A, B, C, D, E, F, G, and H are all integers

[Kevin]

A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A?
-24(5^12)
-23(5^12)
-24(5^6)
23(5^12)
24(5^12)


The distance from G to H is 513 - 512.

The distance between and two consecutive points is constant, so the distance from A to G will be 6 times the distance from G to H or 6(513 - 512).

The value of A, therefore, will be equal to the value of G minus the distance from A to G:

512 - 6(513 - 512) 512 - 6[512(5 - 1)] 512 - 6(512)(4)

512(1 - 24) (-23)512.

The correct answer is B.

I don't see how the second step was achieved. Please explain. Thanks

[Guest]

? Arent the answer choice incorrect ? IF distance between any two consequitive integers is contant , lets say C. So between B and A, D and C etc , the distance is C. And we know the distance between one, so it is 513-512 , which is 1=C

Just subtract one for each and we gt A = 506 ?

I think the question is wrong ? Or the question and answer choices dont match up. Something is missing for sure

[RonPurewal]

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.

--

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

[nocheivyirene]

? Arent the answer choice incorrect ? IF distance between any two consequitive integers is contant , lets say C. So between B and A, D and C etc , the distance is C. And we know the distance between one, so it is 513-512 , which is 1=C

Just subtract one for each and we gt A = 506 ?

I think the question is wrong ? Or the question and answer choices dont match up. Something is missing for sure

Yes true! I was put off by the question too until I realized that it's not 512 but 5^12. Let me repost the question for others who will read this in the future.

Question: A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A?
-24(5^12)
-23(5^12)
-24(5^6)
23(5^12)
24(5^12)

k = 5^13 -5^12 = 5^12(5-1) = 4 (5^12)

Arithmetic Sequence Formula:
an = a1 + (n-1)k
5^12 = A + (7-1) (4) (5^12)
A = 5^12 - 24 (5^12)
A = 5^12 (1 - 24)
A = -23 * 5^12

Answer: B

[jlucero]

Agreed. I edited the original question to make it easier to see.

[nocheivyirene]

A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A?
-24(5^12)
-23(5^12)
-24(5^6)
23(5^12)
24(5^12)

My approach is to draw a number line with A to H lined up.
A B C D E F G H

I calculated the distance between G and H:
k = 5^13 - 5^12
k = 5^12(5-1) = 4*5^12

G looking at the number line is:
G = A + 6k
5^12 = A + 6*4*5^12
A = 5^12 - 24*5^12
A = 5^12 (1 - 24)
A = 5^12 (-23)

Note: this is similar to arithmetic progression formula:
an = a1 + (n-1) k

Answer: B

[tim]

:)

[sfbay]

Ron,

I get the problem but was wondering how you so quickly come to the conclusion that:

* 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.

observing this at a glance is not clear to me

[tim]

Draw it out on a number line, with H = 5^13 five times as big as G = 5^12 :

-----------------------0---G---------------H

because the points are equally spaced, F has to be on the other side of G the same distance away that H is from G:

-------F---------------0---G---------------H

now it should be clear visually. :) let us know if you have any further questions on this one..

[sfbay]

thanks tim. I should have been more clear.

what is the thinking that you so quickly can conclude that 0 is between F and G. being closer to G is easy enough, the problem in itself i get too. Trying to understand what you guys see at a glance that puts 0 between F and G.

distance between points is 5*(5 ^12)

F-------G-------H

since g= 5^12 subtracting something bigger ie 5*(5 ^12) puts you into negative territory. is that what you see?

this is much easier to see if say g=6, distance between 2*6, therefore h = 18, and f must be negative (6-12).

[tim]

to be perfectly honest, i'm quite convinced you didn't read my post. if you did, please paraphrase it back to me and then let me know if you still have questions about how we know this. i think my explanation makes it pretty clear..