Prime Factorization: Practice

Expand the powers and multiply.

$2^2 \times 3 \times 7 = 4 \times 3 \times 7 = 84$

So the number is $84$. The value $42 = 2 \times 3 \times 7$ has only one $2$.

It is even: $286 \div 2 = 143$. Now $143$ is odd with digit sum $8$, so test $7$ (no), then $11$: $143 \div 11 = 13$.

$286 = 2 \times 11 \times 13$

Both $11$ and $13$ are prime, so the factorization is complete.

Read the exponent on the prime $2$.

$2^3 = 2 \times 2 \times 2$

The exponent $3$ counts the copies, so $2$ appears three times (the value $2^3 = 8$ is a different question).

Expand and multiply, smallest primes first.

$2 \times 3^2 \times 5^2 = 2 \times 9 \times 25 = 450$

So the number is $450$. ($150 = 2 \times 3 \times 5^2$ is missing a factor of $3$.)

Take three $2$s, then three $5$s: $1000 \to 500 \to 250 \to 125$, and $125 = 5^3$.

$1{,}000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 = 2^3 \times 5^3$

The forms $10^3$ and $8 \times 125$ are equal in value but not built from primes.

Its digit sum is $18$, so take the $3$s, then the $5$s: $675 \div 3 = 225$, $225 \div 3 = 75$, $75 \div 3 = 25$, $25 = 5^2$.

$675 = 3 \times 3 \times 3 \times 5 \times 5 = 3^3 \times 5^2$

The form $3^2 \times 5^2 = 225$ has one $3$ too few.

Finish each: Jada gets $4 \times 75 = 2^2 \times (3 \times 5^2)$; Theo gets $6 \times 50 = (2 \times 3) \times (2 \times 5^2)$.

$300 = 2 \times 2 \times 3 \times 5 \times 5 = 2^2 \times 3 \times 5^2$

By the Fundamental Theorem of Arithmetic the starting split cannot change the answer, so both reach the same factorization.

It is odd, not a multiple of $3$ or $5$. Test $7$: $539 \div 7 = 77$, and $77 = 7 \times 11$.

$539 = 7 \times 7 \times 11 = 7^2 \times 11$

The form $7 \times 11^2 = 847$ squares the wrong prime.

Test the small primes in turn: $1001 \div 7 = 143$, and $143 = 11 \times 13$.

$1{,}001 = 7 \times 11 \times 13$

Three distinct primes, each once; the forms $7 \times 143$ and $11 \times 91$ are not finished.

Halve three times, then take the $3$s: $1080 \to 540 \to 270 \to 135$, and $135 = 27 \times 5 = 3^3 \times 5$.

$1{,}080 = 2^3 \times 3^3 \times 5$

Three $2$s and three $3$s; $2^3 \times 3^2 \times 5 = 360$ is missing a factor of $3$.

Halve down to an odd number, then factor it: $784 \to 392 \to 196 \to 98 \to 49$, and $49 = 7 \times 7$.

$784 = 2^4 \times 7^2$

That is four $2$s and two $7$s; $16 \times 49$ is the same value but not written with primes.

A perfect square is a number times itself, so in its prime factorization every prime must appear an even number of times (the two equal copies split the primes evenly). Check $2^2 \times 3^4$: two 2s and four 3s, both even, so the primes pair into two identical groups.

$2^2 \times 3^4 = (2 \times 3 \times 3) \times (2 \times 3 \times 3) = 18 \times 18 = 324$

Every exponent is even. Each of the others has a prime with an odd exponent (a leftover that cannot be paired), so they are not perfect squares.