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resilient
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 Posted: Wed Apr 09, 2008 6:35 pm Post subject: consecutive integers! |
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A,B,C,D,F,G,H are all integers, listed in order of increasing size. When these numbers are arranged in a number line, the distance between any two consecutive numbers is constant.If G and H are equal to 3^12 and 5^13, respectively, what is the value of A?
a.-24(5^12)
b.-23(5^12)
c.-24(5^6)
d.23(5^12)
e.24(5^12)
Can someonoe please explain this in simple terms and also ech necessary step? Pretty confused on this? I cannot see the relationship!
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lunarpower
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| Posted: Thu Apr 10, 2008 2:10 am Post subject: Re: consecutive integers! |
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| Enginpasa1 wrote: | A,B,C,D,F,G,H are all integers, listed in order of increasing size. When these numbers are arranged in a number line, the distance between any two consecutive numbers is constant.If G and H are equal to 3^12 and 5^13, respectively, what is the value of A?
a.-24(5^12)
b.-23(5^12)
c.-24(5^6)
d.23(5^12)
e.24(5^12)
Can someonoe please explain this in simple terms and also ech necessary step? Pretty confused on this? I cannot see the relationship! |
WARNING: i think that you made two mistakes in transcribing this problem; i'll describe them below. i've never seen the problem before, but i'd already typed so much that i didn't want to throw the typing away; therefore, i took an educated guess as to the true nature of the problem.
guessed mistake 1: i'm going to assume that there were supposed to be eight points, called A, B, C, D, E, F, G, H; i can't see any reason why they would eliminate E, so i'll just assume you forgot to type it.
guessed mistake 2: i'm going to assume that you wrote 3^12 when you actually meant to write 5^12.
if either of these guesses is incorrect, politely ignore the solution to the problem (although the general advice about number lines, etc. is still valuable).
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if you had 30 seconds to guess on this problem, here's what you'd do. (if you don't understand what i'm talking about, actually get out a sheet of paper and draw it)
* draw the number line with the eight points on it.
* realize that 5^13 is five times as big as 5^12.
* realize that, therefore, zero is going to be between F and G (and is going to be a lot closer to G) on your number line.
* realize that (c) is a tiny tiny fraction of the size of the other numbers in the list, and is nowhere near far enough to the left to be the correct answer. (since your number line shows numbers of the order of magnitude of 5^13, any number like 24*5^6 is so small that it would be indistinguishable from zero if you tried to plot it).
* therefore, you've narrowed the problem down to choices a and b.
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here's the real way to solve the problem:
* this is a problem about an arithmetic sequence. just as you should start by finding the radius in any problem involving circles, you should find the common difference in any problem involving arithmetic sequences. in this case, the common difference is 5^13 - 5^12 (an expression that can't be reduced; you can factor out 5^12, but that won't help things).
* the number you seek is six more common differences below G (look at your number line and count). therefore, the number you seek is
G - 6(common difference)
= (5^12) - 6(5^13 - 5^12)
= 5^12 - 6*5*5^12 + 6*5^12
= (5^12) - 30(5^12) + 6(5^12)
= -23(5^12)
answer = b
sweetness
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luvaduva
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| Posted: Fri May 02, 2008 12:44 pm Post subject: |
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| Nice solution. |
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