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Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
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gmatt1476 wrote:

Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array?

A. 1
B. 2
C. 3
D. 4
E. 5


PS03551.01

Attachment:
2019-09-21_1528.png


Number of dots within nxn array and NOT in kxk array = n^2 - k^2

So, n^2 - k^2 = 48
—> (n-k)(n+k) = 48

Number of possible pairs of (n,k) = Number if even factors of 48 [Note that both n-k and n+k have to be even, else we will get n or k in fraction values which is not allowed, E.g: factor 1x48 is not allowed, since n-k = 1 and n+k = 48 will give (n,k)= (24.5, 23.5) WHICH IS NOT POSSIBLE]

So, feasible factors of 48 = 2x24 [(n,k) = (13,11)], 4x12 [(n,k) = (8,4)] and 6x8 [(n,k) = (7,1)]
= 3 Possible pairs

IMO Option C

Pls Hit kudos if you like the solution

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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
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gmatt1476 wrote:

Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array?

A. 1
B. 2
C. 3
D. 4
E. 5


PS03551.01

Attachment:
2019-09-21_1528.png


In general, an n x n array will have a total of dots
Likewise, an k x k array will have a total of dots
So, - = the number of dots in the n × n array that are NOT in the k × k array

If there are 48 dots in the n × n array but NOT in the k × k array, we can write: - = 48
This means we're looking for integer pairs (n, k) such that k ≤ n that meet the following condition: - = 48
This task is made easier if we recognize that - is a difference of squares, which can be easily factored

Factoring a difference of squares: \(x^2 - y^2 = (x+y)(x-y)\)
This means the equation above can be factored as follows: (n + k)(n - k) = 48
There are five pairs of integers that multiply together to get 48 (48 & 1, 24 & 2, 16 & 3, 12 & 4, 8 & 6)

So, there are five possible cases to consider:
case i) (n + k) = 48 and (n - k) = 1
case ii) (n + k) = 24 and (n - k) = 2
case iii) (n + k) = 16 and (n - k) = 3
case iv) (n + k) = 12 and (n - k) = 4
case v) (n + k) = 8 and (n - k) = 6

Important: Before we select answer choice E, we must recognize that two of 5 cases do not meet the given conditions.
Take case i for example.
If n + k = 48 and n - k = 1, then we can add the two equations to get: 2n = 49, which means n = 24.5 and k = 23.5
Since the question tells us that n and k are positive integers, case i doesn't work

We'll use the same procedure to test the remaining 4 cases
case ii) n + k = 24 and n - k = 2. Add the equations to get 2n = 26, which means n = 13 and k = 11. So, case ii works
case iii) n + k = 16 and n - k = 3. Add the equations to get 2n = 19, which means n = 9.5. So, case iii doesn't work
case iv) n + k = 12 and n - k = 4. Add the equations to get 2n = 16, which means n = 8 and k = 4. So, case iv works
case v) n + k = 8 and n - k = 6. Add the equations to get 2n = 14, which means n = 7 and k = 1. So, case v works

Answer: C
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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
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Alternatively you can trial and error:

n^2 -k^2 = 48
n^2 needs to be greater than 48, so start testing numbers from 7 onward.
7^2 -k^2 =48
-k^2=-1
k^2=1
k=1 .... that's one

Test 8, then 13.
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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
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Dillesh4096 wrote:
gmatt1476 wrote:

Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array?

A. 1
B. 2
C. 3
D. 4
E. 5


PS03551.01

Attachment:
2019-09-21_1528.png


Number of dots within nxn array and NOT in kxk array = n^2 - k^2

So, n^2 - k^2 = 48
—> (n-k)(n+k) = 48

Number of possible pairs of (n,k) = Number if even factors of 48 [Note that both n-k and n+k have to be even, else we will get n or k in fraction values which is not allowed, E.g: factor 1x48 is not allowed, since n-k = 1 and n+k = 48 will give (n,k)= (24.5, 23.5) WHICH IS NOT POSSIBLE]

So, feasible factors of 48 = 2x24 [(n,k) = (13,11)], 4x12 [(n,k) = (8,4)] and 6x8 [(n,k) = (7,1)]
= 3 Possible pairs

IMO Option C

Pls Hit kudos if you like the solution

Posted from my mobile device

Ok. So why n+k & n-k should be even?
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Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
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shaonkarim wrote:
Dillesh4096 wrote:
gmatt1476 wrote:

Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array?

