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at least two consecutive integers

by anant03 » Tue Sep 01, 2015 7:51 am
A list contains twenty integers, not necessarily distinct. Does the list contain at least two consecutive integers?

(1) If any single value in the list is increased by 1, the number of different values in the list does not change.

(2) At least one value occurs more than once in the list.

OAC
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by [email protected] » Tue Sep 01, 2015 9:01 am
Hi anant03,

This question can be solved by TESTing VALUES, but you have to be careful to consider how those 20 integers COULD be 'spaced out.'

We're told that a list contains 20 integers, not necessarily distinct (meaning that there COULD be duplicates). We're asked if the list contains at least two consecutive integers. This is a YES/NO question.

Fact 1: If any single value in the list is increased by 1, the number of different values in the list does not change.

IF....
We have a list of consecutive EVEN integers (0, 2, 4, 6, 8,.....etc.), then the answer to the question is NO.

IF....
We have a list with nineteen 1s and one 2, then the answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: At least one value occurs more than once in the list.

IF....
We have ten 0s and ten 2s, then the answer to the question is NO.

IF....
We have a list with nineteen 1s and one 2, then the answer to the question is YES.
Fact 2 is INSUFFICIENT

Combined, we know:
If any single value in the list is increased by 1, the number of different values in the list does not change.
At least one value occurs more than once in the list.

We already have one 'overlapping' TEST...

IF....
We have a list with nineteen 1s and one 2, then the answer to the question is YES.

This example serves to prove a point - in the first Fact, we have to account for the possibility that ANY single value could be increased by 1, BUT this would not change the number of different values in the list. To account for THAT 'restriction', we have to think about whatever duplicate numbers exist....

We CANNOT have nineteen 1s and one 3....if one of the 1s was increased, then we would have three different numbers (1, 2 and 3), which goes against the restriction. The ONLY way to keep the number of terms from changing is if the "+1" term was already there. The only way to account for THAT is with a pair of consecutive numbers. Thus, there will ALWAYS be at least one pair of consecutive integers and the answer to the question is ALWAYS YES.
Combined, SUFFICIENT

Final Answer: C

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by GMATGuruNY » Tue Sep 01, 2015 10:41 am
anant03 wrote:A list contains twenty integers, not necessarily distinct. Does the list contain at least two consecutive integers?

(1) If any single value in the list is increased by 1, the number of different values in the list does not change.

(2) At least one value occurs more than once in the list.

OAC
Each of the cases tested below is composed of integers in ASCENDING ORDER.
{1, 1, 1...1, 1, 2} implies the following 20 integers:
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2}.

Statement 1: If any single value in the list is increased by 1, the number of different values in the list does not change.
Test one case that also satisfies statement 2.
Case 1: {1, 1, 1...1, 1, 2}, for a total of two distinct values
Here, if 1 increases to 2, or if 2 increases to 3, the list will still contain two distinct values.
In this case, the list contains at least two consecutive integers, so the answer to the question stem is YES.

Test one case that doesn't also satisfy statement 2.
Case 2: {1, 3, 5...35, 37, 39}, for a total of 20 distinct values
Here, if any integer in the list increases by 1, the list will still contain 20 distinct values.
In this case, the list does not contain at least two consecutive integers, so the answer to the questions stem is NO.
INSUFFICIENT.

Statement 2: At least one value occurs more than once in the list.
Case 1 also satisfies statement 2.
In Case 1, the list contains at least two consecutive integers, so the answer to the question stem is YES.
Case 3: {1, 1, 1...1, 1, 1}
In this case, the list does not contain at least two consecutive integers, so the answer to the questions stem is NO.
INSUFFICIENT.

Statements combined:
Case 1 satisfies both statements.

Try adjusting Case 1 so that it does NOT contain at least two consecutive integers.
Case 4: {1, 1, 1...1, 1, 3}, for a total of two distinct values
Not viable:
If one of the 1's increases to 2, then the number of distinct values will increase from two to three, violating the constraint in statement 1.
Implication:
To satisfy both statements, the list MUST contain at least two consecutive integers, so the answer to the question stem is YES.
SUFFICIENT.

The correct answer is C.
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by Max@Math Revolution » Wed Sep 02, 2015 9:20 pm
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


A list contains twenty integers, not necessarily distinct. Does the list contain at least two consecutive integers?

(1) If any single value in the list is increased by 1, the number of different values in the list does not change.

(2) At least one value occurs more than once in the list.

