The surface area and the volume of a sphere are both 3 digit integers times π. If r is the radius of the sphere, how many integral values can be found for r?
4 pi r^2 = (xyz)pi
(4/3) pi r^3 = (ijk) pi
4 r^2 = xyz
(4/3) r^3 = ijk
(4/3) r^3 is an integer less than 1000
and greater than 99
if (4/3) r^3 = 1000 r is 9.08 so try 9
if r = 9, (4/3)r^3 = 972 , the biggest we can have for r^3 criterion
now if (4/3) r^3 = 100 r is 4.21
so smallest r for this criterion is
r = 5
so far we have
5 , 6, 7, 8 , 9
BUT we need to do the area thing
4 r^2 = xyz
if xyz = 1000, r = 15.8
So
15 is the biggest for the area
if xyz = 100, r = 5 exactly
so between 5 and 15
so
5,6,7,8,9 seem to work, five of them
But 4/3 r^3 must be an integer, so r must be a multiple of 3.
I'd say only 6 and 9 are available
Yes, of course, sorry
To solve this problem, we need to use the formulas for the surface area and volume of a sphere and then determine how many integral values of the radius can satisfy the given conditions.
The surface area of a sphere is given by the formula:
A = 4πr²
The volume of a sphere is given by the formula:
V = (4/3)πr³
We are given that both the surface area and volume of the sphere are integers times π. This means that the values for A and V must be whole numbers or multiples of π.
Let's express the given conditions mathematically:
A = k₁π (where k₁ is an integer)
V = k₂π (where k₂ is an integer)
Substituting the formulas for surface area and volume, we get:
4πr² = k₁π
(4/3)πr³ = k₂π
Simplifying the equations, we have:
4r² = k₁
(4/3)r³ = k₂
For an integer value of r, k₁ and k₂ must also be integers.
Now, let's analyze the conditions for k₁ and k₂ so that both 4r² and (4/3)r³ give integer values:
1. For 4r² to be an integer, r must be either an integer or a fraction with a denominator that is a power of 2.
2. For (4/3)r³ to be an integer, r must be either an integer or a fraction with a denominator that is a power of 3.
Since r must satisfy both conditions, it must be an integer or a fraction with a denominator that is both a power of 2 and a power of 3. In other words, r must be a rational number of the form p/q, where p is an integer, q is a power of 2, and q is also a power of 3.
To count the number of integral values that can be found for r, we need to find the number of possible values for p and q.
Since r is constrained by both powers of 2 and powers of 3, we can narrow down our analysis by considering the prime factorization of q. Let's consider the possible values of q:
- q = 2^0 * 3^0 (when q = 1): In this case, r = p, where p can be any integer. So, there is an infinite number of integral values for r.
- q = 2^1 * 3^0 (when q = 2): In this case, r = p/2, where p can be any even integer. So, there are infinite integral values for r.
- q = 2^0 * 3^1 (when q = 3): In this case, r = p/3, where p can be any integer divisible by 3. So, there are infinite integral values for r.
- q = 2^1 * 3^1 (when q = 6): In this case, r = p/6, where p can be any integer divisible by 6. So, there are infinite integral values for r.
We can continue this analysis for larger powers of 2 and 3, but since each additional power will introduce additional constraints, we can conclude that there is an infinite number of integral values for r.
Therefore, the answer is that there are infinitely many integral values that can be found for r.