How many unique rectangles are possible that have a perimeter of not m

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

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 case
l+w=129
128,1
127,2
126,3
..
..
65,64
64 such pairs
Second case
l+w=128
127,1
126,2
125,3
..
..
65,63
64,64
64 such pairs

for (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

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: 1-1, 1-2, 1-3 and so on up to 1-127
one side - 2 and with this side we can make 126 another variants: 2-1, 2-2, 2-3 and so on up to 2-126. But we already have rectangular 2-1 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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It's ok, but correct answer is 4096, so your question still have wrong answer.
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Please search before posting!
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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 short-cuts for this one ???

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.

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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apologies for the confusion , i have not properly worded the question .
edited the question with correct answer options and question stem .

The answer should be 4096.

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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Agree . Everyone has solved the question correctly.
+1 Kudos .

2(L+B) =258
L+B=129
combinations can be
128- 1
127-1,2
.
.
.
65-1,2,3,4.........64
64-1,2,3............64
63-1,2,3..........63
.
.
2-1
1-0

=> total possibility => 2x sum of integers from 1-64
=> 2x 64x65/2
=> 64x65 => 4160

OPTION D

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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