1) \(12 = \color{red}{2 \times 2 \times 3}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(12 \div 2 = 6\)
\(6 \div 2 = 3\)
\(3 \div 3 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(12 = \color{red}{2 \times 2 \times 3}\).
2) \(18 = \color{red}{2 \times 3 \times 3}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(18 \div 2 = 9\)
\(9 \div 3 = 3\)
\(3 \div 3 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(18 = \color{red}{2 \times 3 \times 3}\).
3) \(25 = \color{red}{5 \times 5}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(25 \div 5 = 5\)
\(5 \div 5 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(25 = \color{red}{5 \times 5}\).
4) \(36 = \color{red}{2 \times 2 \times 3 \times 3}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(36 \div 2 = 18\)
\(18 \div 2 = 9\)
\(9 \div 3 = 3\)
\(3 \div 3 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(36 = \color{red}{2 \times 2 \times 3 \times 3}\).
5) \(49 = \color{red}{7 \times 7}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(49 \div 7 = 7\)
\(7 \div 7 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(49 = \color{red}{7 \times 7}\).
6) \(54 = \color{red}{2 \times 3 \times 3 \times 3}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(54 \div 2 = 27\)
\(27 \div 3 = 9\)
\(9 \div 3 = 3\)
\(3 \div 3 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(54 = \color{red}{2 \times 3 \times 3 \times 3}\).
7) \(64 = \color{red}{2 \times 2 \times 2 \times 2 \times 2 \times 2}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(64 \div 2 = 32\)
\(32 \div 2 = 16\)
\(16 \div 2 = 8\)
\(8 \div 2 = 4\)
\(4 \div 2 = 2\)
\(2 \div 2 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(64 = \color{red}{2 \times 2 \times 2 \times 2 \times 2 \times 2}\).
8) \(75 = \color{red}{3 \times 5 \times 5}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(75 \div 3 = 25\)
\(25 \div 5 = 5\)
\(5 \div 5 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(75 = \color{red}{3 \times 5 \times 5}\).
9) \(96 = \color{red}{2 \times 2 \times 2 \times 2 \times 2 \times 3}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(96 \div 2 = 48\)
\(48 \div 2 = 24\)
\(24 \div 2 = 12\)
\(12 \div 2 = 6\)
\(6 \div 2 = 3\)
\(3 \div 3 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(96 = \color{red}{2 \times 2 \times 2 \times 2 \times 2 \times 3}\).
10) \(121 = \color{red}{11 \times 11}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(121 \div 11 = 11\)
\(11 \div 11 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(121 = \color{red}{11 \times 11}\).
11) \(144 = \color{red}{2 \times 2 \times 2 \times 2 \times 3 \times 3}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(144 \div 2 = 72\)
\(72 \div 2 = 36\)
\(36 \div 2 = 18\)
\(18 \div 2 = 9\)
\(9 \div 3 = 3\)
\(3 \div 3 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(144 = \color{red}{2 \times 2 \times 2 \times 2 \times 3 \times 3}\).
12) \(168 = \color{red}{2 \times 2 \times 2 \times 3 \times 7}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(168 \div 2 = 84\)
\(84 \div 2 = 42\)
\(42 \div 2 = 21\)
\(21 \div 3 = 7\)
\(7 \div 7 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(168 = \color{red}{2 \times 2 \times 2 \times 3 \times 7}\).
13) \(210 = \color{red}{2 \times 3 \times 5 \times 7}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(210 \div 2 = 105\)
\(105 \div 3 = 35\)
\(35 \div 5 = 7\)
\(7 \div 7 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(210 = \color{red}{2 \times 3 \times 5 \times 7}\).
14) \(225 = \color{red}{3 \times 3 \times 5 \times 5}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(225 \div 3 = 75\)
\(75 \div 3 = 25\)
\(25 \div 5 = 5\)
\(5 \div 5 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(225 = \color{red}{3 \times 3 \times 5 \times 5}\).
15) \(256 = \color{red}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(256 \div 2 = 128\)
\(128 \div 2 = 64\)
\(64 \div 2 = 32\)
\(32 \div 2 = 16\)
\(16 \div 2 = 8\)
\(8 \div 2 = 4\)
\(4 \div 2 = 2\)
\(2 \div 2 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(256 = \color{red}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}\).
16) \(315 = \color{red}{3 \times 3 \times 5 \times 7}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(315 \div 3 = 105\)
\(105 \div 3 = 35\)
\(35 \div 5 = 7\)
\(7 \div 7 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(315 = \color{red}{3 \times 3 \times 5 \times 7}\).
17) \(420 = \color{red}{2 \times 2 \times 3 \times 5 \times 7}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(420 \div 2 = 210\)
\(210 \div 2 = 105\)
\(105 \div 3 = 35\)
\(35 \div 5 = 7\)
\(7 \div 7 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(420 = \color{red}{2 \times 2 \times 3 \times 5 \times 7}\).
18) \(512 = \color{red}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(512 \div 2 = 256\)
\(256 \div 2 = 128\)
\(128 \div 2 = 64\)
\(64 \div 2 = 32\)
\(32 \div 2 = 16\)
\(16 \div 2 = 8\)
\(8 \div 2 = 4\)
\(4 \div 2 = 2\)
\(2 \div 2 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(512 = \color{red}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}\).
19) \(693 = \color{red}{3 \times 3 \times 7 \times 11}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(693 \div 3 = 231\)
\(231 \div 3 = 77\)
\(77 \div 7 = 11\)
\(11 \div 11 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(693 = \color{red}{3 \times 3 \times 7 \times 11}\).
20) \(840 = \color{red}{2 \times 2 \times 2 \times 3 \times 5 \times 7}\)
Solution:
Step 1: Start with the smallest prime divisor and keep dividing by primes until the quotient is \(1\).
\(840 \div 2 = 420\)
\(420 \div 2 = 210\)
\(210 \div 2 = 105\)
\(105 \div 3 = 35\)
\(35 \div 5 = 7\)
\(7 \div 7 = 1\)
Step 2: The divisors used are all prime, so the prime factorization is \(840 = \color{red}{2 \times 2 \times 2 \times 3 \times 5 \times 7}\).