please solve this one.
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- Brent@GMATPrepNow
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This isn't really a GMAT-style question.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.
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?
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Brent
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In my opinion, this question is too hard/time-consuming for the GMAT (unless I'm missing something quite basic).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
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
- DavidG@VeritasPrep
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Agreed. (That the question is too time-consuming for the GMAT, not that you're missing something basic.)In my opinion, this question is too hard/time-consuming for the GMAT (unless I'm missing something quite basic).
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You guys ARE missing something (kinda) basic here :p
To illustrate, let's take a smaller number, such as 30. To find the product of its factors, think of them in pairs:
1 * 30
2 * 15
3 * 10
5 * 6
So the product of the factors of 30 is 30�, where n is the number of factors divided by 2.
Let's say our number is x. We know that
x� = x�
So n = 9, and our number has 18 factors. From there, we find the smallest three digit number with 18 factors. Using our number of factors trick, we know that
18 = 3 * 3 * 2
so a number that's expressible as p² * q² * r, with p, q, and r prime, should do it. 2² * 3² * 5 is the smallest such number, so that's our guy.
(Properly speaking, we should consider other cases, such as p² * q� or p * q�, but if you've gotten this far you can quickly see why those will be give larger integers than 180.)
To illustrate, let's take a smaller number, such as 30. To find the product of its factors, think of them in pairs:
1 * 30
2 * 15
3 * 10
5 * 6
So the product of the factors of 30 is 30�, where n is the number of factors divided by 2.
Let's say our number is x. We know that
x� = x�
So n = 9, and our number has 18 factors. From there, we find the smallest three digit number with 18 factors. Using our number of factors trick, we know that
18 = 3 * 3 * 2
so a number that's expressible as p² * q² * r, with p, q, and r prime, should do it. 2² * 3² * 5 is the smallest such number, so that's our guy.
(Properly speaking, we should consider other cases, such as p² * q� or p * q�, but if you've gotten this far you can quickly see why those will be give larger integers than 180.)
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Whoops, addendum to that last post: I should've said "with p, q, and r distinct primes" ... but hopefully it was clear from the context/inspiration.