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Senior Manager  P
Joined: 04 Sep 2017
Posts: 318
Let n and k be positive integers with k ≤ n. From an n × n array of do  [#permalink]

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53 00:00

Difficulty:   95% (hard)

Question Stats: 37% (03:04) correct 63% (02:56) wrong based on 468 sessions

### HideShow timer Statistics 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 [ 15.95 KiB | Viewed 8429 times ]
Math Expert V
Joined: 02 Aug 2009
Posts: 8757
Let n and k be positive integers with k ≤ n. From an n × n array of do  [#permalink]

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1
10
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
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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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12
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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

Posted from my mobile device
##### General Discussion
Manager  B
Joined: 09 Nov 2018
Posts: 93
Schools: ISB '21 (A)
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

Hey gmatt1476,
Please... Thanks VP  D
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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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1
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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Manager  B
Joined: 25 Sep 2018
Posts: 65
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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1
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 Dillesh4096 on 06 Nov 2019, 01:11.
Last edited by Dillesh4096 on 07 Dec 2019, 22:36, edited 1 time in total.
Manager  B
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Posts: 65
Re: Let n and k be positive integers with k ≤ n. From an n × n array of do  [#permalink]

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Now its clear! Thanks ☺

Posted from my mobile device
Manager  B
Joined: 18 Aug 2017
Posts: 130
Re: Let n and k be positive integers with k ≤ n. From an n × n array of do  [#permalink]

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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.

Thank you.
Math Expert V
Joined: 02 Aug 2009
Posts: 8757
Re: Let n and k be positive integers with k ≤ n. From an n × n array of do  [#permalink]

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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.

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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1
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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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do  [#permalink]

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k<=n
when k=1 n=3 ( 8 dots not included)
when k=3 n=5 ( 16 dots not included)
when k=5 n=7 (24 dots not included)

The total number of dots not included = 8+16+24= 48

ANS 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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3
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

We can create the equation:

n x n - k x k = 48

n^2 - k^2 = 48

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

Since n and k are positive integers (and n ≥ k), n + k and n - k are also positive integers. Since the product of n + k and n - k is 48, both n + k and n - k are factors of 48, with n + k being the larger factor and n - k being the smaller factor. Therefore, we can have the following cases:

1) n + k = 48 and n - k = 1

2) n + k = 24 and n - k = 2

3) n + k = 16 and n - k = 3

4) n + k = 12 and n - k = 4

5) n + k = 8 and n - k = 6.

It seems there are 5 possible ordered pairs of (n, k); however, we have to make sure that n and k are integers. On the other hand, since their sum and difference are integers, we need only to make sure one of the values of n and k is an integer. Let’s solve for n in each set of equations above. By adding the two equations in the cases above, we have:

1) 2n = 49 → n = 49/2

2) 2n = 26 → n = 13

3) 2n = 19 → n = 19/2

4) 2n = 16 → n = 8

5) 2n = 14 → n = 7

We see that only 3 of the 5 cases yield an integer value for n; therefore, there are only 3 ordered pairs of (n, k) that are integer values.

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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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I did this the long way and not practical for the test. I took differences of squares
1-7 yes
2-nothing =48
...
4-8=48
11-13=48
Admittedly hard to tell when to stop. only stop when k^2-(k-1)^2>48 I guess

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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

48 dots are not to be selected from a n*n and the balance number of dots has to be a square number-> This is how I understood the question.
Basically- n^2-48=k^2. I was able to get just 2 values- 49-48=1^2 and 64-48=4^2.
What is the other possible solution. Re: Let n and k be positive integers with k ≤ n. From an n × n array of do   [#permalink] 30 May 2020, 10:03

# Let n and k be positive integers with k ≤ n. From an n × n array of do  