Number of Divisors: 10 Divisors of 21600 Divisible by 10

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The discussion focuses on determining the number of divisors of 21600 that are divisible by 10 but not by 15. The prime factorization of 21600 is established as 25 × 33 × 52. A divisor is divisible by 10 if it contains at least one factor of 5 and 2, while it is not divisible by 15 if it lacks at least one factor of 3 and 5. The approach involves using the fundamental counting principle to evaluate the choices for the parameters in the divisor's prime factorization.

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How many divisors of $21600$ are divisible by $10$ but not by $15$?
 
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In order to assist our helpers in knowing just where you are stuck, can you show what you have tried or what your thoughts are on how to begin?
 
Prime factor of $21600 = 2^5 \times 3^3 \times 5^2$

Now No. is Divisible by $10$ If It Contain at least one factor of $5$ and $2$

and No. is Non Divisible If It not Contain at least one $3$ and $5$

Now How Can I proceed after that

Thanks
 
I think you are on the right track with the prime factorization. I would write it as:

$$21600=2\cdot5\left(2^4\cdot3^3\cdot5 \right)= 10\cdot2^4\cdot3^3\cdot5$$

Now looking at the factor to the right of 10, consider a divisor of 21600 of the form:

$$2^{n_1}\cdot3^{n_2}\cdot5^{n_3}$$

What are the number of choices we have for the parameters $n_i$ such that this factor is not divisible by 3? Then apply the fundamental counting principle. What do you find?
 

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