What is the smallest number by which 243 is multiplied to get a perfect cube?

Let's break down this math puzzle and find the solution together. You're looking to take the number 243 and multiply it by the smallest possible whole number to make the result a "perfect cube." Don't worry if "perfect cube" sounds a little intimidating; we'll explain it clearly.

Understanding Perfect Cubes

First, what exactly is a perfect cube? A perfect cube is a number that can be obtained by multiplying a whole number by itself three times. In other words, it's a number that is the result of cubing an integer (raising it to the power of 3).

Here are some examples of perfect cubes:

1 x 1 x 1 = 1 (1 cubed)
2 x 2 x 2 = 8 (2 cubed)
3 x 3 x 3 = 27 (3 cubed)
4 x 4 x 4 = 64 (4 cubed)
5 x 5 x 5 = 125 (5 cubed)
10 x 10 x 10 = 1000 (10 cubed)

When we talk about making a number a perfect cube by multiplying, we need to look at its prime factors. Prime factors are the smallest prime numbers that multiply together to give you that number. Think of them as the building blocks of a number.

Finding the Prime Factors of 243

Our starting number is 243. Let's find its prime factors. We can do this by repeatedly dividing by the smallest possible prime numbers.

Is 243 divisible by 2? No, because it's an odd number.

Is 243 divisible by 3? Yes. Let's divide:

243 ÷ 3 = 81

Now let's look at 81. Is it divisible by 3?

81 ÷ 3 = 27

We have 27. Is it divisible by 3?

27 ÷ 3 = 9

And 9 is also divisible by 3:

9 ÷ 3 = 3

Finally, 3 is a prime number, so it's only divisible by itself.

3 ÷ 3 = 1

So, the prime factorization of 243 is: 3 x 3 x 3 x 3 x 3.

We can write this using exponents as 3⁵.

What We Need for a Perfect Cube

Remember, for a number to be a perfect cube, each of its prime factors must appear a number of times that is a multiple of 3 (like 3, 6, 9, etc.).

In our prime factorization of 243 (3 x 3 x 3 x 3 x 3, or 3⁵), the prime factor 3 appears 5 times.

To make this a perfect cube, we need the exponent of 3 to be a multiple of 3. The next multiple of 3 after 5 is 6.

So, we want our prime factor 3 to appear 6 times. Currently, it appears 5 times. How many more times do we need to multiply by 3?

We need 6 factors of 3, and we have 5. That means we need 1 more factor of 3.

6 - 5 = 1

The Smallest Number to Multiply By

To get that additional factor of 3, we need to multiply 243 by 3.

Let's check our work:

Original number: 243 (which is 3⁵)

Number we're multiplying by: 3 (which is 3¹)

New number: 243 x 3 = 729

In terms of exponents, this is 3⁵ x 3¹ = 3⁽⁵⁺¹⁾ = 3⁶.

Is 3⁶ a perfect cube? Yes! It means 3 x 3 x 3 x 3 x 3 x 3. We can group these factors into sets of three:

(3 x 3 x 3) x (3 x 3 x 3)

This is 27 x 27, which equals 729. Or, in terms of cubes, it's (3²)³ = 9³, which is 9 x 9 x 9 = 729.

So, the smallest number by which 243 is multiplied to get a perfect cube is 3.

Summary of the Process:

Understand what a perfect cube is: a number that is the result of an integer multiplied by itself three times (e.g., 8 is 2x2x2).
Find the prime factorization of the given number (243).
The prime factorization of 243 is 3 x 3 x 3 x 3 x 3, or 3⁵.
For a number to be a perfect cube, each of its prime factors must have an exponent that is a multiple of 3.
In 3⁵, the exponent 5 is not a multiple of 3. The next multiple of 3 is 6.
To reach an exponent of 6, we need one more factor of 3 (3⁶).
Therefore, we must multiply 243 by 3.

The resulting number is 729, which is a perfect cube (9 x 9 x 9).

Key takeaway: To find the smallest number to multiply by to make a number a perfect cube, identify its prime factors. For each prime factor, count how many times it appears. If the count is not a multiple of 3, add just enough factors of that prime number to make the count a multiple of 3. The product of these added factors is your answer.

Frequently Asked Questions (FAQ)

How do I find the prime factorization of a number?

To find the prime factorization, start by dividing the number by the smallest prime number (2) as many times as possible. Then, move to the next prime number (3) and divide as many times as possible. Continue this process with successive prime numbers (5, 7, 11, and so on) until you are left with 1. The prime numbers you used as divisors are the prime factors.

Why do prime factors need to have exponents that are multiples of 3 for a perfect cube?

A perfect cube is formed by cubing an integer. When you cube an integer, say 'n', you get n x n x n. If the prime factorization of 'n' is p₁^a * p₂^b ..., then 'n' cubed becomes (p₁^a * p₂^b ...)³ = p₁^3a * p₂^3b .... As you can see, the exponents of the prime factors in a perfect cube are always multiples of 3.

What if a prime factor appears 3 times already?

If a prime factor already appears 3 times (or any multiple of 3, like 6 or 9), you don't need to add any more of that factor to make it a perfect cube. For example, if you had the number 8 (which is 2³), it's already a perfect cube, so you would multiply it by 1.

Can you give another example?

Certainly! Let's say we want to make the number 54 a perfect cube. First, find its prime factorization: 54 = 2 x 3 x 3 x 3 = 2¹ x 3³. The prime factor 3 already has an exponent of 3 (a multiple of 3), so it's good. However, the prime factor 2 has an exponent of 1, which is not a multiple of 3. The next multiple of 3 after 1 is 3. So, we need two more factors of 2 (2³). Therefore, we need to multiply 54 by 2 x 2 = 4. The new number will be 54 x 4 = 216, which is 6³ (2 x 3)³ = 2³ x 3³.