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Math Expert V
Joined: 02 Sep 2009
Posts: 58453
A, B, C, D, E, F, G, and H are all integers, listed in order of increa  [#permalink]

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52 00:00

Difficulty:   75% (hard)

Question Stats: 59% (02:26) correct 41% (02:36) wrong based on 548 sessions

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Tough and Tricky questions: Sequences.

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?

A. -24(5^12)
B. -23(5^12)
C. -24(5^6)
D. 23(5^12)
E. 24(5^12)

Kudos for a correct solution.

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A, B, C, D, E, F, G, and H are all integers, listed in order of increa  [#permalink]

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6
$$H = 5^{13}$$

$$G = 5^{12}$$

$$H - G = 5^{13} - 5^{12} = 4 * 5^{12}$$

A ......1....... B ........2....... C ........3......... D ..........4........... E ......5....... F ........6......... G ................... H

From point A, $$4 * 5^{12}$$ has to be added 6 times to reach point $$G = 5^{12}$$

$$A + 6* 4 * 5^{12} = 5^{12}$$

$$A = -23 * 5^{12}$$

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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa  [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Sequences.

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?

A. -24(5^12)
B. -23(5^12)
C. -24(5^6)
D. 23(5^12)
E. 24(5^12)

Kudos for a correct solution.

Ans : B

Assume that the numbers appear as shown below on the number line

A-----B-----C-----D-----E-----F-----G-----H
| |
(5^12) (5^13)

As the values for G and H are given , we can calculate the difference between any two terms of the series .

Common Difference ,d = (5^13) - (5^12)
= (5^12) *[ 5-1 ]
= (5^12)*(4)

Also F + d = G as the terms are in equidistant and in increasing order.

So F + (5^12)*(4) = (5^12).

That is , F = (5^12) - (5^12)*(4)
= (5^12)[ 1-4]
= (5^12) (-3)

Similarly , E = F - d
= (5^12)[-3-4]
= (5^12)*(-7)

You can see a -4 getting added to the non-exponent part of the values . That is , according to the pattern , D SHOULD BE (5^12)*(-7-4)= (5^12)*(-11)
Following this pattern , A = (5^12)*(-23)

Ans : B
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A, B, C, D, E, F, G, and H are all integers, listed in order of increa  [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Sequences.

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?

A. -24(5^12)
B. -23(5^12)
C. -24(5^6)
D. 23(5^12)
E. 24(5^12)

H = 5^13
G = 5^12

Distance between H and G = 5^13 - 5^12
= 5^12(5^1 - 1)
= (5^12)(4)
= (4)(5^12)

So, F = 5^12 - (4)(5^12)
E = 5^12 - (4)(5^12) - (4)(5^12)
D = 5^12 - (4)(5^12) - (4)(5^12) - (4)(5^12)
C = 5^12 - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12)
B = 5^12 - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12)
A = 5^12 - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12)

So, A = 5^12 - [(4)(5^12) + (4)(5^12) + (4)(5^12) + (4)(5^12) + (4)(5^12) + (4)(5^12)]
= 5^12 - (24)(5^12)
= (1)(5^12) - (24)(5^12)
= (-23)(5^12)
= Answer choice B

Cheers,
Brent
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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa  [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Sequences.

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?

A. -24(5^12)
B. -23(5^12)
C. -24(5^6)
D. 23(5^12)
E. 24(5^12)

We are given that A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size, and that the distance between any two consecutive numbers, which we can denote by the variable n, is constant.

Since G and H are equal to 5^12 and 5^13, respectively, then the distance between them is:

n = 5^13 - 5^12

n = 5^12(5 - 1)

n = 5^12(4)

Since A is 6 numbers ahead of G, A is 6n less than G, and therefore:

A = G - 6n

A = 5^12 - 6[5^12(4)]

A = 5^12[1 - 6(4)]

A = 5^12[-23] = -23(5^12)

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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa  [#permalink]

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How I solved it is:
A, B, C, D, E, F, G, H (Total of 8 numbers = n)
Since its a sequence and each term is a certain constant more than the other.
H - G = 5^13 - 5^12 = (5^12) (5-1) = 4 x (5^12) = Our common difference, d.

Also unrelated,
G = A + (n-1) d
5^12 = A + (7-1) 4 (5^12)
As n=7, because its the 7th term in the sequence

Therefore,
A = (5^12) - 24 (5^12)
= (5^12) (1-24) Taking 5^12 common
= - 23 (5^12)
= Answer choice B
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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa  [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Sequences.

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?

A. -24(5^12)
B. -23(5^12)
C. -24(5^6)
D. 23(5^12)
E. 24(5^12)

Kudos for a correct solution.

Values of G and H are 5^12 and 5^13, respectively, and there is only one gap between them

so each gap between any two adjacent numbers = 5^13 - 5^12 = 5^12 (5-1) = 4* 5^12

order of numbers is A___B___C___D___E___F___G___H

A is 6 gaps smaller than G

i.e. A = 5^12 - 6*(4* 5^12) = -23* 5^12

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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa  [#permalink]

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Let distance between any two consecutive numbers is d.
Distance between G and H is $$d= 5^{13} - 5^{12} = 4*5^{12}$$

Sequence {A, B, C, D, E, F, G, H}
A is the first term of sequence, G is the 7th term of sequence, thus: $$A +6d = G$$
-->$$A + 6*4*5^{12} = 5^{12}$$ --> $$A + 24*5^{12} = 5^{12}$$-->$$A = (-23)*5^{12}$$

Bunuel wrote:

Tough and Tricky questions: Sequences.

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?

A. -24(5^12)
B. -23(5^12)
C. -24(5^6)
D. 23(5^12)
E. 24(5^12)

Kudos for a correct solution.

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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa  [#permalink]

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Diff between last and second last term = 5^13 - 5^12 = 4 * 5^12
Since difference is constant , we can use decreasing AP. (changing this order from ascending to descending)
So 1st term is 5^13 and difference = - 4 * 5^12
= a + (n-1) d
= 5^13 + 7 * (- 4 * 5^12)
= 5^13 - 28* 5^12
= 5^12 * -23
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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa  [#permalink]

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PareshGmat wrote:
$$H = 5^{13}$$

$$G = 5^{12}$$

$$H - G = 5^{13} - 5^{12} = 4 * 5^{12}$$

A ......1....... B ........2....... C ........3......... D ..........4........... E ......5....... F ........6......... G ................... H

From point A, $$4 * 5^{12}$$ has to be added 6 times to reach point $$G = 5^{12}$$

$$A + 6* 4 * 5^{12} = 5^{12}$$

$$A = -23 * 5^{12}$$

Quick question , could you please explain how you got 4*5^12? Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa   [#permalink] 05 Jul 2019, 14:17
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