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How many unique rectangles are possible that have a perimeter of not m [#permalink]
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08 Mar 2015, 09:41
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How many unique rectangles are possible that have a perimeter of not more than 258cm and integers for length of each side? A. 65 B. 129 C. 4158 D. 4160 E. 130
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Re: How many unique rectangles are possible that have a perimeter of not m [#permalink]
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08 Mar 2015, 21:53
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Start with lengths 1,1 1,1; 1,2;1,3.....1,128(128 combinations) 2,2;2,3;2,4....2,127(126 combinations) 3,3;3,4;3,5....3,126(124 combinations) 4,4;4,5;4,6....4,125(122 combinations) 62,62;62,63;62,64;62,65;62,66;62,67(6 combinations) 63,63;63,64;63,65;63,66,(4 combinations) 64,64; 64,65......(2 combinations) sum= n(a1+an)/2 n= 128/2=64 a1=2 an=128 sum= 64*(128+2)/2= 4160 hence answer is D Kudos if this helps



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How many rectangles can be formed such that their perimeter is less [#permalink]
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04 Apr 2015, 06:07
How many unique rectangles can be formed such that their perimeter of not more than 258 and integers for the length of each sides ? a) 129 b) 65 c) 4158 d) 4160 e) None of the above. Apologies, adding exact question and correct options.
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Last edited by Lucky2783 on 04 Apr 2015, 16:44, edited 2 times in total.



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Re: How many rectangles can be formed such that their perimeter is less [#permalink]
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04 Apr 2015, 07:26
Lucky2783 wrote: How many rectangles can be formed such that their perimeter is less than 258 ?
a) 178 b) 284 c) 1094 d) 4160 e) 4095 Author of this task say nothing about integers, but I think it is only possible variant that sides can't be decimals. So our max perimeter should be 256 and sum of two opposite sizes should be 256 / 2 = 128 Let's take some first variant and look for pattern one side  1 and with this side we can make 127 another variants: 11, 12, 13 and so on up to 1127 one side  2 and with this side we can make 126 another variants: 21, 22, 23 and so on up to 2126. But we already have rectangular 21 so we should subtract this variant and we have 125 possible variants one side  3 and we have 123 another possible variants so now we can make a formula for calculating possible variants for side n: variants = 129  2n Our max size for side before repetion wil be equal 129 / 2 = 64.5. (we can use ony 64) And side 64 will be have 1 variant. 129  2*64 = 1 And now we should calculate sum of all this variants from 127 to 1 Formula for calcualting sum of sequence ((1 number + last number) / 2) * quantity of numbers ((127 + 1) / 2) * 64 = 4096 And we don't have such answer in our variants, so it'll be really interesting to know where I have made mistake.
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Re: How many rectangles can be formed such that their perimeter is less [#permalink]
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04 Apr 2015, 09:15
Lucky2783 wrote: How many rectangles can be formed such that their perimeter is less than 258 ?
a) 178 b) 284 c) 1094 d) 4160 e) 4095 hi Harley1980 and Lucky2783, i too think answer is 4096.. although it is necessary that the question mentions that the sides are integers to get a finite numbers, otherwise the answer can be infinite with a pair of sides .1 or .001, etc now if we consider that sides are integer, lets look at the solution.. perimeter has to be an even number so largest perimeter is 256 the possible perimeters are 4,6,8,...256.. if perimeter is 4.. sides are 1 and 1.. possibility 1 if perimeter is 6.. sides are 2 and 1..possibility 1 if perimeter is 8.. sides are 2 and 2 or 1 and 3..possibilities 2 and so on.. it follows a pattern.. each succeeding pair has one extra possiblity.. so total=1+1+2+2+3+3+.....63+63+64=2(1+2+3+...+63)+64=2*64*63/2+64=64*63+64=64*64=4096... if the perimeter is equal to less than 258 then 258 will also add up and it will have 64 posssiblities and the answer than will be 4160 D
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Re: How many rectangles can be formed such that their perimeter is less [#permalink]
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04 Apr 2015, 09:34
chetan2u wrote: Lucky2783 wrote: How many rectangles can be formed such that their perimeter is less than 258 ?
a) 178 b) 284 c) 1094 d) 4160 e) 4095 hi Harley1980 and Lucky2783, i too think answer is 4096.. although it is necessary that the question mentions that the sides are integers to get a finite numbers, otherwise the answer can be infinite with a pair of sides .1 or .001, etc now if we consider that sides are integer, lets look at the solution.. perimeter has to be an even number so largest perimeter is 256 the possible perimeters are 4,6,8,...256.. if perimeter is 4.. sides are 1 and 1.. possibility 1 if perimeter is 6.. sides are 2 and 1..possibility 1 if perimeter is 8.. sides are 2 and 2 or 1 and 3..possibilities 2 and so on.. it follows a pattern.. each succeeding pair has one extra possiblity.. so total=1+1+2+2+3+3+.....63+63+64=2(1+2+3+...+63)+64=2*64*63/2+64=64*63+64=64*64=4096... if the perimeter is equal to less than 258 then 258 will also add up and it will have 64 posssiblities and the answer than will be 4160 D apologies for the confusion , i have not properly worded the question . edited the question with correct answer options and question stem .
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Re: How many rectangles can be formed such that their perimeter is less [#permalink]
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04 Apr 2015, 12:28
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Lucky2783 wrote: chetan2u wrote: Lucky2783 wrote: How many rectangles can be formed such that their perimeter is less than 258 ?
a) 178 b) 284 c) 1094 d) 4160 e) 4095 hi Harley1980 and Lucky2783, i too think answer is 4096.. although it is necessary that the question mentions that the sides are integers to get a finite numbers, otherwise the answer can be infinite with a pair of sides .1 or .001, etc now if we consider that sides are integer, lets look at the solution.. perimeter has to be an even number so largest perimeter is 256 the possible perimeters are 4,6,8,...256.. if perimeter is 4.. sides are 1 and 1.. possibility 1 if perimeter is 6.. sides are 2 and 1..possibility 1 if perimeter is 8.. sides are 2 and 2 or 1 and 3..possibilities 2 and so on.. it follows a pattern.. each succeeding pair has one extra possiblity.. so total=1+1+2+2+3+3+.....63+63+64=2(1+2+3+...+63)+64=2*64*63/2+64=64*63+64=64*64=4096... if the perimeter is equal to less than 258 then 258 will also add up and it will have 64 posssiblities and the answer than will be 4160 D apologies for the confusion , i have not properly worded the question . edited the question with correct answer options and question stem . It's ok, but correct answer is 4096, so your question still have wrong answer.
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Re: How many rectangles can be formed such that their perimeter is less [#permalink]
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04 Apr 2015, 13:35
The answer should be 4096.