A. 1
B. 2
C. 3
D. 4
E. 5


PS03551.01

Attachment:
2019-09-21_1528.png


Number of dots within nxn array and NOT in kxk array = n^2 - k^2

So, n^2 - k^2 = 48
—> (n-k)(n+k) = 48

Number of possible pairs of (n,k) = Number if even factors of 48 [Note that both n-k and n+k have to be even, else we will get n or k in fraction values which is not allowed, E.g: factor 1x48 is not allowed, since n-k = 1 and n+k = 48 will give (n,k)= (24.5, 23.5) WHICH IS NOT POSSIBLE]

So, feasible factors of 48 = 2x24 [(n,k) = (13,11)], 4x12 [(n,k) = (8,4)] and 6x8 [(n,k) = (7,1)]
= 3 Possible pairs

IMO Option C

Pls Hit kudos if you like the solution

Posted from my mobile device

Ok. So why n+k & n-k should be even?


Let take a case when not both are NOT even
If you see the highlighted part,
A possible value of (n+k)(n-k) = 48x1
—> n + k = 48 &
n - k = 1

Adding both we get (n + k) + (n - k) = 48 + 1
—> 2n = 49
—> n = 24.5
So, we will get the values of n and k as fractions which are not possible as we are talking about nxn matrix. We can’t have a 24.5x24.5 matrix, can we?

Hope I’m clear.

Originally posted by CareerGeek on 06 Nov 2019, 02:11.
Last edited by CareerGeek on 07 Dec 2019, 23:36, edited 1 time in total.
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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
Hi Experts,

Is the OA include all possibilities?

I find that (n,k) = (8,4) is also a possible answer.
But it is not in the OA.

If it is further included, then there will we 3+1 = 4 possible pairs.

Please help

Thank you.
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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
Expert Reply
ballest127 wrote:
Hi Experts,

Is the OA include all possibilities?

I find that (n,k) = (8,4) is also a possible answer.
But it is not in the OA.

If it is further included, then there will we 3+1 = 4 possible pairs.

Please help

Thank you.


It is included in the 3 cases
13,11 ; 8,4 ; 7,1
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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
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gmatt1476 wrote:

Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array?

A. 1
B. 2
C. 3
D. 4
E. 5


PS03551.01

Attachment:
2019-09-21_1528.png


n^2 - k^2 = 48
(n+k)(n-k) = 48 = 2^4*3
n+k = 48; n-k = 1; n = 49/2; k = 47/2; Not feasible
n+k = 24; n-k =2; n = 13; k = 11; Feasible solution
n+k = 12; n-k = 4; n = 8; k = 4; Feasible solution
n+k = 16; n-k = 3; n=19/2; k = 13/2; Not feasible
n+k = 8; n-k = 6; n=7; k = 1; Feasible solution

Since k<n; 3 solutions are feasible

IMO C
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Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
#Easy 60 sec solution, Please hit kudos if you like the approach.

Number of dots within nxn array = n^2, Number of dots within kxk array = k^2
Asked?? n^2-k^2 = 48 ( Given the constraint that both n & k are integers and k<=n)

Approach:

--> ( n+k)(n-k) = 48

Since 48, a number, that can be easily worked upon is a small number to factorize, Lets do it:

48 x 1 =48
24 x 2 =48
16 x 3 =48
12 x 4 =48
8 x 6 =48

On taking the first combination, We get
n + k =48
n - k = 1
--------------(Adding and solving for n)
2n = 49 (Odd) --> Which means n will not be an integer ( but that's against the question constraints).

Hence I will leave all those options where 2n= Odd i.e. the Sum of the above pairs of factors is ODD



Upon quickly observing the available factors, we get to know that 2n= Even only in 3 conditions ( 24 x 2 =48 , 12 x 4 =48 , 8 x 6 =48 ).

Hence only 3 pairs are possible :
(n,k) belongs to [ (13,11), (8,4),(7,1) ]..... Hence Option C is correct
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Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
If you examine the given arrays, they actually give a huge clue.

For each array, the number of dots NOT included = (n)^2 - (k)^2 or the difference of squares.

When N = 3 and K = 5, the amount of dots that are in the N by N array but NOT in the K by K array is = 16

(5)^2 - (3)^2 = 25 - 9 = 16

When N = 3 and K = 1 the number of dots that are in the N by N array but NOT in the K by K array = 8

(3)^2 - (1)^2 = 9 - 1 = 8

Thus, we are looking for how many “difference of squares” we can make with Integers such that the result = 48

(n)^2 - (k)^2 = 48

(n + k) (n - k) = 48

the Number N must be an Integer Value that lies between the 2 numbers that make up a Factor Pair that multiplies to 48. We will be able to find such a number for N when:

(1)the Factor Pair consists of 2 ODD Numbers

OR

(2)the Factor Pair Consists of 2 Even Numbers

There are 3 possible cases that work:

(N+ K) (N - K) = 48
_______________
(24) * (2) ———> N will = 13 and K will = 11 ——-> 13^2 - 11^2 = 48 dots not included in the array


(12) * (4) ———> N will = 8 and K will = 4 ——-> 8^2 - 4^2 = 64 - 16 = 48 dots not included in the array



(8) * (6) ———> N will = 7 and K will = 1 ———> 7^2 - 1^2 = 48 dots not included in the array


There are 3 such solutions to (n, k)

-C-

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Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
gmatt1476 wrote:

Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array?

