Problem Solving

aarzoo wrote:the product of the factors of a three digit number is equal to the ninth power of the number. find the least such number.

This isn't really a GMAT-style question.

It's missing the 5 answer choices. If there were 5 answer choices, we could just check them to see which one satisfies the given conditions.

What's the source of this question?

Cheers,
Brent

aarzoo wrote:the product of the factors of a three digit number is equal to the ninth power of the number. find the least such number.

A) 2^17
B) 768
C) 288
D) 180
E) 140

In my opinion, this question is too hard/time-consuming for the GMAT (unless I'm missing something quite basic).

If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) =(5)(4)(2) = 40

If factors of the number must equal to the ninth power of the number, we want the number of divisors to be divisible by 9.

E) 140 = (2¹)(5¹)(7¹)
So, the number of positive divisors of 140 = (1+1)(1+1)(1+1) =(2)(2)(2) = 8
Since 8 is NOT divisible by 9, we can ELIMINATE A

D) 180 = (2²)(3²)(5¹)
So, the number of positive divisors of 180 = (2+1)(2+1)(1+1) =(3)(3)(2) = 18
18 IS divisible by 9, so B seems like a good contender.
To be sure, let's look at the factors of 180

So, the factors of 180 are: 2�, 2¹, 2², (3¹)(2�), (3¹)(2¹), (3¹)(2²), (3²)(2�), (3²)(2¹), (3²)(2²), (2�)(5), (2¹)(5), (2²)(5), (3¹)(2�)(5), (3¹)(2¹)(5), (3¹)(2²)(5), (3²)(2�)(5), (3²)(2¹)(5), and (3²)(2²)(5)

So, the product of all of the factors = (2�)(2¹)(2²)(3¹)(2�)(3¹)(2¹)(3¹)(2²)(3²)(2�)(3²)(2¹)(3²)(2²)(2�)(5)(2¹)(5)(2²)(5)(3¹)(2�)(5)(3¹)(2¹)(5)(3¹)(2²)(5)(3²)(2�)(5)(3²)(2¹)(5)(3²)(2²)(5)

The product of all of the factors = (2^18)(3^18)(5^9)

Notice that the product (2^18)(3^18)(5^9) is equal to [(2²)(3²)(5¹)]�

Answer: D