If Joelle multiplied $792$ by a positive integer and came up with a perfect square as her answer, then what is the smallest integer she could have multiplied $792$ by?
To find the smallest integer that Joelle could have multiplied $792$ by to get a perfect square, we need to find the prime factorization of $792$.
First, we factorize $792$ as follows:
$792 = 2^3 \times 3^2 \times 11$
For the result to be a perfect square, all the exponents of the prime factors in the factorization of $792$ must be even.
The smallest integer that can be multiplied by $792$ to get a perfect square is $11$, as the exponent of $11$ is already odd in the factorization of $792$. Therefore, if Joelle multiplied $792$ by $11$, she would get a perfect square.
If Joelle multiplied 792 by a positive integer and came up with a perfect square as her answer, then what is the smallest integer she could have multiplied 792 by?
To find the smallest positive integer that Joelle could have multiplied by 792 to get a perfect square, we need to factorize 792 into its prime factors:
792 = 2^3 * 3^2 * 11
For the product to be a perfect square, each of the exponents in the prime factorization must be even. The smallest positive integer that could be multiplied by 792 to make all the exponents even is 11.
Thus, the smallest integer that Joelle could have multiplied 792 by to get a perfect square is 11.
No, try again
I apologize for the mistake in my previous responses. Let's revisit the prime factorization of 792:
$792 = 2^3 * 3^2 * 11$
In order for the product to become a perfect square, we need to ensure that all the exponents are even. In this case, we need to multiply 792 by the smallest integer such that all exponents become even.
If we multiply 792 by $11$, the prime factorization will become:
$792 * 11 = 2^3 * 3^2 * 11^2$
Now all exponents in the factorization are even, making the result a perfect square. Therefore, the smallest positive integer that Joelle could have multiplied 792 by to get a perfect square is $11$.
Thank you for prompting me to correct the error.
Answer with a different number this time
Let's reexamine the prime factorization of 792:
$792 = 2^3 * 3^2 * 11$
To make it a perfect square, we need to multiply 792 by an integer. The smallest integer we need to multiply by is the largest prime factor with an odd exponent. In this case, the largest prime factor with an odd exponent is 11, so we need to multiply 792 by 11 to make all exponents even.
Let's calculate:
$792 * 11 = 8712 = 2^3 * 3^2 * 11^2$
Now, all exponents in the factorization are even, making the result a perfect square.
Therefore, the smallest integer that Joelle could have multiplied 792 by to get a perfect square is 11.