heureka

What is the largest integer less than 2018 that cannot be written as a sum of two or more consecutive integers?

Two or more consecutive integers:

\(\small{ \begin{array}{|r|r|r|l|} \hline \text{consecutive} \\ \text{integers} \\ \hline 2 & n+(n+1) & = 2n + 1 & \text{all odd numbers} \\ \hline 3 & (n-1) + n + (n+1) & = 3n & \text{all multiples of}~3 \\ \hline 4 & (n-1) + n + (n+1)+(n+2) & = 2(2n+1) & \text{all odd numbers} \times 2 \\ \hline 5 & (n-2)+(n-1) + n + (n+1)+(n+2) & = 5n & \text{all multiples of}~5 \\ \hline 6 & & = 3(2n+1) & \text{all odd numbers} \times 3 \\ \hline 7 & & = 7n & \text{all multiples of}~7 \\ \hline 8 & & = 4(2n+1) & \text{all odd numbers} \times 4 \\ \hline 9 & & = 9n & \text{all multiples of}~9 \\ \hline 10 & & = 5(2n+1) & \text{all odd numbers} \times 5 \\ \hline 11 & & = 11n & \text{all multiples of}~11 \\ \hline 12 & & = 6(2n+1) & \text{all odd numbers} \times 6 \\ \hline \ldots & \\ \hline \end{array} }\)

Numbers can be written as a sum of two or more consecutive integers:

\(\begin{array}{|l|} \hline \text{Rule $1$: All odd numbers without $1$ }\\ \text{Rule $2$: All multiples of an odd number without $1$ }\\ \text{Rule $3$: All odd numbers $\times$ an even number without $1$ }\\\\ \text{In general all multiples of an odd number can be written }\\ \text{as sum of two or more consecutive integers (without 1)} \\ \hline \end{array}\)

Example:

\(\begin{array}{|rcll|} \hline 3 & 5 & 7 & 9 & 11 & 13 & 15 & \ldots \\ 6 & 10 & 14 & 18 & 22 & 26 & 30 & \ldots \\ 9 & 15 & 21 & 27 & 33 & 39 & 41 & \ldots \\ 12 & 20 & 28 & 36 & 44 & 52 & 60 & \ldots \\ 15 & 25 & 35 & 45 & 55 & 65 & 75 & \ldots \\ 18 & 30 & 42 & 54 & 66 & 78 & 90 & \ldots \\ 21 & 35 & 49 & 64 & 77 & 91 &105 & \ldots \\ 24 & 40 & 56 & 72 & 99 &104 &120 & \ldots \\ \ldots &\ldots &\ldots &\ldots &\ldots &\ldots &\ldots &\ldots \\ \hline \end{array}\)

So all numbers without an odd prime number in their factorisation cannot be written

as a sum of two or more consecutive integers also the number 1.

Thank you, CPhill

What is the smallest positive integer n for which

9n-2 and 7n+3

share a common factor greater than 1?

\(\text{Euclidean algorithm:}~ \boxed{gcd(a-nb,b)=gcd(a,b)} \)

\(\begin{array}{|rcll|} \hline && gcd(9n-2,7n+3) \\ &=& gcd(9n-2-1\cdot(7n+3) ,7n+3)=gcd(2n-5,7n+3) \\ &=& gcd(2n-5,7n+3-3\cdot(2n-5))=gcd(2n-5,n+18) \\ &=& gcd(2n-5-2\cdot(n+18),n+18)=gcd(-41,n+18) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline gcd(-41,\underbrace{n+18}_{=41~(\Rightarrow n>0!)}) \qquad -41~ \text{is prime} \\\\ n+18 &=& 41 \\ n &=& 41-18 \\ n &=& 23 \\ \hline \end{array}\)

The common factor greater than 1 is \(\mathbf{41}\).

and the smallest positive integer n is \(\mathbf{23}\)

Hint: \(gcd(-41,41) = 41\)

if f(x) = -x^3 +3x^2 -2, find the coordinates of the point(s) where the tangent line is horizontal

\(\begin{array}{|lrcllcl|} \hline & f(x) &=& -x^3+3x^2-2 \\\\ & \boxed{f'(x) = 0\ ?} \\\\ & \boxed{f'(x) = -3x^2+6x } \\ & -3x^2+6x &=& 0 \\ & x(-3x+6) &=& 0 \\ 1. & x &=& 0, \qquad & y &=& -0^3+3\cdot 0^2 - 2 \\ & && & y &=& -2 \\ & \text{Point}_1~ (0,-2) \\\\ 2. & -3x+6 &=& 0 \\ & 3x &=& 6 \\ & x &=& \dfrac{6}{3} \\ & x &=& 2, \qquad & y &=& -2^3+3\cdot 2^2 - 2 \\ & && & y &=& -8+12-2 \\ & && & y &=& 2 \\ & \text{Point}_2~ (2,2) \\ \hline \end{array}\)

