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honchos
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Total combinations possible are 5

a ............... b
36 .............. 1
18............... 2
12 .............. 3
9 ............... 4
6 ................. 6

Answer = B
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honchos
Certain rectangle has the length of a centimeters and the width of b centimeters, where a and b are integers and a≥b. If the area of the rectangle is 36 square centimeters, then how many values of a are possible?

A. 4
B. 5
C. 6
D. 7
E. 8

M28-06

The \(area = ab=36=36*1=18*2=12*3=9*4=6*6\), thus \(a\) can take 5 values: 36, 18, 12, 9, and 6.

Answer: B.


Bunuel is there any other method based on some Prime Factorization Concept?

Not really. We should break 36 into the product of two multiples and see in which pairs the first number is more than or equal to the second number.
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Why not a rectangle of sides 7.2 cm and 5 cm or 8.5x4 cm ? I dont see any point highlighting that dimensions have to be pure integers.


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Why not a rectangle of sides 7.2 cm and 5 cm or 8.5x4 cm ? I dont see any point highlighting that dimensions have to be pure integers.


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Certain rectangle has the length of a centimeters and the width of b centimeters, where a and b are integers and a≥b.
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Oh sorry missed that bit.. Damm this small screen of mobile !!


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Question, even if it says the figure is a rectangle, we should still consider a>=b?

I wasn't sure if a=b would be counter, because then it would be a square instead of a rectangle
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I adopted a different approach though, a-b>=0...a*b=36....b=36/a,,,,,,,a-36/a>=0,,,,,a^2-36>=0,,, draw the wavy curve,,, u will get a>=6 or a<=-6,,, a can't be negative, indeed,, hence all the factors of 36 greater than or equal to 6,,,, 6,9,12,18,36,,,,
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