Factorization, GCF and LCM

Prime factorization, the number of divisors, GCF, LCM and word problems. 6 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution.

Problem 1 Prime factorization

a) Find the number whose prime factorization is \(2^3\cdot 3^2\cdot 7\).

b) Find the prime factorization of 504, 2310 and 9900.

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a) \(8\cdot 9\cdot 7 = 504\).

b) \(504 = 2^3\cdot 3^2\cdot 7\); \(2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11\); \(9900 = 2^2\cdot 3^2\cdot 5^2\cdot 11\).

Problem 2 Number of divisors

a) Find the number of positive divisors of 72, 180 and 1000.

b) The number \(6\cdot 10^n\) has 60 positive divisors. Find \(n\).

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If \(n = p^a q^b r^c\), it has \((a + 1)(b + 1)(c + 1)\) divisors.

a) \(72 = 2^3\cdot 3^2\): 12. \(180 = 2^2\cdot 3^2\cdot 5\): 18. \(1000 = 2^3\cdot 5^3\): 16.

b) \(6\cdot 10^n = 2^{n + 1}\cdot 3\cdot 5^n\): \((n + 2)\cdot 2\cdot(n + 1) = 60\), \((n + 1)(n + 2) = 30\), \(n = 4\).

Problem 3 GCF and LCM

a) Find \(\gcd(84, 126)\) and \(\operatorname{lcm}(84, 126)\).

b) Find the GCF and the LCM of 252, 360 and 420.

c) Use the Euclidean algorithm to find \(\gcd(1071, 462)\).

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a) \(84 = 2^2\cdot 3\cdot 7\), \(126 = 2\cdot 3^2\cdot 7\): GCF \(= 2\cdot 3\cdot 7 = 42\), LCM \(= 2^2\cdot 3^2\cdot 7 = 252\).

b) \(252 = 2^2 3^2 7\), \(360 = 2^3 3^2 5\), \(420 = 2^2\cdot 3\cdot 5\cdot 7\). GCF \(= 2^2\cdot 3 = 12\), LCM \(= 2^3\cdot 3^2\cdot 5\cdot 7 = 2520\).

c) \(1071 = 2\cdot 462 + 147\), \(462 = 3\cdot 147 + 21\), \(147 = 7\cdot 21\). GCF \(= 21\).

Problem 4 Word problems with GCF

a) A shop has 48 red, 72 blue and 120 green marbles. They are put in bags with the same number in each bag and only one colour per bag. What is the least number of bags?

b) At a tournament there are 24 football players, 18 volleyball players and 30 basketball players. Rooms must have the same number of players from one team. What is the least number of rooms?

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a) Bag size \(= \gcd(48, 72, 120) = 24\). Bags: \(2 + 3 + 5 = 10\).

b) Room size \(= \gcd(24, 18, 30) = 6\). Rooms: \(4 + 3 + 5 = 12\).

Problem 5 Word problems with LCM

a) Two buses leave a station together. One returns every 45 minutes, the other every 60 minutes. After how many minutes do they meet again at the station?

b) A number leaves remainder 3 when divided by 4, remainder 4 when divided by 5 and remainder 5 when divided by 6. Find the smallest such number.

c) A number leaves remainder 1 when divided by 2, 3, 4, 5 and 6, and it is divisible by 7. Find the smallest such number.

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a) \(\operatorname{lcm}(45, 60) = 180\) minutes.

b) The number plus 1 is divisible by 4, 5 and 6: \(\operatorname{lcm} = 60\). Smallest: 59.

c) The number minus 1 is a multiple of 60: 61, 121, 181, 241, 301, … The first divisible by 7 is 301.

Problem 6 GCF · LCM

a) Find \(\gcd(18, 42)\cdot\operatorname{lcm}(18, 42)\).

b) Two numbers have GCF 12 and LCM 360. One of them is 72. Find the other.

c) Show that \(60^n\) is divisible by 12 for every \(n \in \mathbb{N}\).

Show solution

a) \(\gcd\cdot\operatorname{lcm} = 18\cdot 42 = 756\).

b) \(\frac{12\cdot 360}{72} = 60\).

c) \(60 = 12\cdot 5\), so \(60^n = 12\cdot 5\cdot 60^{n - 1}\).