heureka

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:

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

What is the remainder of 5^2010 when it is divided by 7?

\(\begin{array}{rcll} 5^{2010} \pmod {7} &=& \ ? \end{array}\)

\(\small{ \begin{array}{|lrcll|} \hline 1. & gcd(7,5) &=& 1 \qquad | \qquad 7 \text{ and } 5 \text{ are relatively prim } \\ 2. & 7 \text{ is a prim number } \\ 3. & \phi() \text{ is Euler's totient function, Euler's phi function }\\ & \phi(p) &=& p-1 \qquad p \text{ is a prim number} \\ & \phi(7) &=& 7-1 \\ & \phi(7) &=& {\color{red}6} \\ 4. & 5^{\phi(7)} &\equiv& 1 \pmod{7} \\ & 5^{{\color{red}6}} &\equiv& 1 \pmod{7} \\ \hline &\text{ Let } \phi(n) \text{ denote the totient function. } \\ &\text{Then } a^{\phi(n)} \equiv 1 \pmod {n} \text{ for all } a \text{ relatively prime to } n. \\ \hline \end{array} }\)

\(\begin{array}{lrcll} 5. & 2010 &=& {\color{red}6}\cdot 335 \\ & 5^{2010 } \pmod{7} &=& 5^{ {\color{red}6}\cdot 335 } \pmod{7} \\ & &=& ( 5^{ {\color{red}6} } )^{335} \pmod{7} \quad & | \quad 5^{{\color{red}6}} \equiv {\color{blue}1} \pmod{7} \\ & &\equiv& ( {\color{blue}1} )^{335} \pmod{7} \\ & &\equiv& 1 \pmod{7} \\ \end{array}\)

The remainder of \(\frac{5^{2010}} {7}\) is \(\mathbf{1}\)

How can the fact that 24^2-5^2=551 be used to find the factors of 551 (not including 1 or 551)?

\(\begin{array}{|rcll|} \hline \text{Binomial Theorem} \\ (a+b)^2 &=& a^2+2ab+b^2 \\ (a-b)^2 &=& a^2-2ab+b^2 \\ (a-b)(a+b) &=& a^2-b^2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (a-b)(a+b) &=& a^2-b^2 \quad & | \quad a = 24 \qquad b = 5 \\ (24-5)(24+5) &=& 24^2-5^2 \\ 19\cdot 29 &=& 24^2-5^2 \\ \mathbf{19\cdot 29} & \mathbf{=} & \mathbf{551} \\ \hline \end{array}\)

solve for y ?

(4 + 2/3)/6 + y/2 = 2

\(\begin{array}{|rcll|} \hline \frac{ 4 + \frac{2}{3} } {6} + \frac{y}{2} &=& 2 \\\\ \frac{1}{6}\cdot \Big( 4 + \frac{2}{3} \Big) + \frac{y}{2} &=& 2 \quad & | \quad \cdot 6 \\\\ \frac{6}{6}\cdot \Big( 4 + \frac{2}{3} \Big) + \frac{6 y}{2} &=& 2\cdot 6 \\\\ 4 + \frac{2}{3} + 3 y &=& 12 \quad & | \quad \cdot 3 \\\\ 4\cdot 3 + \frac{2\cdot 3}{3} + 3 y\cdot 3 &=& 12 \cdot 3 \\\\ 12 + 2 + 9 y &=& 36 \\ 14 + 9 y &=& 36 \quad & | \quad -14 \\ 9y &=& 36 -14 \\ 9y &=& 22 \quad & | \quad :9 \\\\ \mathbf{y} & \mathbf{=} & \mathbf{\frac{22}{9} } \\ \hline \end{array}\)

37 degree celsius to Fahrenheit

Let Celsius  \(^\circ C = 37\)

Let Fahrenheit \( ^\circ F =\ ?\)

\(\begin{array}{|rcll|} \hline ^\circ F &=& ^\circ C \times 1.8 + 32 \\\\ ^\circ F &=& 37^{\circ} \times 1.8 + 32 \\ ^\circ F &=& 66.6 + 32 \\ \mathbf{^\circ F} & \mathbf{=} & \mathbf{98.6} \\ \hline \end{array}\)

37 degree celsius are 98.6 Fahrenheit

P(x)= (x-a)(x-10)(x-20)+15. Given that p(7)=15....

what is the value of a?

\(\begin{array}{|rcll|} \hline P(x) &=& (x-a)(x-10)(x-20)+15 \quad & | \quad x = 7 \\\\ P(7) &=& (7-a)(7-10)(7-20)+15 \quad & | \quad P(7) = 15 \\ 15 &=& (7-a)(7-10)(7-20)+15 \quad & | \quad -15 \\ 0 &=& (7-a)(7-10)(7-20) \\\\ 0 &=& (7-a)(-3)(-13) \\ 0 &=& (7-a)\cdot 3\cdot 13 \\ 0 &=& (7-a)\cdot 39 \quad & | \quad :39 \\ 0 &=& (7-a)\cdot 1 \\ 0 &=& 7-a \\ a &=& 7 \\ \hline \end{array}\)

Fully expand (x^4+x^3+x^2+x+1)^4

Use this calculator web2.0calc:

Points D, E, and F are the midpoints of sides BC, CA, and AB, respectively, of triangle ABC.
Points X, Y, and Z are the midpoints of EF, FD, and DE, respectively.
If the area of triangle XYZ is 21, then what is the area of triangle CXY?

