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22 Jan 2026, 03:26
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D
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Difficulty:
85%
(hard)
Question Stats:
38% (02:00) correct
62%
(01:56)
wrong
based on 60
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If S is a set of 3r distinct positive integers , and the range of S is 15 – r , how many possible values of r are there ?
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
22 Jan 2026, 08:42
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Conditions
r must be a positive integer where 15 - r ≥ 0, so r ≤ 15. For a set S of n = 3r distinct positive integers, the minimum possible range is n - 1, achieved by consecutive integers like {1, 2, ..., n} .
Feasibility Check
The range 15 - r must satisfy 15 - r ≥ 3r - 1, simplifying to r ≤ 4 . For r ≥ 5, the required range falls below the minimum possible, making it impossible .
There are 4 possible values of r: 1, 2, 3, and 4.
Verification
For r=1 (n=3), min range=2 ≤14: possible, e.g., {1,2,15}.
For r=2 (n=6), min range=5 ≤13: possible, e.g., {1,2,3,4,5,14}.
For r=3 (n=9), min range=8 ≤12: possible, e.g., {1 to 9} range=8.
For r=4 (n=12), min range=11=11: possible, e.g., {1 to 12} .
ANS - D
r must be a positive integer where 15 - r ≥ 0, so r ≤ 15. For a set S of n = 3r distinct positive integers, the minimum possible range is n - 1, achieved by consecutive integers like {1, 2, ..., n} .
Feasibility Check
The range 15 - r must satisfy 15 - r ≥ 3r - 1, simplifying to r ≤ 4 . For r ≥ 5, the required range falls below the minimum possible, making it impossible .
There are 4 possible values of r: 1, 2, 3, and 4.
Verification
For r=1 (n=3), min range=2 ≤14: possible, e.g., {1,2,15}.
For r=2 (n=6), min range=5 ≤13: possible, e.g., {1,2,3,4,5,14}.
For r=3 (n=9), min range=8 ≤12: possible, e.g., {1 to 9} range=8.
For r=4 (n=12), min range=11=11: possible, e.g., {1 to 12} .
ANS - D
03 May 2026, 18:58
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If S is a set of 3r distinct positive integers and the range of S is 15 – r, how many possible values of r are there?
The minimum possible value of the range is 0. So, 15 - r ≥ 0. Thus, r ≤ 15.
Also, as r increases, the range decreases.
Meanwhile, since the number of integers in the set is 3r, as r increases, the number of integers in the set increases.
So, as the range decreases, the number of integers in the set increases. We see that this dynamic limits the possible number of integers in the set to a relatively small value. So, using that information along with the answer choices, we can tell that we can quickly answer this question by just testing values.
r = 0
Range = 15 - 0 = 15
Number of Integers in the Set = 3 × 0 = 0
Impossible (because a set of zero distinct integers cannot have a range of 15)
r = 1
Range = 15 - 1 = 14
Number of Integers in the Set = 3 × 1 = 3
Possible
r = 2
Range = 15 - 2 = 13
Number of Integers in the Set = 3 × 2 = 6
Possible
r = 3
Range = 15 - 3 = 12
Number of Integers in the Set = 3 × 3 = 9
Possible
r = 4
Range = 15 - 4 = 11
Number of Integers in the Set = 3 × 4 = 12
Possible (because the first and last of 12 consecutive integers can be 11 apart)
We can see that since, if we go higher, 3r will be much greater than 15 - r, we can stop at r = 4.
So, there are four possible values of r: 1, 2, 3, and 4.
A. 1
B. 2
C. 3
D. 4
E. 5
Correct answer: D
The minimum possible value of the range is 0. So, 15 - r ≥ 0. Thus, r ≤ 15.
Also, as r increases, the range decreases.
Meanwhile, since the number of integers in the set is 3r, as r increases, the number of integers in the set increases.
So, as the range decreases, the number of integers in the set increases. We see that this dynamic limits the possible number of integers in the set to a relatively small value. So, using that information along with the answer choices, we can tell that we can quickly answer this question by just testing values.
r = 0
Range = 15 - 0 = 15
Number of Integers in the Set = 3 × 0 = 0
Impossible (because a set of zero distinct integers cannot have a range of 15)
r = 1
Range = 15 - 1 = 14
Number of Integers in the Set = 3 × 1 = 3
Possible
r = 2
Range = 15 - 2 = 13
Number of Integers in the Set = 3 × 2 = 6
Possible
r = 3
Range = 15 - 3 = 12
Number of Integers in the Set = 3 × 3 = 9
Possible
r = 4
Range = 15 - 4 = 11
Number of Integers in the Set = 3 × 4 = 12
Possible (because the first and last of 12 consecutive integers can be 11 apart)
We can see that since, if we go higher, 3r will be much greater than 15 - r, we can stop at r = 4.
So, there are four possible values of r: 1, 2, 3, and 4.
A. 1
B. 2
C. 3
D. 4
E. 5
Correct answer: D
