Prime factorization
Factor a number into primes and write it with powers, from two-digit to five-digit numbers. If the number is prime, say so.
How to solve it
Every natural number greater than one is a product of primes, in exactly one way up to the order of the factors. That is the fundamental theorem of arithmetic. The factorization is a number's passport: all its divisors, the gcd and the lcm can be read from it.
Step by step
- Divide by the smallest prime divisor as long as it divides: first $2$, then $3$, $5$, $7$, $11$, …
- Divisibility tests help: by $2$, the last digit is even; by $3$, the digit sum is divisible by $3$; by $5$, it ends in $0$ or $5$; by $11$, the alternating sum of the digits is divisible by $11$.
- Try divisors only while their square is not larger than what is left. If none divides, the rest is prime.
- Collect equal factors into powers: $2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 = 2^3 \cdot 3^3$.
Common mistakes
- Leaving a composite number in the answer: $2^2 \cdot 21$ is not done yet, $21 = 3 \cdot 7$.
- Dividing by $2$ once and moving on although it divides again: $216 = 2^3 \cdot 27$, not $2 \cdot 108$.
- Stopping before the square root and calling $91 = 7 \cdot 13$ or $221 = 13 \cdot 17$ prime.
- Including $1$ in the factorization: it is not a prime.