Areas = ?

\(\begin{array}{|rclcrcl|} \hline \triangle ABC &=& A \\ \triangle DEF &=& \frac14 A \\ \triangle XYZ &=& \frac14 \triangle DEF \\ &=& \frac14 \cdot \frac14 A \\ &=& \frac{1}{16} A = 21 \\ \frac{1}{16} A &=& 21 \\ A &=& 16\cdot 21 \\\\ \triangle CXY &=& \frac{AB}{4} \cdot \frac{h}{2} \quad & | & \sin(A) &=& \frac{h}{ \frac{AC}{2} +\frac{AC}{4} } \\ & & & | & \sin(A) &=& \frac{h}{ \frac34 AC } \\ & & & | & h &=& \frac34 AC \sin(A) \\ \triangle CXY &=& \frac{AB}{4} \cdot \frac{ \frac34 AC \sin(A) }{2} \\ &=& \frac{3}{32} \cdot AB\cdot AC\cdot \sin(A) \quad & | & AB\cdot AC\cdot \sin(A) &=& 2A \\ &=& \frac{3}{32} \cdot 2A \\ &=& \frac{6}{32} A \quad & | & A &=& 16\cdot 21 \\ &=& \frac{6}{32} \cdot 16\cdot 21 \\ &=& \frac{6}{2} \cdot 21 \\ &=& 3 \cdot 21 \\ \triangle CXY &=& 63 \\ \hline \end{array}\)

two vertices of right triangle ABC are A(-2,6) and C(7,3).

If the right angle is at vertex A and vertex B is on the x-axis,

identify the coordinates of point B

Let \(\vec{A} = \binom{-2}{6}\)
Let \(\vec{B} = \binom{x}{0}\)
Let \(\vec{C} = \binom{7}{3}\)

\(\begin{array}{rcll} \vec{AC} = \ ? \\ \vec{AC} &=& \vec{C} - \vec{A} \\ \vec{AC} &=& \binom{7}{3} - \binom{-2}{6} \\ \vec{AC} &=& \binom{7+2}{3-6} \\ \vec{AC} &=& \binom{9}{-3} \\ \end{array}\)

\(\begin{array}{rcll} \vec{AB} = \ ? \\ \vec{AB} &=& \vec{B} - \vec{A} \\ \vec{AB} &=& \binom{x}{0} - \binom{-2}{6} \\ \vec{AB} &=& \binom{x+2}{-6} \\ \end{array}\)

\(\triangle ABC\) is a right triangle then \(\vec{AC}\cdot \vec{AB} = 0\)

\(\begin{array}{rcll} \mathbf{ \vec{AC}\cdot \vec{AB} } & \mathbf{=} & \mathbf{0} \\ \binom{9}{-3} \cdot \binom{x+2}{-6} &=& 0 \\ 9\cdot(x+2) + (-3)\cdot (-6) &=& 0 \\ 9x+18+18 &=& 0 \\ 9x+ 36 &=& 0 \quad & \quad :9 \\ x+ 4 &=& 0 \\ \mathbf{x} & \mathbf{=} & \mathbf{-4} \\ \end{array} \)

\(\mathbf{\vec{B} = \binom{-4}{0} } \)

how do i prove triangleABC is a right triangle?

vertices: A(-3, 5), B(4,7), and C(-1,-2)

Let \(\vec{A} = \binom{-3}{5}\)
Let \(\vec{B} = \binom{4}{7}\)
Let \(\vec{C} = \binom{-1}{-2}\)

\(\begin{array}{rcll} \vec{CA} = \ ? \\ \vec{CA} &=& \vec{A} - \vec{C} \\ \vec{CA} &=& \binom{-3}{5} - \binom{-1}{-2} \\ \vec{CA} &=& \binom{-3+1}{5+2} \\ \vec{CA} &=& \binom{-2}{7} \\ \end{array}\)

\(\begin{array}{rcll} \vec{AB} = \ ? \\ \vec{AB} &=& \vec{B} - \vec{A} \\ \vec{AB} &=& \binom{4}{7} - \binom{-3}{5} \\ \vec{AB} &=& \binom{4+3}{7-5} \\ \vec{AB} &=& \binom{7}{2} \\ \end{array}\)

\(\triangle ABC\) is a right triangle if \(\vec{CA}\cdot \vec{AB} = 0\)

\(\begin{array}{rcll} \vec{CA}\cdot \vec{AB} = \ ? \\ \vec{CA}\cdot \vec{AB} &=& \binom{-2}{7} \cdot \binom{7}{2} \\ \vec{CA}\cdot \vec{AB} &=& (-2)\cdot 7 + 7 \cdot 2 \\ \vec{CA}\cdot \vec{AB} &=& -14+14 \\ \vec{CA}\cdot \vec{AB} &=& 0 \\ \end{array}\)

\(\mathbf{\triangle ABC}\) is a right triangle