If k is an integer such that 56 < k < 66, what is the value

From (1) k can be: 57, 59, 61, 63, or 65.
From (2) k can be: 59, 62, or 65.

Thus k can be 59, or 65. Two different answers. Hence not sufficient.

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 independent equations ensures a solution.

If k is an integer such that 56 < k < 66, what is the value of k ?

(1) If k were divided by 2, the remainder would be 1.
(2) If k + 1 were divided by 3, the remainder would be 0.


In the original condition, there is 1 variable(k), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), k=2t+1=57,59,61...., which is not unique and not sufficient.
For 2), k+1=3m, k=3m-1=59, 62, 65, which is not unique and not sufficient.
When 1) & 2), n=59,65, which is not unique and not sufficient. Therefore, the answer is E.


 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

k is an integer such that 56 < k < 66 which means k=(57,58,59,60,61,62,63,64,65)
Now
Statement 1
If k were divided by 2, the remainder would be 1. which means k is odd so possible values of k=(57,59,61,63,65), not sufficient

Statement 2
If k + 1 were divided by 3, the remainder would be 0 so the possible values of k+1 are 60,63,66 so possible values of k=(59,62,65), insufficient

Combining 1 and 2
we still get 2 values 59 and 65 so insufficient

Answer is E

Please consider a kudos if you get the explanation

the most common mistake while solving this question is -

frm stat 1 and stat 2
people will compare the values of k = 57 59 61 63 65 (from stat 1 )
to the values of k+1 = 60 63 66 (from stat 2 )

and will end up choosing 63 as the value of K which is WRONG
but from stat 2
the values of k will be 59 62 and 65.

so from 1 and 2 the values will be 59 and 65...not sufficient.
answer - E

Solution:

We are given that k is between 56 and 66. We must determine the value of k.

Statement One Alone:

If k were divided by 2, the remainder would be 1.

This means k must be an odd number. However, since 56 < k < 66, k can be any odd integer between 56 and 66, which means that k could be 57, 59, 61, 63, or 65. Statement one is not sufficient to determine a value of k. We can eliminate answer choices A and D.

Statement Two Alone:

If k + 1 were divided by 3 the remainder would be 0.

This means k + 1 is a multiple of 3. However, since 56 < k < 66, there is more than one value between 56 and 66 that, when increased by 1, will be a multiple of 3. For example, since 60, 63 and 66 are all multiples of 3, we see that k could be 59, 62 or 65. Since we have three possible values for k, statement two is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

From statement one we know that k can be 57, 59, 61, 63, or 65, and from statement two we know that k could be 59, 62 or 65. From the combined statements, then, k could be either 59 or 65; thus, we can’t determine a unique value for k. Statements one and two together are not sufficient to answer the question.

The answer is E.

k can be 57, 58,...........65

(1) K is odd because on division with 2, the remainder is 1.
So k is of the form 2n+1.

K can be 57, 59, 61, 63, or 65

Not sufficient

BCE

(2) That means k is not divisible by 3.
k can be 59, 62, or 65

Not sufficient

Both statement together:

Statement 1: k=57, 59, 61, 63, or 65
Statement 2: k= 59,62, or 65


No unique value for k.

E

The 2 options state that K and K+1 being consecutive numbers are not divisible by 2 and 3 respectively. K not divisible by 2 but K+1 is divisible by 3. This is not possible and hence the answer is straight E.

Hi All,

We're told that K is an INTEGER and that 56 < K < 66. We're asked for the value of k. This question can be solved by TESTing VALUES.

1) If K were divided by 2, the remainder would be 1.

Based on the information in Fact 1, K could be 57, 59, 61, 63 or 65.
Fact 1 is INSUFFICIENT

2) If (K+1) were divided by 3, the remainder would be 0.

Based on the information in Fact 1, (K+1) could be 57, 60, 63 or 66.... so K could be 56, 59, 62 or 65
Fact 2 is INSUFFICIENT

Combined, we can see that there are two possibilities that 'fit' both Facts: 59 and 65
Combined, INSUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi Somesh86

The answer is indeed E but I have doubt in your explanation. Why is it not possible to have an integer K that is not divisible by 2, but simultaneously (K+1) to be divisible by 3. Eg, we can take K=5, here K is not divisible by 2, but K+1 = 6, IS divisible by 3.

Bunuel hello, Is it not possible to use the rebuilding the remainder equation on this question?

If k is an integer such that 56 < k < 66, what is the value of k ?

(1) If k were divided by 2, the remainder would be 1.
57,59 both work
Insufficient
(2) If k + 1 were divided by 3, the remainder would be 0.
59,62,65 all work
Insufficient

C: 59 and 65 work

E.

Answer: Option E

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Nice question. Got it wrong tho, had to hurry because I passed the 2.40 mark up. but soon realised where I went wrong:

my 2 cents:

(1) If k were divided by 2, the remainder would be 1.

\(k=2q+1\), clearly K can be multiple values between 56 and 66 that can satisfy this. ( \(57,59, 61, 63\)..)

Insuff.

(2) If \(k + 1\) were divided by 3, the remainder would be 0.

\(K+1= 3p + 0\) ----- \(k= 3p-1\) . again, K can take multiple values between 56 and 66 ( they just have to be one less than multiple of 3: \(59, 62, 65\))

1+ 2 combined.

\(k= 3p-1\) ;\( k=2q+1\) ----> \(3p-1 = 2q+1\) ----> \(2p= 3p-2\) ( basically, could K be UNIQUE a multiple of 2 such that it K = multiple of \(3 - 2\) ). Here's where I went wrong, 58 is a possibility, but it's not the only one, unfortunately. so, I marked C, which is wrong.

64 is another possibility. hence, even together they're insufficient.

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