sqrt{n}$ — so the final recorded pair is always the most balanced. ($84$ isn't a perfect square, so no exact $\sqrt{84}$ pair exists; $(7,12)$ straddles $\sqrt{84}\approx 9.17$.)

Read the set: the largest factor of any number is the number itself. So $n=12$. Verify: $\lfloor \sqrt{12}\rfloor =3$; pairs $(1,12),(2,6),(3,4)$ → $\{1,2,3,4,6,12\}$. Exactly the target set ✅. Why prime factorisation locks it in: $12=2^2\times 3$. Its factors are every product $2^a\times 3^b$ with $a\in \{0,1,2\}$ and $b\in \{0,1\}$ — that's $3\times 2=6$ combinations, matching the six-element set. No smaller number produces this exact set, because the set contains $12$, which pins the answer down completely.

Factor count of $12$? ::: $6$. Smallest number with exactly $3$ factors? ::: $4$ (it is $2^2$). HCF of $18$ and $24$? ::: $6$. Distinct rectangles with area $72$? ::: $6$. Most balanced factor pair of $84$? ::: $(7,12)$.