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Joined: 19 May 2004
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if <a<b<c<d and a,b,c,d all are all positive [#permalink]
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06 Aug 2004, 07:17
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if <a<b<c<d and a,b,c,d all are all positive multiple of 2, what is the value of d =?
1)a+b+c+d=170
2)abcd=2^16
If you have a different way than just picking numbers please explain it.
(Originally posted by Praetorian)



Director
Joined: 20 Jul 2004
Posts: 592

I don't know how this was solved when posted before.
1)a+b+c+d=170
Insufficient, because it can be 2+4+6+158 or 2+4+8+156
2)abcd=2^16
States that a, b, c, d are not only multiples of 2, but are simple powers of 2, like 2, 4, 8, 16, 32, 64, 128, etc
Insufficient, since it can be 2^1, 2^2, 2^3, 2^10 or 2^1, 2^2, 2^4, 2^9
3) together
2, 4, 8, 16, 32, 64, 128 = 2^1, 2^2, 2^3, 2^4, 2^5, 2^6, 2^7
to get 16 in powers and still get a sum of 170, it should be 2, 8, 32, 128
So, C.



CIO
Joined: 09 Mar 2003
Posts: 463

Dookie,
The answer is C. This problem is definately a little confusing, but it could be done like this (according to your request of not plugging in):
1) should be seen as not enough information right off the bat. There are tons of 4 even numbers that add up to 170.
2) If their product is 2^16, then they all must be a power of 2. Taken alone, though, there are multiple ways to get 16 adding up different powers, all the while making d, the last one, the highest. One thing to note with it is that D must be at least 2^6. Anything less than that won't work because at 2^5 the max would be (2^2)(2^3)(2^4)(2^5)=2^14.
But together it gets interesting. Obviously, 170 has nothing to do with a power of 2. But if they add up to 170, the highest power we could possibly have is 2^7, since that's 128, and 2^8 (which would work in the second statement alone) is definately too high.
So d is some power of 2, either 128 or 64.
170128=32: a power of 2 that can be divided into three smaller powers of 2.
17064=106: not a power of 2 so it can't work.
So D must be 128.



Senior Manager
Joined: 19 May 2004
Posts: 291

Thanks for your answer Ian.
It seems like there is no fast solution for this one.
I hope i won't meet such a time consumer in the test!










