What is the smallest perfect cube that has a factor of 4725?
Factors of 4725 are 3,3,3,5,5,7
I have 3^3 and 5^2 and 7. So I need 1 more 5 and 2 more of 7
So: 3^3*5^3*7^3
= 1157625
When I cube root 1157625 it is 105. So is this the answer?
The smallest perfect cube that has a factor of 4725 is 1157625
looks good to me.
Ok thank you!
Yes, you are correct. The smallest perfect cube that has a factor of 4725 is 1157625.
To find the smallest perfect cube that has a factor of 4725, we need to prime factorize 4725. We can start by observing that the prime factorization of 4725 is 3 * 3 * 3 * 5 * 5 * 7.
To create a perfect cube, we need each prime factor to have an exponent that is a multiple of 3. Currently, the exponents are 1, 1, 1, 2, 2, and 1, respectively. We have excess factors of 5 and 7, so we need to increase their exponents to be multiples of 3.
Since we need an additional factor of 5, we can increase the exponent to 3. This means we now have 3 * 3 * 3 * 5 * 5 * 5 * 7.
Additionally, we need to increase the exponent of 7 by 2 to make it a multiple of 3. This gives us 3 * 3 * 3 * 5 * 5 * 5 * 7 * 7 * 7.
Now we have a perfect cube with each prime factor having an exponent of 3. To find the value of this cube, we multiply the prime factors together: 3 * 3 * 3 * 5 * 5 * 5 * 7 * 7 * 7 = 1157625.
To determine if this is the smallest perfect cube that has a factor of 4725, we can take the cube root of 1157625. The cube root of 1157625 is approximately 105. Since this is the smallest integer cube root, we can conclude that the smallest perfect cube that has a factor of 4725 is 1157625.