Identify prime numbers and write prime factorizations.
Free, timed SHSAT practice sets on primes & prime factorization— every question includes a full worked explanation, and there's no account required.
Loading questions…
A prime has exactly two factors, 1 and itself: 2, 3, 5, 7, 11, 13, 17, 19, 23, ... The number 1 is not prime, and 2 is the only even prime.
To find a prime factorization, divide by the smallest prime that works, again and again, until the last number is prime: \(60 = 2 \times 30 = 2 \times 2 \times 15 = 2 \times 2 \times 3 \times 5 = 2^{2} \times 3 \times 5\).
Use the divisibility tests to choose the next prime: even means 2, digit sum divisible by 3 means 3, last digit 0 or 5 means 5.
To check if a number is prime, test only the primes up to its square root. For 47, test 2, 3 and 5 (\(7^2 = 49\) is already too big). None divide it, so 47 is prime.
Common traps:
Divide by primes repeatedly: 80 = \(2^{4} \times 5\). \(2^{4}\) leaves out a factor of 5, so its product is 16. \(2^{3} \times 10\) stops early: 10 is not prime. \(2^{5} \times 5\) has an extra factor of 2, so its product is 160.
59 has no divisors other than 1 and itself, so it is prime. The other choices are odd and do not end in 5, so they can look prime, but each one factors: \(39 = 3 \times 13\), \(49 = 7^{2}\), \(69 = 3 \times 23\).
468 = \(2^{2} \times 3^{2} \times 13\), which has 5 prime factors counting repeats. Counting each different prime only once gives \(3\). Counting 1 as a prime factor gives \(6\). Counting every factor of 468 gives \(18\).