Prime Factorization: Additional Practice (Set 2)

Take the $2$s, then the $3$s, then the $5$: $360 \to 180 \to 90 \to 45$, then $45 = 3^2 \times 5$.

$360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 2^3 \times 3^2 \times 5$

Three $2$s and two $3$s; the other options miss one of those counts.

Its digit sum is $9$, so take the $3$s, then factor: $441 \div 3 = 147$, $147 \div 3 = 49$, $49 = 7 \times 7$.

$441 = 3 \times 3 \times 7 \times 7 = 3^2 \times 7^2$

The forms $9 \times 49$ and $3^2 \times 49$ keep a composite factor.

Its digit sum is $6$, so divide by $3$: $231 \div 3 = 77$, and $77 = 7 \times 11$.

$231 = 3 \times 7 \times 11$

Each prime appears once; the forms $3 \times 77$ and $7 \times 33$ leave a composite factor.

Digit sum $9$, so take the $3$s, then the $5$, then factor: $2205 \div 3 = 735$, $735 \div 3 = 245$, $245 \div 5 = 49$, $49 = 7^2$.

$2{,}205 = 3^2 \times 5 \times 7^2$

The other forms leave a composite factor such as $49$ or $245$.

It ends in $5$, so divide by $5$: $385 \div 5 = 77$, and $77 = 7 \times 11$.

$385 = 5 \times 7 \times 11$

The forms $5 \times 77$ and $7 \times 55$ stop before reaching primes.

Halve twice, then take the $3$s: $1188 \div 2 = 594$, $594 \div 2 = 297$, $297 \div 3 = 99$, $99 \div 3 = 33$, $33 \div 3 = 11$.

$1{,}188 = 2^2 \times 3^3 \times 11$

That is two $2$s, three $3$s, and one $11$; the other forms stop before primes.

Halve twice, then take the $3$s: $612 \div 2 = 306$, $306 \div 2 = 153$, $153 \div 3 = 51$, $51 \div 3 = 17$.

$612 = 2 \times 2 \times 3 \times 3 \times 17 = 2^2 \times 3^2 \times 17$

Since $17$ is prime, the tree ends; $2^2 \times 3 \times 17 = 204$ drops a $3$.

Finish each: Liam gets $8 \times 63 = 2^3 \times (3^2 \times 7)$; Nora gets $9 \times 56 = 3^2 \times (2^3 \times 7)$.

$504 = 2^3 \times 3^2 \times 7$

By the Fundamental Theorem of Arithmetic both routes reach the same primes, so the factorization is shared.

A factor must be built only from the primes that appear: $2, 3, 5, 7$. The number is $210$.

$11 \nmid 210, \text{ since } 11 \text{ is not among the primes } 2, 3, 5, 7$

So $11$ is not a factor, while $35 = 5 \times 7$, $21 = 3 \times 7$, and $30 = 2 \times 3 \times 5$ all are.

Keep halving: $4096 \to 2048 \to 1024 \to 512 \to 256 \to 128 \to 64 \to 32 \to 16 \to 8 \to 4 \to 2 \to 1$, twelve halvings in all.

$4{,}096 = 2^{12}$

So $4{,}096$ is a power of $2$ alone, with the prime $2$ used twelve times.

Peel off small primes: $2310 \div 2 = 1155$, $1155 \div 3 = 385$, $385 \div 5 = 77$, $77 = 7 \times 11$.

$2{,}310 = 2 \times 3 \times 5 \times 7 \times 11$

This is the product of the first five primes, each once.

Divide by small primes: $858 \div 2 = 429$, $429 \div 3 = 143$, $143 = 11 \times 13$.

$858 = 2 \times 3 \times 11 \times 13$

Four distinct primes, each once; the other forms leave a composite factor such as $143$ or $429$.