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# How many distinct rectangles are there with integer side lengths such that the numerical value of area of the rectangle in square units is e

+2
715
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How many distinct rectangles are there with integer side lengths such that the numerical value of area of the rectangle in square units is equal to $5$ times the numerical value of the perimeter in units? (Two rectangles are considered to be distinct if they are not congruent.)

Aug 8, 2017

#1
+94545
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We have the following :

xy  = 5 [ 2 (x + y) ]      where x, y are the dimensions of each rectangle

xy  =  10x + 10y

Solving for y, we have that

y =  [ 10x ] / [ x - 10  ]

It's clear that  x > 10   and as  x→ infinity, y → 10

And because y decreases as x increases....the smallest integer value possible for y is when y = 11 and x  = 110

And y will have an integer value of 60   whenever x = 12

And when x = 14, y = 35

And when x = 15, y = 30

And when x= 20, y = 20

And no x integer values from 21 - 29   result in y being an integer

And when x = 30, y = 15...but these are just the same dimensions as when x = 15 and  y = 30

So....there are 5  distinct rectangles possible which fit the criteria :

20 x 20

15 x 30

14 x 35

12 x 60

11 x 110

Aug 9, 2017
#2
+20847
+1

How many distinct rectangles are there with integer side lengths such that the numerical value of area of the rectangle in square units is equal to 5 times the numerical value of the perimeter in units?

Formula:
$$\begin{array}{rcll} \text{area of the rectangle } &=& xy \\ 5 \times \text{ the numerical value of the perimeter } &=& 5\times [ 2(x+y) ] \\ xy &=& 5\times [ 2(x+y) ] \\ xy &=& 10\times (x+y) \\ xy &=& 10x+ 10y \\ xy - 10x - 10y &=& 0 \quad & | \quad + 100 \\ xy - 10x - 10y +100 &=& 100 \\ (x-10)\times (y-10) &=& 100 \\\\ \mathbf{(x-10)\times (y-10)} & \mathbf{=} & \mathbf{100} \\ \end{array}$$

So x-10 and y-10 are integers, whose product is 100
How many divisors does 100 have?

Divisors:
1 | 2 | 4 | 5 | 10 | 20 | 25 | 50 | 100 (9 divisors)

$$\begin{array}{|rrcll|} \hline \text{or } & 1\times 100 &=& 100 \\ \text{or } & 2\times 50 &=& 100 \\ \text{or } & 4\times 25 &=& 100 \\ \text{or } & 5\times 20 &=& 100 \\ \text{or } & 10\times 10 &=& 100 \\ \hline \end{array}$$

Solution:

$$\begin{array}{|rclcl|} \hline \underbrace{(x-10)}_{=1} &\times& \underbrace{(y-10)}_{=100} & = & \mathbf{1\times 100} \\\\ x-10 = 1 && y-10 = 100 \\ x = 11 && y = 110 \\ &\mathbf{(11\times 110)} \\ \hline \underbrace{(x-10)}_{=2} &\times& \underbrace{(y-10)}_{=50} & = & \mathbf{2\times 50} \\\\ x-10 = 2 && y-10 = 50 \\ x = 12 && y = 60 \\ &\mathbf{(12\times 60)} \\ \hline \underbrace{(x-10)}_{=4} &\times& \underbrace{(y-10)}_{=25} & = & \mathbf{4\times 25} \\\\ x-10 = 4 && y-10 = 25 \\ x = 14 && y = 35 \\ &\mathbf{(14\times 35)} \\ \hline \underbrace{(x-10)}_{=5} &\times& \underbrace{(y-10)}_{=20} & = & \mathbf{5\times 20} \\\\ x-10 = 5 && y-10 = 20 \\ x = 15 && y = 30 \\ &\mathbf{(15\times 30)} \\ \hline \underbrace{(x-10)}_{=10} &\times& \underbrace{(y-10)}_{=10} & = & \mathbf{10\times 10} \\\\ x-10 = 10 && y-10 = 10 \\ x = 20 && y = 20 \\ &\mathbf{(20\times 20)} \\ \hline \end{array}$$

There are 5  distinct rectangles:

$$\mathbf{(11\times 110)} \\ \mathbf{(12\times 60)} \\ \mathbf{(14\times 35)} \\ \mathbf{(15\times 30)} \\ \mathbf{(20\times 20)}$$

Aug 9, 2017