Answer:
Trish is correct, since [tex]4, 6, 8, 12[/tex] are divisors of [tex]48[/tex].
Step-by-step explanation:
Trish is correct, since [tex]4, 6, 8, 12[/tex] are divisors of [tex]48[/tex]. By factor decomposition we find the following product of prime numbers:
[tex]48 = 2\times 2 \times 2 \times 2 \times 3[/tex]
[tex]48 = 2^{4}\times 3[/tex]
The factor decomposition of [tex]4, 6, 8, 12[/tex] are, respectively:
[tex]4 = 2^{2}[/tex]
[tex]6 = 2\times 3[/tex]
[tex]8 = 2^{3}[/tex]
[tex]12 = 2^{2}\times 3[/tex]
Since each number is contained in the factor decomposition of 48, we can re-arrange [tex]48[/tex] in the following two ways:
Donna
[tex]48 = (2\times 3) \times (2^{3})[/tex]
[tex]48 = 6 \times 8[/tex]
Larry
[tex]48 = 2^{2} \times (2^{2}\times 3)[/tex]
[tex]48 = 4\times 12[/tex]