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How many rectangles can be formed such that their perimeter is less [#permalink]
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04 Apr 2015, 16:42
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Harley1980 wrote: It's ok, but correct answer is 4096, so your question still have wrong answer.
nopes the answer is correct as i mentioned in the spoiler, the question stem was not correct . 2(l+w) <=258 (l+w) <=129 first casel+w=129 128,1 127,2 126,3 .. .. 65,64 64 such pairsSecond casel+w=128 127,1 126,2 125,3 .. .. 65,63 64,64 64 such pairsfor (l+w)=127 and 126 we will get 63 such pairs for each . finally for (l+w)=3 and 2 we will get only 1 pair. so SUM = \(\frac{(1+64)}{2} * 2*64\) = 64*65 = Answer D
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How many rectangles can be formed such that their perimeter is less [#permalink]
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04 Apr 2015, 17:59
Lucky2783 wrote: Harley1980 wrote: It's ok, but correct answer is 4096, so your question still have wrong answer.
nopes the answer is correct as i mentioned in the spoiler, the question stem was not correct . 2(l+w) <=258 (l+w) <=129 first casel+w=129 128,1 127,2 126,3 .. .. 65,64 64 such pairsSecond casel+w=128 127,1 126,2 125,3 .. .. 65,63 64,64 64 such pairsfor (l+w)=127 and 126 we will get 63 such pairs for each . finally for (l+w)=3 and 2 we will get only 1 pair. so SUM = \(\frac{(1+64)}{2} * 2*64\) = 64*65 = Answer D hi , how can the answer be correct if the question itself is wrong? if the question says not more than 258 , then 64 is added to 4096 and answer becomes 4160 D...
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Re: How many rectangles can be formed such that their perimeter is less [#permalink]
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04 Apr 2015, 18:06
chetan2u wrote: hi , how can the answer be correct if the question itself is wrong? if the question says not more than 258 , then 64 is added to 4096 and answer becomes 4160 D... Agree . Everyone has solved the question correctly. +1 Kudos .
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Re: How many unique rectangles are possible that have a perimeter of not m [#permalink]
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Re: How many unique rectangles are possible that have a perimeter of not m [#permalink]
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12 Apr 2015, 21:46
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quite a few good explanations above there but can someone enlighten me how can we solve this question in less than 2 min?? any recomended shortcuts for this one ???



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Re: How many unique rectangles are possible that have a perimeter of not m [#permalink]
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15 Sep 2017, 07:56
2(L+B) =258 L+B=129 combinations can be 128 1 1271,2 . . . 651,2,3,4.........64 641,2,3............64 631,2,3..........63 . . 21 10 => total possibility => 2x sum of integers from 164 => 2x 64x65/2 => 64x65 => 4160 OPTION D
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Re: How many unique rectangles are possible that have a perimeter of not m [#permalink]
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15 Sep 2017, 09:29
anshul2014 wrote: quite a few good explanations above there but can someone enlighten me how can we solve this question in less than 2 min?? any recomended shortcuts for this one ??? I know, it's too late now but still here you go. 2(length + breadth) ≤ 258 (length + breadth) ≤ 129 Now, neither of them can be zero. So, let's give them 1 each from 129. It reduces to (length + breadth) ≤ 127 This is a distribution question now. We will have many scenarios where length and breadth values will interchange which are repeated solutions. We will have to have those cases. But, there will be scenarios where interchanging the values of length and breadth won't affect as in when they are equal (talking about squares ). How many squares will be formed. Since, they will be equal, their sum will be even. So, we have a total of 64 such square cases. Total repeated rectangles formed = [C(127+2, 2) + 64]/2 = 4160. The Variable : I change but remain constant
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Re: How many unique rectangles are possible that have a perimeter of not m [#permalink]
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15 Sep 2017, 09:30
The.Variable wrote: anshul2014 wrote: quite a few good explanations above there but can someone enlighten me how can we solve this question in less than 2 min?? any recomended shortcuts for this one ??? I know, it's too late now but still here you go. 2(length + breadth) ≤ 258 (length + breadth) ≤ 129 Now, neither of them can be zero. So, let's give them 1 each from 129. It reduces to (length + breadth) ≤ 127 This is a distribution question now. We will have many scenarios where length and breadth values will interchange which are repeated solutions. We will have to have those cases. But, there will be scenarios where interchanging the values of length and breadth won't affect as in when they are equal (talking about squares ). How many squares will be formed. Since, they will be equal, their sum will be even. So, we have a total of 64 such square cases. Total repeated rectangles formed = [C(127+2, 2) + 64]/2 = 4160. The Variable : I change but remain constant Edit: Read: We will have to remove such cases.The Variable : I change but remain constant
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