2x is a perfect square, 3x a perfect cube, and 5x a perfect 5th power. Find the sum of the exponents in the prime factorization of the smallest such positive integer x. Thank you.
2,3,5 are coprime to each other.
Say the prime factorization is
x=(2^p)(3^q)(5^r)
where p>0,q>0,r>0 for x>0
To satisfy "2x is a perfect square" =>
mod(p+1,2)=0
mod(q,3)=0
mod(r,5)=0
=> the smallest solution is therefore p=15
Similarly for "3x is a perfect cube" =>
mod(q,2)=0
mod(q+1,3)=0
mod(q,5)=0
=> the smallest solution is q=20
For "5x is a perfect 5th power",
mod(r,2)=0
mod(r,3)=0
mod(r,5+1)=0
=> the smallest solution is 24
So
x=(2^15)(3^20)(5^24)
=6810125783203125000000000000000
and p+q+r=15+20+24=59
Thank you.
you're welcome!
To find the value of x that satisfies the given conditions, we need to factorize 2x, 3x, and 5x into their prime factors.
Let's start with 2x. We know that 2x is a perfect square, which means it must be in the form of (a^2) where a is a positive integer. So, we can write 2x as (a^2). Since 2 is already a prime number, there are no additional factors to consider.
Next, let's consider 3x. We are told that 3x is a perfect cube, so it can be expressed as (b^3) where b is a positive integer. So, we can write 3x as (b^3). Similar to 2x, 3 is a prime number and no additional factors are needed.
Moving on to 5x, we are told that 5x is a perfect 5th power. Therefore, it can be written as (c^5) where c is a positive integer. So, we can express 5x as (c^5). Again, 5 is itself a prime number and no extra factors are required.
Now, let's find the smallest integer value of x that satisfies all three conditions. In order for x to be the smallest positive integer, we need to find the smallest possible values for a, b, and c.
For a, b, and c to be the smallest possible values, we take their prime factors as 2, 3, and 5 respectively.
So, a = 2^1, b = 3^1, and c = 5^1.
Now, to find x, we multiply a, b, and c together:
x = (2^1) * (3^1) * (5^1) = 2 * 3 * 5 = 30.
The smallest positive integer value of x that satisfies the given conditions is 30.
To get the sum of the exponents in the prime factorization of x, we add the exponents of 2, 3, and 5: 1 + 1 + 1 = 3.
Therefore, the sum of the exponents in the prime factorization of the smallest possible integer x is 3.