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difficult question

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difficult question

by CITI29 » Mon Jul 21, 2008 6:45 am
For any integer k > 1, the term “length of an integer” refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

a- 5
b- 6
c- 15
d- 16
e- 18
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by reachac » Mon Jul 21, 2008 7:07 am
IMO D
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by CITI29 » Mon Jul 21, 2008 7:27 am
yes, thats correct...can u explain pls
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by reachac » Mon Jul 21, 2008 7:39 am
Length is being determined by the no. of prime factors. Now we wana maximize the length(and hence the no. of factors) under the given constraint x+3y<1000. i.e the size of x or Y( and hence the product of the prime factors) is the limiting force here. To achieve both the above( max number factors within the given size constraint), we should take the smallest prime number ie 2

Now 2^8 = 256 at this point , 3Y< 1000-256. . So Y can also be take to be 256.

Hence max length = 8+8 =16.

U can try other combinations of x and Y but the one above is the optimum value combination. Also as per the options it is the second highest value to this should assure you a lil while marking the option on the test
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by kris610 » Mon Jul 21, 2008 7:44 am
I went with D.

This was my approach (took 5 minutes though):

Try to find X and Y such that both are powers of 2.

X=512 and Y=128. 3Y=384. Sum is less than 1000. The length of X=9 and that of Y=7.
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by ramyaravindran » Mon Jul 21, 2008 8:22 am
If you chose x=512 (2 ^ 9) and y = 128 (2 ^ 7) then the maximum possible length will be 16. So the answer should be D. Chose any other number greater than this will not satisfy the condition x + 3y < 1000. What is the OA??
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