To find the least integral value of [tex]\( k \)[/tex] that will make [tex]\( C = 216k \)[/tex] a perfect square, we need to analyze the prime factorization of 216 and then determine the necessary multiplicative factor [tex]\( k \)[/tex] that makes [tex]\( 216k \)[/tex] a perfect square.
1. Prime Factorization of 216:
Start by factorizing 216 into its prime factors:
[tex]\[
216 = 2^3 \times 3^3
\][/tex]
2. Characteristics of Perfect Squares:
A perfect square must have even powers for all its prime factors. Current factorization of [tex]\( 216 \)[/tex] has odd exponents (both [tex]\( 2^3 \)[/tex] and [tex]\( 3^3 \)[/tex]). Therefore, to convert them into even exponents, each needs to be increased by 1.
3. Adjusting the Exponents:
In order to make [tex]\( C = 216k \)[/tex] a perfect square:
- Multiply [tex]\( 216 \)[/tex] by [tex]\( 2 \)[/tex] to account for the odd exponent of [tex]\( 2 \)[/tex] (raising [tex]\( 2^3 \)[/tex] to [tex]\( 2^4 \)[/tex]).
- Multiply [tex]\( 216 \)[/tex] by [tex]\( 3 \)[/tex] to account for the odd exponent of [tex]\( 3 \)[/tex] (raising [tex]\( 3^3 \)[/tex] to [tex]\( 3^4 \)[/tex]).
4. Combining the Factors:
To ensure both prime exponents are even, multiply [tex]\( 216 \)[/tex] by the least common multiple of the factors required, which is [tex]\( 2 \times 3 = 6 \)[/tex].
Therefore, the least integral value of [tex]\( k \)[/tex] that will make [tex]\( C = 216k \)[/tex] a perfect square is:
[tex]\[
\boxed{6}
\][/tex]
When [tex]\( k = 6 \)[/tex],
[tex]\[
C = 216 \times 6 = 1296 = 36^2
\][/tex]
1296 is indeed a perfect square, confirming our [tex]\( k \)[/tex] value. Thus, the correct answer is option D: 6.