Raten und Rechnen

\(\begin{array}{rrcrcr} & I && II && III \\ IV & abc &-& def &=& ad \\ & - && - && * \\ V & ghi & : & eg &=& bj \\ \\ \hline \\ VI & eha &+& dhc &=& fbj \\ \end{array}\)

Lösung:

\(a = 7 \\ b = 1 \\ c= 5 \\ d = 6 \\ e = 3 \\ f = 9 \\ g = 4 \\ h = 0 \\ i = 8 \\ j = 2\)

\(\begin{array}{rrcrcr} & I && II && III \\ IV & 715 &-& 639 &=& 76 \\ & - && - && * \\ V & 408 & : & 34 &=& 12 \\ \\ \hline \\ VI & 307 &+& 605 &=& 912 \\ \end{array}\)

if x=sin(x+y), find dy/dx

\(\begin{array}{|rcl|} \hline &&\text{Differentiate both sides of the equation, getting: } \\ D(x) &=& D(\sin(x+y)) \\ &&\text{Use the chain rule}: \\ D(x) &=& \cos(x+y)\cdot D(x+y) \quad | \quad D(x) = 1 \\ 1 &=& \cos(x+y)\cdot D(x+y) \\ 1 &=& \cos(x+y)\cdot(1+y') \\ \dfrac{1}{\cos(x+y)} &=& 1+y' \\ \mathbf{ y' } & \mathbf{=} & \mathbf{ \dfrac{1}{\cos(x+y) }-1 } \\\\ \mathbf{ y' } & \mathbf{=} & \mathbf{ \sec(x+y)-1 } \\ \hline \end{array} \)

if xy³=8, find dy/dx ?

1. Solution:

Differentiate both sides of the equation, getting : \(\begin{array}{|lll|} \hline \text{Differentiate both sides of the equation, getting: } \\ \qquad D(xy^3) &= D(8) \\ \text{Use the product rule}: \\ \qquad \{ xD(y^3)+D(x)y^3\} &= 0 \quad & | \quad D(8) = 0 \\ \qquad \{ x(3y^2y') + (1)\cdot y^3 \} &= 0 \quad & | \quad : y^2 \\ \qquad 3xy' + y &= 0 \quad & | \quad : -y \\ \qquad 3xy' &= -y \quad & | \quad : 3x \\ \qquad \mathbf{ y' } & \mathbf{=-\dfrac{ y } {3x} } \\ \hline \end{array} \)

2. Solution:

\(\text{General formula for derivative of implicit function} \\ \text{ If ${\displaystyle R(x,y)=0,}$ the derivative of the implicit function $y(x)$ is given by } \\ \text{$\dfrac{dy}{dx} = -\dfrac{R_x}{R_y}$} \\ \text{with $R_x = \dfrac{\partial R}{\partial x}$ and $R_y = \dfrac{\partial R}{\partial y}$ } \)

\(\begin{array}{|lrcll|} \hline \text{Begin with: } \\ & xy^3 -8 &=& 0 \\ & R_x &=& 1\cdot y^3 \\ & R_y &=& x\cdot 3y^2 \\ & \dfrac{dy}{dx} &=& -\dfrac{R_x}{R_y} \\ & &=& -\dfrac{y^3}{x\cdot 3y^2} \\ & \mathbf{ \dfrac{dy}{dx} } & \mathbf{=} & \mathbf{-\dfrac{y}{3x}} \\ \hline \end{array}\)

In a trapezoid $ABCD$ with $AB$ parallel to $CD$, the diagonals $AC$ and $BD$ intersect at $E$.

If the area of triangle $ABE$ is 50 square units, and the area of triangle$ADE$ is 20 square units,

what is the area of trapezoid $ABCD$?

