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Re: Given a positive integer c, how many integers are greater [#permalink]
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answer: C
Consider c as 1, then c = 1 and 2c = 2: 0 integer is between 1 and 2, Thus C is correct.
Consider c as 2, then c = 2 and 2c = 4: 1 integer is between 2 and 4 (3), Thus C is correct.
*We know that 2c = c + c. All the values between c and 2c are c+1, c+2, …, 2c-1. So there are (2c-1)-(c+1)+1 = c-1 integers between c and 2c.

Originally posted by Fatemeh on 22 Feb 2018, 17:38.
Last edited by Fatemeh on 23 Feb 2018, 12:24, edited 1 time in total.
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Re: Given a positive integer c, how many integers are greater [#permalink]
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Very nicely explained. Thanks.

FatemehAsgarinejad wrote:
answer: C
Consider c as 1, then c = 1 and 2c = 2: 0 integer is between 1 and 2, Thus C is correct.
Consider c as 2, then c = 2 and 2c = 4: 1 integer is between 2 and 4 (3), Thus C is correct.
*We know that 2c = c + c. All the values between c and 2c are c+1, c+2, …, 2c-1. So there are integers between c and 2c.
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Re: Given a positive integer c, how many integers are greater [#permalink]
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



Given a positive integer \(c\), how many integers are greater than c and less than \(2c\)?

A. \(\frac{c}{2}\)

B. \(c\)

C. \(c - 1\)

D. \(c - 2\)

E. \(c + 1\)


Recall that the number of integers between two integers a and b, inclusive, is b - a + 1. However, here we want the numbers of integers between c and 2c, excluding themselves. Therefore, the number of integers is 2c - c + 1 - 2 = c - 1. Note that we subtracted 2 on the left side of the equation because we needed to exclude the endpoint values c and 2c.

Alternate Solution:

We can express the integers between c and 2c as c + 1, c + 2, … , c + (c - 1). Since c + 1 is the first integer, c + 2 is the second integer and so on, c + (c - 1) is the c - 1st integer.

Answer: C
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Re: Given a positive integer c, how many integers are greater [#permalink]
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