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# Number Theory

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If $n = 2^{10} \cdot 3^{18} \cdot 5^{4}$, how many of the natural-number factors of n are multiples of 150?

Jul 8, 2022

#1
+1158
+10

can you properly write latex?

just copy the text inside your dollar signs and go to the latex button. then, paste the text you copied into the text box for latex and then click the OK button and voila!

Is this what you mean? and if so, what did you do to start this problem? where is your work?

$$n = 2^{10} \cdot 3^{18} \cdot 5^{4}$$

Jul 8, 2022
#2
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Note that 150 is equal to 2 * 3 * 5^2.

This means that our divisors need at least 1 2, 1 3, and 2 5s.

That means that we can select any power of 2 from 1 - 10. This gives us 10 ways.

In addition, we can select any power of 3 from 1 - 18. This gives us 18 ways.

Finally, since we need 2 5s, we can choose any power of 5 from 2-4. This counts as 3 ways.

We multiply all of these together to receive 540 ways.

Jul 8, 2022
#3
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[2^10 * 3^18 * 5^4] / 150 ==1,652,994,086,400==2^9 × 3^17 × 5^2

Number of divisors ==[9 +1] * [17 +1] * [2 + 1] ==10 x 18 x 3 ==540 divisors.

Jul 8, 2022
#4
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Guest, I don't think computing the avlues would be a great idea. We could take a more less bashy approach with C and P. But that works too! :D

Jul 8, 2022