Prime Factorization: Additional Practice (Set 1)

Halve twice, then take the $3$: $156 \div 2 = 78$, $78 \div 2 = 39$, $39 \div 3 = 13$, and $13$ is prime.

$156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13$

The form $2 \times 3 \times 13 = 78$ has one $2$ too few.

Expand the powers and multiply.

$2^4 \times 3^2 = 16 \times 9 = 144$

So the number is $144$. ($72 = 2^3 \times 3^2$ has one $2$ too few.)

It is odd, not a multiple of $3$ or $5$, and $7, 11, 13$ all leave a remainder. Test $17$: $323 \div 17 = 19$.

$323 = 17 \times 19$

Both are prime, so the factorization is complete; $17^2 = 289$, not $323$.

Read the exponent on the prime $2$.

$2^2 = 2 \times 2$

The exponent $2$ means $2$ appears twice (the value $2^2 = 4$ is not what is asked).

Expand and multiply.

$2^3 \times 7 = 8 \times 7 = 56$

So the number is $56$. ($28 = 2^2 \times 7$ has one $2$ too few, and $112 = 2^4 \times 7$ has one too many.)

It is odd, not a multiple of $3$ or $5$. Test $7$: $847 \div 7 = 121$, and $121 = 11 \times 11$.

$847 = 7 \times 11 \times 11 = 7 \times 11^2$

The form $7^2 \times 11 = 539$ squares the wrong prime.

Halve four times, then take the $3$s: $2160 \to 1080 \to 540 \to 270 \to 135$, and $135 = 3^3 \times 5$.

$2{,}160 = 2^4 \times 3^3 \times 5$

Four $2$s and three $3$s; $2^3 \times 3^3 \times 5 = 1080$ is missing a $2$.

Digit sum $15$, so take the $3$, then the $5$s, then factor: $1725 \div 3 = 575$, $575 \div 5 = 115$, $115 \div 5 = 23$.

$1{,}725 = 3 \times 5 \times 5 \times 23 = 3 \times 5^2 \times 23$

Since $23$ is prime, the factorization is complete; the others leave a composite factor.

Expand and multiply.

$3^2 \times 11 = 9 \times 11 = 99$

So the number is $99$. ($297 = 3^3 \times 11$ has one $3$ too many.)

A perfect cube is a number times itself three times, so in its prime factorization every prime must appear a number of times that is a multiple of 3 (the three equal copies split the primes evenly). Check $2^3 \times 3^3$: three 2s and three 3s, so the primes split into three identical groups.

$2^3 \times 3^3 = (2 \times 3) \times (2 \times 3) \times (2 \times 3) = 6 \times 6 \times 6 = 216$

Every exponent is a multiple of 3. Each of the others has a prime whose count is not a multiple of 3, so they are not perfect cubes.

Digit sum $6$, so divide by $3$: $3003 \div 3 = 1001$, and $1001 = 7 \times 11 \times 13$.

$3{,}003 = 3 \times 7 \times 11 \times 13$

Four distinct primes, each once; the form $3 \times 1001$ stops before primes.

The number is $2 \times 5 \times 7 = 70$, so every factor must be built from the primes $2, 5, 7$.

$21 = 3 \times 7, \text{ and } 3 \text{ is not among } 2, 5, 7$

So $21$ is not a factor, while $14 = 2 \times 7$, $35 = 5 \times 7$, and $70 = 2 \times 5 \times 7$ all are.