MHB Number of Divisors: 10 Divisors of 21600 Divisible by 10

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To determine how many divisors of 21600 are divisible by 10 but not by 15, the prime factorization of 21600 is essential: 21600 = 2^5 × 3^3 × 5^2. A divisor is divisible by 10 if it includes at least one factor of both 2 and 5. However, to ensure the divisor is not divisible by 15, it must exclude the factor of 3. By analyzing the combinations of the remaining factors, particularly focusing on the constraints for n_1, n_2, and n_3, one can apply the fundamental counting principle to find the total number of valid divisors. This approach leads to the solution for the problem posed.
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How many divisors of $21600$ are divisible by $10$ but not by $15$?
 
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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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