\(\text{Let Area $A_1=[ADE] = 20 $}\\ \text{Let Area $A_2=[ABE] = 50 $}\\ \text{Let Area $A_3=[DEC] $}\\ \text{Let Area $A_4=[BEC] $}\\ \text{Let Area $A_5=[ADC] $}\\ \text{Let Area $A_6=[BDC] $}\)

\(\begin{array}{|rcll|} \hline \dfrac{A_2}{A_1} &=& \dfrac{50}{20} \\ &=& \dfrac52 \\\\ \hline A_1 &=& \dfrac{DE\cdot AH}{2} \\\\ A_2 &=& \dfrac{BE\cdot AH}{2} \\\\ A_3 &=& \dfrac{DE\cdot CK}{2} \\\\ A_4 &=& \dfrac{BE\cdot CK}{2} \\ \hline \\ \dfrac{A_2}{A_1} &=& \dfrac{\dfrac{BE\cdot AH}{2}}{\dfrac{DE\cdot AH}{2}} \\\\ &=& \dfrac{BE\cdot AH}{DE\cdot AH} \\\\ \dfrac{A_2}{A_1} &=& \dfrac{BE}{DE} \quad & | \quad \dfrac{A_2}{A_1} = \dfrac52 \\\\ \mathbf{\dfrac52} & \mathbf{=}& \mathbf{\dfrac{BE}{DE}} \\\\ \hline \dfrac{A_4}{A_3} &=& \dfrac{\dfrac{BE\cdot CK}{2}}{\dfrac{DE\cdot CK}{2}} \\\\ &=& \dfrac{BE\cdot CK}{DE\cdot CK} \\\\ \dfrac{A_4}{A_3} &=& \dfrac{BE}{DE} \quad & | \quad \dfrac{BE}{DE} = \dfrac52 \\\\ \mathbf{\dfrac{A_4}{A_3}} &\mathbf{=}& \mathbf{\dfrac52} \qquad \text{or} \qquad \mathbf{A_3 = \dfrac{2}{5}\cdot A_4 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline AM &=& BN \\ \hline \\ A_5 &=& \dfrac{DC\cdot AM}{2} \\\\ A_6 &=& \dfrac{DC\cdot BN}{2} \\ \hline \\ \dfrac{A_5}{A_6} &=& \dfrac{\dfrac{DC\cdot AM}{2}}{\dfrac{DC\cdot BN}{2}} \\\\ &=& \dfrac{DC\cdot AM}{DC\cdot BN} \quad & | \quad AM=BN \\\\ &=& \dfrac{DC\cdot BN}{DC\cdot BN} \\\\ \dfrac{A_5}{A_6} &=& 1 \\\\ \mathbf{ A_5 } & \mathbf{=} & \mathbf{A_6} \quad & | \quad A_5 = A_1+A_3,\quad A_6 =A_4+A_3 \\ A_1+A_3 &=& A_4+A_3 \quad & | \ -A_3\\ A_1 &=& A_4 \quad & | \quad A_1 = 20\\ \mathbf{20} & \mathbf{=}& \mathbf{ A_4 } \\ \hline \\ \mathbf{A_3} & \mathbf{=} & \mathbf{ \dfrac{2}{5}\cdot A_4 } \quad | \quad A_4 = 20 \\ A_3 &=& \dfrac{2}{5}\cdot 20 \\ \mathbf{A_3} & \mathbf{=} & \mathbf{ 8 } \\ \hline \end{array}\)

\(\text{The area of trapezoid $ABCD = A_1+A_2+A_3+A_4 = 20+50+8+20=\mathbf{98}$ }\)

Drag and drop the expressions into the boxes to correctly complete the proof of the polynomial identity.

Hi Melody

For Example:

\(\begin{array}{|lrclrcl|} \hline (1) & \sin(x+y) &=& \sin(x)\cos(y) +\cos(x)\sin(y) \\ (2) & \sin(x-y) &=& \sin(x)\cos(y) -\cos(x)\sin(y) \\ \\ \hline \\ (1)-(2): & \sin(x+y)-\sin(x-y) &=& \sin(x)\cos(y) +\cos(x)\sin(y) \\ &&&-\Big(\sin(x)\cos(y) -\cos(x)\sin(y)\Big) \\ & \sin(x+y)-\sin(x-y) &=& \sin(x)\cos(y) +\cos(x)\sin(y) \\ &&& -\sin(x)\cos(y) +\cos(x)\sin(y) \\ & \sin(x+y)-\sin(x-y) &=& \cos(x)\sin(y) +\cos(x)\sin(y) \\ & \sin(x+y)-\sin(x-y) &=& 2\cos(x)\sin(y) \\ \\ \hline \\ (3): \mathbf{x+y = A} \\ (4): \mathbf{x-y = B} \\ \\ (3)+(4): & 2x = A+B \\ & \mathbf{x = \dfrac{A+B}{2}} \\\\ (3)-(4): & 2y = A-B \\ & \mathbf{ y = \dfrac{A-B}{2}} \\\\ \hline & \sin(x+y)-\sin(x-y) &=& 2\cos(x)\sin(y) \\ & \sin(A)-\sin(B) &=& 2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2} \right) \\ \hline \end{array}\)