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by yvonne12 » Sun Apr 15, 2007 5:40 pm
for any positive N, the length of N is defined as the number of prime factors where the product is n. For example the length of 75 is 3 since 75=3x5x5. How many 2 digit pos. intehers has length 6?

Could someone show me step by step on how to approach this kind of question? I cant come up with a strategy to figure this out quickly and less than a minute.
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by Tame the CAT » Sun Apr 15, 2007 6:09 pm
There should only be 2 numbers with a length of 6 as described in the question.

I tried 2^6 and got 64
Then I tried 2^5*3 and got 96

There are no more combinations of 6 prime numbers that would give me a two digit number.
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by rajesh_ctm » Sun Apr 15, 2007 7:35 pm
From the answer choices, you can figure out that there are not many such numbers possible. So listing down all the possible numbers would be a good strategy.

First find out what is the minimum number which has 6 prime factors. That will be 2X2X2X2X2X2 = 64

Now you can get the next greater number which will have 6 prime factors by increasing one factor from 2 to 3 (next possible prime number). So the next number is 2X2X2X2X2X3 = 96

If you further increase any factor (2 to 3 or 3 to 4), the result will cross 100, so 2 is the answer.
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by Cybermusings » Mon Apr 16, 2007 4:48 am
2*2*2*2*2*2 = 64

and 2*2*2*2*2*3 = 96

If you try susbstituting 3 for 2 any further we will move on to 3 digit numbers.

Hence answer is only 2 such double digit numbers exist
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