To express \( 2625 \) in terms of its prime factors, we will start with factorization.
First, notice that \( 2625 \) is an odd number, so it is not divisible by \( 2 \). Now, let's check for divisibility by \( 3 \):
To see if \( 2625 \) is divisible by \( 3 \), we can sum the digits: \[ 2 + 6 + 2 + 5 = 15 \] Since \( 15 \) is divisible by \( 3 \), \( 2625 \) is also divisible by \( 3 \). Now divide \( 2625 \) by \( 3 \): \[ 2625 \div 3 = 875 \]
Next, we need to factor \( 875 \). Since \( 875 \) ends in \( 5 \), it is divisible by \( 5 \): \[ 875 \div 5 = 175 \]
Now, we continue with \( 175 \), which also ends in \( 5 \): \[ 175 \div 5 = 35 \]
Next, we factor \( 35 \), which again ends in \( 5 \): \[ 35 \div 5 = 7 \]
Now, we have reached \( 7 \), which is a prime number. Therefore, the complete prime factorization of \( 2625 \) is: \[ 2625 = 3^1 \times 5^3 \times 7^1 \]
With the prime factorization: \[ 2625 = 3^1 \times 5^3 \times 7^1 \]
Perfect Cube Requirement
For \( 2625M \) to be a perfect cube, all prime factors in the factorization of \( 2625M \) must have exponents that are multiples of \( 3 \).
Let us identify the required adjustments to the exponents:
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The prime factor \( 3 \) has an exponent of \( 1 \). We need it to become \( 3 \): \[ 1 \rightarrow 3 \quad \text{(add \( 2 \))} \]
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The prime factor \( 5 \) has an exponent of \( 3 \). It is already a multiple of \( 3 \): \[ 3 \rightarrow 3 \quad \text{(add \( 0 \))} \]
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The prime factor \( 7 \) has an exponent of \( 1 \). We need it to become \( 3 \): \[ 1 \rightarrow 3 \quad \text{(add \( 2 \))} \]
Constructing \( M \)
Now, we can express \( M \) in terms of the prime factors needed to achieve these adjustments:
- For \( 3 \), we need \( 3^2 \),
- For \( 5 \), we need \( 5^0 \) (no additional \( 5 \)),
- For \( 7 \), we need \( 7^2 \).
Thus, \[ M = 3^2 \times 5^0 \times 7^2 = 9 \times 1 \times 49 = 441 \]
Therefore, the value of \( M \) is: \[ \boxed{441} \]
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