Carcass wrote:
If a = 105 and a^3 = 21 × 25 × 45 × b, what is the value of b?
A. 35
B. 42
C. 45
D. 49
E. 54
This question calls for some
prime factorizationa = 105 = (3)(5)(7)
So, a³ = [(3)(5)(7)]³ = (3)(3)(3)(5)(5)(5)(7)(7)(7)
We're told that
a³ = 21 × 25 × 45 × bTake: a³ = (3)(3)(3)(5)(5)(5)(7)(7)(7)
Rewrite as: a³ = (3)(3)(
3)(5)(5)(5)(7)(7)(
7) = (3)(3)(
21)(5)(5)(5)(7)(7)
Now rewrite as: a³ = (3)(3)(
21)(
5)(
5)(5)(7)(7) = (3)(3)(
21)(
25)(5)(7)(7)
Now rewrite as: a³ = (
3)(
3)(
21)(
25)(
9)(7)(7) = (
21)(
25)(
45)(7)(7)
If
21 × 25 × 45 × b = (
21)(
25)(
45)(
7)(
7), then
b = (7)(7) =
49Answer: D