If a = 105 and a^3 = 21 × 25 × 45 × b, what is the value of b?

Solution : a = 7*5*3
==> 7*7*7*5*5*5*3*3*3 = 7*3*5*5*3*3*5*b
==> b = 7*7 = 49
Option D

a = 105 = 3*5*7
a^3 = (3*5*7)^3 = 3^3 * 5^3 * 7^3
Also, a^3 = 21*25*45*b = 3*7*5*5*3*3*5*b = 3^3*5^3*7*b
--> 3^3*5^3*7^3 = 3^3*5^3*7*b
--> b = 49
Answer: D

KAPLAN OFFICIAL SOLUTION:

The very first step in this problem is substituting 105 for the value of a in the main equation.

This makes our equation

\(105^3= 21 * 25 * 45 * b\).

Whenever we see numbers such as 105^3, which we COULD calculate but would be extremely time consuming, we should consider it a clue that there is likely a faster way to solve the problem. In many cases, this one included, the faster solution involves finding the prime factors of each number. 105 becomes 3 * 5 * 7. 105^3, therefore, becomes:

\((3 * 5 * 7) * (3 * 5 * 7) *(3 * 5 * 7)\)

On the other side of the equation 21 becomes 3 * 7, 25 becomes 5 * 5 and 45 becomes 3 * 3 * 5. When we rewrite the equation with the prime factors we get:

\((3 * 5 * 7) * ( 3 * 5 * 7) * (3 * 5 * 7)\)

=

\((3 * 7) * (5 * 5) * (3 * 3 * 5) * b\)

Once we have this equation, we can cancel out any numbers we see on both sides of the equation. This process leaves us with

7 * 7 = b.

Thus, b = 49, which is the correct answer (D). The challenge was not necessarily in the level of math here, but more in your ability to identify the best and fastest approach for this given problem in the midst of your exam. Always be on the lookout for shortcuts on a time-pressured exam like the GMAT, and for ways that you can use your knowledge of number properties to identify things like prime factors.

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