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Which of the following is the lowest positive integer that is divisible by \(\mathrm{2}\), \(\mathrm{3}\), \(\mathrm{4}\), \(\mathrm{5}\), \(\mathrm{6}\), \(\mathrm{7}\), \(\mathrm{8}\), and \(\mathrm{9}\)?
Let's think about what this question is really asking. Imagine you have 8 different boxes, and you want to find the smallest pile of items that you can divide evenly among all these boxes without any items left over.
When a number is "divisible by" another number, it means you can divide it evenly with no remainder. So we need the smallest positive number that can be divided evenly by 2, 3, 4, 5, 6, 7, 8, and 9.
This is exactly what mathematicians call the Least Common Multiple (LCM) - the smallest number that all our given numbers can divide into evenly.
Process Skill: TRANSLATE - Converting the divisibility requirement into the LCM concept
To find the LCM efficiently, we need to think of each number as being built from smaller building blocks called prime numbers. Just like how any object can be broken down into atoms, any number can be broken down into prime factors.
Let's break down each number:
Now we can see the "building blocks" we need: powers of 2, powers of 3, and the primes 5 and 7.
Here's the key insight: to make sure our answer is divisible by ALL the given numbers, we need enough of each prime factor to satisfy the most demanding requirement.
Looking at our prime factorizations:
Think of it like this: if you need to satisfy 8 different people's requirements, you need to meet the highest standard among all of them.
Process Skill: INFER - Recognizing that we need the highest power of each prime factor
Now we multiply together the highest powers of all prime factors:
\(\mathrm{LCM} = 2^3 \times 3^2 \times 5^1 \times 7^1\)
Let's calculate step by step:
\(\mathrm{LCM} = 8 \times 9 \times 5 \times 7\)
\(= 72 \times 5 \times 7\)
\(= 360 \times 7\)
\(= 2,520\)
The smallest positive integer divisible by 2, 3, 4, 5, 6, 7, 8, and 9 is 2,520.
Looking at our answer choices, this matches option C. 2,520.
We can verify: \(2,520 \div 2 = 1,260\), \(2,520 \div 3 = 840\), \(2,520 \div 4 = 630\), \(2,520 \div 5 = 504\), \(2,520 \div 6 = 420\), \(2,520 \div 7 = 360\), \(2,520 \div 8 = 315\), and \(2,520 \div 9 = 280\). All divisions result in whole numbers, confirming our answer.
1. Misunderstanding the concept of "divisible by all numbers"
Students often think they need to multiply all the given numbers together (\(2\times3\times4\times5\times6\times7\times8\times9\)) to find a number divisible by all of them. This leads to an unnecessarily large number and doesn't leverage the mathematical concept of LCM. The key insight is recognizing that we need the smallest such number, which is the LCM.
2. Attempting to use trial and error with the answer choices
Some students may try to test each answer choice by dividing it by all given numbers (2,3,4,5,6,7,8,9) instead of using the systematic LCM approach. While this might work, it's time-consuming and doesn't build conceptual understanding. More importantly, it's prone to arithmetic errors when checking 8 divisions for each answer choice.
3. Forgetting that some numbers share common factors
Students might not realize that numbers like 4, 6, 8, and 9 can be broken down into smaller prime factors that overlap with other given numbers. For example, if a number is divisible by 8, it's automatically divisible by 4 and 2. This leads them to count factors multiple times unnecessarily.
1. Incorrect prime factorization
Students often make errors when breaking down composite numbers into prime factors. Common mistakes include: writing \(8 = 2^2\) instead of \(2^3\), or writing \(9 = 3^3\) instead of \(3^2\). These errors in the foundation step will carry through and give an incorrect final answer.
2. Taking the wrong power of prime factors
Even with correct prime factorizations, students may take the minimum power instead of the maximum power of each prime factor. For instance, seeing that 2 appears as \(2^1\) (from 2), \(2^2\) (from 4), and \(2^3\) (from 8), they might incorrectly take \(2^1\) instead of \(2^3\). The LCM requires the highest power of each prime to ensure divisibility by all original numbers.
3. Arithmetic calculation errors
When computing \(2^3 \times 3^2 \times 5^1 \times 7^1 = 8 \times 9 \times 5 \times 7\), students may make multiplication errors in the step-by-step calculation. For example: calculating \(8 \times 9 = 81\) instead of 72, or \(360 \times 7 = 2,250\) instead of 2,520.
1. Misreading the calculated result
After correctly calculating 2,520, students might misread their own work and select a similar-looking number from the answer choices, such as 2,504 if it were an option, or confuse the order of digits.
2. Second-guessing the systematic approach
Students may correctly calculate 2,520 but then doubt their systematic LCM method because other answer choices seem "close enough." They might switch to a larger number like 3,024, thinking they made the number "too small" to be divisible by all the given numbers.