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 Post subject: GMAT Number TheoryPosted: Tue Jun 08, 2010 2:14 pm

Joined: Sun May 30, 2010 3:15 am
Posts: 424
If a = 600 and a² = 8² × 3² × 5² × k, which of the following can be a value of 2k?

A. 25
B. 36
C. -9
D. 50
E. -3

(D) The most efficient way to solve this problem is by first breaking a down into its prime components. You can accomplish this by repeatedly dividing a by small prime numbers until a has been exhausted: a = 2 × 2 × 2 × 3 × 5 × 5.

The next step is to calculate a². We can do this by just doubling the number of each factor in the prime string for a. In other words if there are three 2’s in the prime decomposition of a, there will be six of them in the prime decomposition of a². In that way, we have:
a² = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5 × 5.

Notice, there are twice as many of each factor (if it was a³ we would triple the factors). Now, let’s simplify the factors using exponents:
a² = 8² × 3² × 5² × 25.

By comparing our string to the string stated in the problem, it should be clear that k = 25 and that it must be true that 2k = 50; the correct answer is (D).
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In the answer there is only one “3”. It should be a = 2 × 2 × 2 × 3 × 3 × 5 × 5
Or should it be a = 2 × 2 × 2 × 3 × 5 and only one “5”?

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 Post subject: Re: math (test 4, question 14): number theory, prime factorsPosted: Tue Jun 08, 2010 2:16 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Either variant you can check by multiplying the numbers:
2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800
2 × 2 × 2 × 3 × 5 = 120

On the other hand, 2 × 2 × 2 × 3 × 5 × 5 = 600.

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 Post subject: Re: math (test 4, question 14): number theory, prime factorsPosted: Fri Jun 18, 2010 4:39 pm

Joined: Sun May 30, 2010 3:15 am
Posts: 424
A lot easier to take the square root of both sides, then divide by 5, 3, and 8 to get 5 = √k then square both sides to get 25 = k then simply multiply by 2.

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 Post subject: Re: math (test 4, question 14): number theory, prime factorsPosted: Fri Jun 18, 2010 4:49 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Yes, this question can also be solved in this way. However be careful when dealing with roots. Make sure you take roots only of non-negative numbers.

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