Exercises — Prime factorization — factor trees, ladder method

A prime is a natural number $>1$ whose only divisors are $1$ and itself.

• $1$ — has just one divisor, so it is a unit, not prime.
• $2$ — divisors $1,2$ only → prime.
• $9=3\times 3$ → composite.
• $17$ — test $2,3$ ($\sqrt{17}\approx 4.1$, stop at $4$): $17$ odd, digit-sum $8$ not divisible by $3$ → prime.
• $21=3\times 7$ → composite.
• $23$ — test $2,3$ ($\sqrt{23}\approx 4.8$): odd, digit-sum $5$ → prime.
• $51=3\times 17$ (digit-sum $6$) → composite.

Primes: $2,17,23.$

$25=5\times 5$, so the box is $==5==$. Leaves are $2,5,5$. $50=2\cdot 5^2.$

Always divide by the smallest prime that fits, top to bottom.

 2 | 180
 2 |  90
 3 |  45
 3 |  15
 5 |   5
   |   1

• $180$ even → ÷2 = $90$; still even → ÷2 = $45$. Now odd, stop dividing by $2$.
• $45$: digit-sum $9$ → ÷3 = $15$ → ÷3 = $5$. Then ÷5 = $1$. $180=2^2\cdot 3^2\cdot 5.$

 3 | 1155
 5 |  385
 7 |   77
11 |   11
   |    1

• $1155$ is odd (skip $2$). Digit-sum $12$ → ÷3 = $385$.
• $385$ ends in $5$ → ÷5 = $77$.
• $77=7\times 11$ → ÷7 = $11$, ÷11 = $1$. $1155=3\cdot 5\cdot 7\cdot 11.$

Halve repeatedly: $256\to 128\to 64\to 32\to 16\to 8\to 4\to 2\to 1$ — that is eight halvings. $256=2^8.$

Multiply the left-column primes: $2\cdot 2\cdot 2\cdot 3\cdot 5=8\cdot 15=120$, i.e. $120=2^3\cdot 3\cdot 5.$ Using the Number of Divisors rule $d(n)=(a_1+1)(a_2+1)\cdots$: $d(120)=(3+1)(1+1)(1+1)=4\cdot 2\cdot 2=16.$

We need the exponent-plus-one factors to multiply to $12$. List the ways to write $12$ as a product, subtract $1$ from each factor to get exponents, and put the biggest exponent on the smallest prime (that keeps the number small).

The smallest is $==60===2^2\cdot 3\cdot 5$, with $d(60)=(2+1)(1+1)(1+1)=12$. ✅

Divisible by $9=3^2$ means $b\ge 2$. Divisor count $(a+1)(b+1)=6$.

• $b=2\Rightarrow (a+1)\cdot 3=6\Rightarrow a+1=2\Rightarrow a=1$. Gives $n=2^1\cdot 3^2=18$.
• $b=5\Rightarrow (a+1)\cdot 6=6\Rightarrow a=0$. Gives $n=3^5=243$.
• $b=3,4$ give $(b+1)=4,5$, neither divides $6$ → rejected.

Answers: $n=18$ or $n=243.$

From E3 and E4: $180=2^2\cdot 3^2\cdot 5$, and $1155=3\cdot 5\cdot 7\cdot 11$. Line the primes up (missing prime = exponent $0$):

$\text{HCF}=3^1\cdot 5^1=15,\text{LCM}=2^2\cdot 3^2\cdot 5\cdot 7\cdot 11=13860.$ Check: $15\times 13860=207900$, and $180\times 1155=207900.$ ✅ (See HCF and LCM.)

Cancel the shared primes (that's exactly the HCF, $15$ from E9): $\frac{180}{1155}=\frac{2^2\cdot 3^2\cdot 5}{3\cdot 5\cdot 7\cdot 11}=\frac{2^2\cdot 3}{7\cdot 11}=\frac{12}{77}.$ $\gcd (12,77)=1$, so $\frac{12}{77}$ is fully reduced. (See Simplifying Fractions.)

The identity $\text{HCF}\times \text{LCM}=a\times b$ gives $b=\frac{\text{HCF}\times \text{LCM}}{a}=\frac{12\times 360}{72}=\frac{4320}{72}=60.$ Sanity check with primes: $72=2^3\cdot 3^2$, $60=2^2\cdot 3\cdot 5$. Min exponents $2^2\cdot 3^1=12$ ✅; max $2^3\cdot 3^2\cdot 5=360$ ✅.

A trailing zero is one factor of $10=2\times 5$. So the number of trailing zeros is the number of paired $2$s and $5$s, i.e. $\min (\text{count of }2,\text{count of }5)$. $180\times 1155=(2^2\cdot 3^2\cdot 5)(3\cdot 5\cdot 7\cdot 11)=2^2\cdot 3^3\cdot 5^2\cdot 7\cdot 11.$ Count of $2$s $=2$, count of $5$s $=2$ → $\min (2,2)=2$. Two trailing zeros. (Indeed $207900$ ends in $00$.)

A perfect square has every prime exponent even (a square pairs each atom with a twin). Factorize:

 2 | 720   2 | 360   2 | 180   2 | 90   3 | 45   3 | 15   5 | 5   | 1

$720=2^4\cdot 3^2\cdot 5^1.$ Exponents: $2\to 4$ (even ✅), $3\to 2$ (even ✅), $5\to 1$ (odd ✗). Only the $5$ is unpaired, so multiply by one more $5$: $k=5,720\cdot 5=3600=60^2.$

A perfect cube needs every exponent to be a multiple of $3$ (each atom appears in triples). From $720=2^4\cdot 3^2\cdot 5^1$, push each exponent up to the next multiple of $3$:

• $2$: $4\to 6$, need $2^{6-4}=2^2$.
• $3$: $2\to 3$, need $3^{3-2}=3^1$.
• $5$: $1\to 3$, need $5^{3-1}=5^2$. $m=2^2\cdot 3\cdot 5^2=4\cdot 3\cdot 25=300.$ Check: $720\cdot 300=216000=60^3$. ✅

Smallest number with exactly 12 divisors? ::: $60=2^2\cdot 3\cdot 5$. HCF and LCM of $180$ and $1155$? ::: $\text{HCF}=15$, $\text{LCM}=13860$. $\frac{180}{1155}$ in lowest terms? ::: $\frac{12}{77}$. Trailing zeros of $180\times 1155$? ::: $2$. Smallest $k$ making $720k$ a perfect square? ::: $k=5$. Smallest $m$ making $720m$ a perfect cube? ::: $m=300$.