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Let n and k be positive integers with k ≤ n. From an n × n array of do
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21 Sep 2019, 03:30
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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:
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Let n and k be positive integers with k ≤ n. From an n × n array of do
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19 Oct 2019, 05:17
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: 20190921_1528.png The number of dots in n*n array of dots = \(n^2\), and in k*k = \(k^2\), so \(n^2k^2=48\), where n>k. \(n^2k^2=48....(nk)(n+k)=48\), so product of nk and n+k has to be 48 and since the product is even, at least one of nk or n+k should be even. But nk 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*24so nk=2 and n+k=24..Add both, so 2n=26..n=13 and k=11 (b) 48=4*12so nk=4 and n+k=12..Add both, so 2n=16..n=8 and k=4 (c) 48=6*8so nk=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
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21 Sep 2019, 03:45
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: 20190921_1528.png Number of dots within nxn array and NOT in kxk array = n^2  k^2 So, n^2  k^2 = 48 —> (nk)(n+k) = 48 Number of possible pairs of (n,k) = Number if even factors of 48 [Note that both nk 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 nk = 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 solutionPosted from my mobile device




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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do
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19 Oct 2019, 04:45
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: 20190921_1528.png Hey gmatt1476, Could you please upload the Offcial solutions for the Quant part also please?? Please... Thanks



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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do
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27 Oct 2019, 16:10
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
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06 Nov 2019, 01:05
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: 20190921_1528.png Number of dots within nxn array and NOT in kxk array = n^2  k^2 So, n^2  k^2 = 48 —> (nk)(n+k) = 48 Number of possible pairs of (n,k) = Number if even factors of 48 [Note that both nk 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 nk = 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 solutionPosted from my mobile deviceOk. So why n+k & nk should be even?



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Let n and k be positive integers with k ≤ n. From an n × n array of do
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Updated on: 07 Dec 2019, 22:36
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: 20190921_1528.png Number of dots within nxn array and NOT in kxk array = n^2  k^2 So, n^2  k^2 = 48 —> (nk)(n+k) = 48 Number of possible pairs of (n,k) = Number if even factors of 48 [ Note that both nk 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 nk = 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 solutionPosted from my mobile deviceOk. So why n+k & nk 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)(nk) = 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.



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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do
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06 Nov 2019, 01:26
Now its clear! Thanks ☺
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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do
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12 Nov 2019, 07:14
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
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12 Nov 2019, 07:44
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
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12 Nov 2019, 08:58
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: 20190921_1528.png n^2  k^2 = 48 (n+k)(nk) = 48 = 2^4*3 n+k = 48; nk = 1; n = 49/2; k = 47/2; Not feasible n+k = 24; nk =2; n = 13; k = 11; Feasible solution n+k = 12; nk = 4; n = 8; k = 4; Feasible solution n+k = 16; nk = 3; n=19/2; k = 13/2; Not feasible n+k = 8; nk = 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
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14 Nov 2019, 10:13
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 Answer: 3 pairs of (n,k)
ANS C



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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do
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13 Dec 2019, 20:00
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: 20190921_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. Answer: C
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Re: Let n and k be positive integers with k ≤ n. From an n × n array of do
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13 Dec 2019, 20:22
I did this the long way and not practical for the test. I took differences of squares 17 yes 2nothing =48 ... 48=48 1113=48 Admittedly hard to tell when to stop. only stop when k^2(k1)^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
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30 May 2020, 10:03
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: 20190921_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^248=k^2. I was able to get just 2 values 4948=1^2 and 6448=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
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30 May 2020, 10:03