A. 1
B. 2
C. 3
D. 4
E. 5


PS03551.01

Attachment:
2019-09-21_1528.png


This is my Approach:

In the figure on the left we have n=5 and k=3. 5^2 -3^2 = 25 - 9 = 16 dots in the NxN array of dots that are not in the Kxk array of dots.

In order to have exactly 48 dots in the NxN array of dots that are not in the KxK array of dots, we should divide 48 by 16 = 3 Pairs, which means we need 3 figures like the one on the left in the question in order to 48 dots(16*3) in the NxN array that are not in the KxK array of dots
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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
This is a Hard question. Do it only if you have enough time in the window.

The number of dots in n*n array of dots = \(n^2\), and in k*k array = \(k^2\), so \(n^2 − k^2 = 48\), where n > k.

This implies (n − k)(n + k) = 48, so the product of (n - k) and (n + k) has to be 48 and since the product is even, at least one of n - k or n + k should be even.
Check for even numbers whose product is 48.

We can have the following cases:
Case 1: n + k = 48 and n - k = 1
Case 2: n + k = 24 and n - k = 2
Case 3: n + k = 16 and n - k = 3
Case 4: n + k = 12 and n - k = 4
Case 5: n + k = 8 and n - k = 6.

Since n - k and n + k will have the same property, so both have to be even. This part is tricky. Hope you understand the rule.

Now, there are only 3 cases where this is true – Cases 2, 4, and 5.

Thus, the answer is C.

Hope this helps. :)
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Let n and k be positive integers with k ≤ n. From an n × n array of do [#permalink]
Here its a good thing to know all the squares from 1 to 15. It might not be a very elegant solution to write them all down, but it was the fastest way for me.

Only three pairs fit the description:

49-1
64-16
169-121

I do remember though, that if the median between two numbers is an integer, they can be written as the difference of squares:

so if (n+k)(n-k) = 48 and the factor pairs of 48 are:

1*48
2*24 = (13-11)(13+11)
3*16
4*12 = (8-4)(8+4)
6*8 = (7-1)(7+1)

then for the three bolded pairs the median will be an integer and thus we have three possible solutions for the difference of squares to equal 48.
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Re: Let n and k be positive integers with k n. From an n × n array of do [#permalink]
Expert Reply
The algebra formula difference of squares is frequently asked at the GMAT.
Many GMAT questions use the formula to solve fast the problem.
Try always to create a pattern for the questions. Try to recognize as fast as you can the correct subject involved and the proper tool to be used.
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Re: Let n and k be positive integers with k n. From an n × n array of do [#permalink]
n^2 - k^2 = 48
(n+k)(n-k) = 48

Prime factorisation of 48:
1. 1 x 48 = 48
2. 2 x 24 = 48
3. 3 x 16 = 48
4. 4 x 12 = 48
5. 6 x 8 = 48

n+k = a (the larger factor in the five cases above)
n-k = b (the smaller factor in the five cases above)

Add these two equations:
2n = a+b
From here, we can see that for n to be an integer, the sum of a and b must be even (to cancel out the 2)
Pairs 2, 4, 6 (2 x 24, 4 x 12, 6 x 8) satisfy that -- so they will work.

You can solve further to check, but if running out on time, this much should suffice.
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Re: Let n and k be positive integers with k n. From an n × n array of do [#permalink]
chetan2u wrote:
gmatt1476 wrote:

Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array?

A. 1
B. 2
C. 3
D. 4
E. 5


PS03551.01

Attachment:
2019-09-21_1528.png


The number of dots in n*n array of dots = \(n^2\), and in k*k = \(k^2\), so \(n^2-k^2=48\), where n>k.

\(n^2-k^2=48....(n-k)(n+k)=48\), so product of n-k and n+k has to be 48 and since the product is even, at least one of n-k or n+k should be even.
But n-k and n+k will have the same property, so both have to be even...
check for even numbers whose product is 48..

(a) 48=2*24
so n-k=2 and n+k=24..Add both, so 2n=26..n=13 and k=11
(b) 48=4*12
so n-k=4 and n+k=12..Add both, so 2n=16..n=8 and k=4
(c) 48=6*8
so n-k=6 and n+k=8..Add both, so 2n=14..n=7 and k=1

3 cases

C


Hi!
Thank you for your explanation.
Could not understand this- [b]But n-k and n+k will have the same property, so both have to be even...[b]
Can you please elaborate a bit more- wrt the context?

Thank you!
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Re: Let n and k be positive integers with k n. From an n × n array of do [#permalink]
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