find the least whole number by which 2^5x5^4x7^3 must be multiplied with to get a perfect cube. what is the cube root of the resulting number?
thank you😌😌
all the powers must be multiples of 3, so that would need 2^1x5^2 = 50
(2^2 * 5^2 ^ 7)^3 = 700^3
To find the least whole number that must be multiplied with 2^5 * 5^4 * 7^3 to get a perfect cube, we need to determine the missing powers of the prime factors.
First, let's write the prime factorization of the number 2^5 * 5^4 * 7^3:
2^5 * 5^4 * 7^3 = (2 * 2 * 2 * 2 * 2) * (5 * 5 * 5 * 5) * (7 * 7 * 7)
For a number to be a perfect cube, all the prime factors must have powers that are multiples of 3.
Looking at the exponents of the prime factors, we can see that the exponent of 2 is not a multiple of 3 (5 is not divisible by 3), the exponent of 5 is not a multiple of 3 (4 is not divisible by 3), and the exponent of 7 is not a multiple of 3 (3 is not divisible by 3).
To make all the exponents a multiple of 3, we need to increase the exponents of 2 and 5 by 1 each. So, we multiply the original number by (2^1 * 5^1) = 2 * 5 = 10.
The new number becomes:
(2^5 * 5^4 * 7^3) * (2^1 * 5^1) = 2^6 * 5^5 * 7^3
Now, the exponents of all prime factors (2, 5, and 7) are multiples of 3.
Next, we find the cube root of 2^6 * 5^5 * 7^3:
cube root of (2^6 * 5^5 * 7^3) = cube root of 2^6 * cube root of 5^5 * cube root of 7^3
The cube root of each prime factor raised to some power is obtained by dividing the exponent by 3:
cube root of 2^6 = 2^(6/3) = 2^2 = 4
cube root of 5^5 = 5^(5/3)
cube root of 7^3 = 7^(3/3) = 7^1 = 7
Therefore, the cube root of 2^6 * 5^5 * 7^3 is 4 * (5^(5/3)) * 7.
Please note that the exact value of the cube root of (5^(5/3)) cannot be determined without a calculator or approximation techniques.
Ah, math questions! I must be honest, numbers make me feel a little dizzy. But fear not, I shall do my best to provide you with an answer that is both amusing and informative!
To find the least whole number by which 2^5 x 5^4 x 7^3 must be multiplied to get a perfect cube, we need to analyze the powers of each prime factor.
Let's break it down:
2^5 = 32
5^4 = 625
7^3 = 343
To get a perfect cube, we need each of these powers to be divisible by 3.
So, for 2^5 to be divisible by 3, we need to multiply it by 2^1:
32 x 2 = 64
Similarly, for 5^4 to be divisible by 3, we need to multiply it by 5^2:
625 x 25 = 15625
Lastly, for 7^3 to be divisible by 3, we need to multiply it by 7^1:
343 x 7 = 2401
To find the least whole number, we simply multiply them together:
64 x 15625 x 2401 = 6,400,000
Now, to calculate the cube root of this number, we have:
∛6,400,000 ≈ 169.115
So, the cube root of the resulting number is approximately 169.115. Voila!
Please keep in mind that I am just a clown bot, prone to err on the side of humor. Use mathematical verifications to ensure the accuracy of the answer.
To find the least whole number by which 2^5 x 5^4 x 7^3 must be multiplied to get a perfect cube, we need to determine the powers of each prime factor and ensure that all the exponents are divisible by 3.
Prime factorization:
2^5 x 5^4 x 7^3
Now let's break down each prime factor's exponent and determine how to make them divisible by 3:
2^5: Already divisible by 3, no additional multiple of 2 is needed.
5^4: We need to multiply it by 5^2 to make it divisible by 3.
5^4 x (5^2) = 5^6
7^3: We need to multiply it by 7 to make it divisible by 3.
7^3 x 7 = 7^4
Now we have the prime factorization:
2^5 x 5^6 x 7^4
To find the least whole number, we take the product of all the primes raised to their respective exponents:
2^5 x 5^6 x 7^4 = 2^5 x 5^6 x (2^4 x 7^4) = 2^9 x 5^6 x 7^4
The cube root of this resulting number will be:
cube root of (2^9 x 5^6 x 7^4) = 2^3 x 5^2 x 7^2 = 8 x 25 x 49 = 9800
Therefore, the least whole number by which 2^5 x 5^4 x 7^3 must be multiplied to get a perfect cube is 9800, and the cube root of the resulting number is 9800.