In the original condition we have 20 variables and we need 20 equations to match the number of variables and equations. Since we only have 1 each in 1) and 2), E is likely the answer. Using both 1) & 2) together (<-- instead of solving things with 1), 2) separately, using them together saves us time)

Even if any integer is increased by 1,the total number of different integers do not change, and at there is at least one pair of same integer values. Therefore, there is at least 1 pair of consecutive integers. Assuming there isn't any consecutive pair, for example {1,1,.....,1,3}, we have two different integers (1, 3). When the value 1 is increased by 1, we get {1,,....,1,2,3} and thus we get three different integers 1,2,3 which goes against cond 1). Therefore, C is the answer.

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by nikhilgmat31 » Thu Sep 03, 2015 5:35 am
GMATGuruNY wrote:
anant03 wrote:A list contains twenty integers, not necessarily distinct. Does the list contain at least two consecutive integers?

(1) If any single value in the list is increased by 1, the number of different values in the list does not change.

(2) At least one value occurs more than once in the list.

OAC
Each of the cases tested below is composed of integers in ASCENDING ORDER.
{1, 1, 1...1, 1, 2} implies the following 20 integers:
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2}.

Statement 1: If any single value in the list is increased by 1, the number of different values in the list does not change.
Test one case that also satisfies statement 2.
Case 1: {1, 1, 1...1, 1, 2}, for a total of two distinct values
Here, if 1 increases to 2, or if 2 increases to 3, the list will still contain two distinct values.
In this case, the list contains at least two consecutive integers, so the answer to the question stem is YES.

Test one case that doesn't also satisfy statement 2.
Case 2: {1, 3, 5...35, 37, 39}, for a total of 20 distinct values
Here, if any integer in the list increases by 1, the list will still contain 20 distinct values.
In this case, the list does not contain at least two consecutive integers, so the answer to the questions stem is NO.
INSUFFICIENT.

Statement 2: At least one value occurs more than once in the list.
Case 1 also satisfies statement 2.
In Case 1, the list contains at least two consecutive integers, so the answer to the question stem is YES.
Case 3: {1, 1, 1...1, 1, 1}
In this case, the list does not contain at least two consecutive integers, so the answer to the questions stem is NO.
INSUFFICIENT.

Statements combined:
Case 1 satisfies both statements.

Try adjusting Case 1 so that it does NOT contain at least two consecutive integers.
Case 4: {1, 1, 1...1, 1, 3}, for a total of two distinct values
Not viable:
If one of the 1's increases to 2, then the number of distinct values will increase from two to three, violating the constraint in statement 1.
Implication:
To satisfy both statements, the list MUST contain at least two consecutive integers, so the answer to the question stem is YES.
SUFFICIENT.

The correct answer is C.

But Mitch if we combine both statements - I have reduced items to 5 for simplicity
and take a set as ( 1,1,1,1,5)

if we increase 5 by 1 number of distinct numbers will not change from 2.
but the set don't have any 2 consecutive numbers.
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by GMATGuruNY » Thu Sep 03, 2015 5:52 am
nikhilgmat31 wrote:=But Mitch if we combine both statements - I have reduced items to 5 for simplicity
and take a set as ( 1,1,1,1,5)

if we increase 5 by 1 number of distinct numbers will not change from 2.
but the set don't have any 2 consecutive numbers.
The set in red contains two different values (1 and 5).
If one of the 1's increases to 2, the resulting set -- {1, 1, 1, 2, 5} -- will contain THREE different values, violating the constraint that if any single value in the list is increased by 1, the number of different values in the list does not change.
Thus, the set in red does not satisfy statement 1.
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by Matt@VeritasPrep » Fri Sep 04, 2015 6:06 pm
Another way of thinking about this:

S1::

Suppose that we have two consecutive integers in the set, n and (n - 1). Adding 1 to (n - 1) gives us n, so now instead of n and (n - 1), we have n and n.

This means that (n - 1) is no longer in the list unless we have ANOTHER (n - 1) somewhere else in the list. This would contradict S1, since the number of different values has dropped. So we have two possibilities here:

Possibility 1:: We have (n - 1), another (n - 1), and n in the list. In this case, we CAN have two consecutive integers, since adding 1 to (n - 1) does NOT change the number of different values in the list.

Possibility 2:: We don't have two (n - 1)s. Hence we CANNOT have (n - 1) and n BOTH in the list, since adding 1 to (n - 1) would give us two n's and REDUCE the number of different values in the list. Thus we do NOT have two consecutive integers in the list.

While S1 is not sufficient, it does allow us to rephrase the question as: "Does the list contain three values (n - 1), (n - 1), and n?"

S2 by itself is obviously insufficient.

Taking the two together, we DO have two of the same integer. We know that adding 1 to this integer does NOT change the number of different values in the list, so we must also already have n in the list. (If we didn't already have n, adding 1 to (n - 1) would give us n, a NEW number, and INCREASE the number of different numbers in the list.) Hence we have two consecutive integers in the list, and C is the answer.

Great question!
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