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2 questions from ETS practice tests

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2 questions from ETS practice tests [#permalink] New post 18 Jul 2006, 08:55
Hi everyone. I have two questions for you:

1. For any positive integer n, the "length" is defined as the number of prime factors whose product is n. E.g. The length of 75 is 3 because 3 x 5 x5.

How many two-digit positive integers have length 6?

1) None
2) One
3) Two
4) Three
5) Four

(I thought it was one--96 was the only one I counted, but I suppose 64 is also an answer. What's the trick? What are you supposed to do/see quickly to get this in 2 minutes or less? Is there a more efficient way of approaching this besides picking random numbers and factoring them?)

2nd question:

Is the positive integer j divisible by a greater number of different prime numbers than the positive integer k?

(1) j is divisible by 30

(2) k = 1,000

(I chose E thinking that they were looking for the number of prime factors that j has....but is the catch the word "different"? No matter what j is, because it is divisible by 30, at least 2, 3 and 5 will be factors. The only unique prime factors 1000 has are 2 and 5.

Could someone please confirm that choice "E" would in fact be the answer if the word "different" was removed from the stem.

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Re: 2 questions from ETS practice tests [#permalink] New post 18 Jul 2006, 09:36
First One
According to me the way tackle this is take the six prime factors rather than factoring all 2 digit numbers.

2 being the smallest prime no, the smallest 2 digit no of length 6 is
2*2*2*2*2*2 = 2^6 = 64

The next biggest no will be
2^5 * 3 = 32* 3 = 96

Any other combination of 6 prime nos will result in a no having more than 2 digits .
So Answer is TWO.

For the second one I think the answer should be C
The question is to find if there is a no J more having ‘different’ prime factors than K

(1) – This tells that j has 2, 3, 5 as factors, i.e min 3 diff prime factors. This does not talk about whether j has more diff prime factors, and neither does it say anything about K
(2) - k = 1000, has only (2 ‘diff’ prime factors, 2 and 5). Again this alone is not suff.

If we combine the two the min ‘diff’ prime factors that j has is 3 which is > than that of k (1000). So both together are sufficient.
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 [#permalink] New post 18 Jul 2006, 10:35
Q1: C

First take smallest 6 numbers
2,2,2,2,2,2 = 64 then
2,2,2,2,2,3 = 96 then
If we increase any number then surely the product will be greater than 99.

St1: Clearly INSUFF
St2: Clearly INSUFF
j is for sure a multiple of 2,3 and 5 prime numbers
but k = 1000 is a multiple of only 2 and 5 prime number: SUFF

In case "different" removed then surely E. j could be 30,000, which include all factors of 1000 and could be 30, which have very few factors than 1000.


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 [#permalink] New post 19 Jul 2006, 00:28
1) C

I dont know whether what I did was the fastest but it sure took me less than 2 mins

take the least prime number.
We know the length is 6 hence

2*2*2*2*2*2 = 64
Now replace the last 2 with other prime numbers
2*2*2*2*2*3 = 96

Now if we replace 3 by 5 we get a three digit number.
Hence the numbers are 64 and 96

2) C

Cond1) We can guess what numbers can satisfy J but dont know K hence Insuff
Cond2) we know K but not J hence not suff.

K has two different prome numbers 2 and 5
And J is divisible by 30 hence it can be divisble by atleast 2, 3 and 5
Hence C
  [#permalink] 19 Jul 2006, 00:28
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2 questions from ETS practice tests

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