heureka

Find the largest positive integer n such that \(12^n\) divides \(1000!\).

\(\begin{array}{|rcll|} \hline \mathbf{2^k =\ ?} \\ \hline k &=& \lfloor\dfrac{1000}{2}\rfloor + \lfloor\dfrac{1000}{2^2}\rfloor+\lfloor\dfrac{1000}{2^3}\rfloor \\ && +\lfloor\dfrac{1000}{2^4}\rfloor+ \lfloor\dfrac{1000}{2^5}\rfloor+\lfloor\dfrac{1000}{2^6}\rfloor \\ && +\lfloor\dfrac{1000}{2^7}\rfloor+ \lfloor\dfrac{1000}{2^8}\rfloor+\lfloor\dfrac{1000}{2^9}\rfloor \\\\ k &=& 500+ 250 +125 \\ && +\lfloor62.5\rfloor + \lfloor31.25\rfloor + \lfloor15.625\rfloor \\ && +\lfloor7.8125\rfloor + \lfloor3.90625\rfloor + \lfloor1.953125\rfloor \\\\ k &=& 500+ 250 +125 +62 +31 \\ && + 15 +7 + 3 +1 \\ \mathbf{k} &=& \mathbf{994} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{3^m =\ ?} \\ \hline m &=& \lfloor\dfrac{1000}{3}\rfloor + \lfloor\dfrac{1000}{3^2}\rfloor+\lfloor\dfrac{1000}{3^3}\rfloor \\ && +\lfloor\dfrac{1000}{3^4}\rfloor+ \lfloor\dfrac{1000}{3^5}\rfloor+\lfloor\dfrac{1000}{3^6}\rfloor \\\\ m &=& +\lfloor333.\bar{3}\rfloor + \lfloor111.\bar{1}\rfloor + \lfloor37.\bar{037}\rfloor \\ && +\lfloor12.34\ldots\rfloor + \lfloor4.11\ldots\rfloor + \lfloor1.37\ldots\rfloor \\\\ m &=& 333+ 111 + 37 \\ && +12 +4 + 1 \\ \mathbf{m} &=& \mathbf{498} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 1000! &=& 2^{994} \times 3^{498}\times \ldots \times 997 \quad | \quad \text{prime factors} \\\\ 1000! &=& \left(2^2\right)^{497} \times 3^{498}\times\ldots \\\\ 1000! &=& \left(2^2\right)^{497} \times 3^{497}3 \times\ldots \\\\ 1000! &=& 4^{497} \times 3^{497}3 \times\ldots \\\\ 1000! &=& (4+3)^{497}3 \times\ldots \\\\ 1000! &=& 12^{\mathbf{497}}3 \times\ldots \\ \hline \end{array}\)

The largest positive integer n such that \(12^n\) divides \(1000!\) is 497

Erich lists the sequence of non-perfect squares:

\(2,\ 3,\ 5,\ 6,\ 7,\ 8,\ 10,\ 11,\ 12,\ 13,\ 14,\ 15,\ 17,\ \ldots\)
What is the 1000th term in this sequence?

Formula: \(a_n = n + \lfloor 0.5 + \sqrt{n} \rfloor \)

\(\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{n + \lfloor 0.5 + \sqrt{n} \rfloor } \quad | \quad n=1000 \\\\ a_{1000} &=& 1000 + \lfloor 0.5 + \sqrt{1000} \rfloor \\ a_{1000} &=& 1000 + \lfloor 0.5 + 31.6\ldots \rfloor \\ a_{1000} &=& 1000 + \lfloor 32.1\ldots \rfloor \\ a_{1000} &=& 1000 + 32 \\ \mathbf{a_{1000}} &=& \mathbf{1032} \\ \hline \end{array}\)

In parallelogram ABCD, M is the midpoint of CD.  
The area of triangle BNC is 42.  
Find the area of triangle MNC.

\(\text{Let the area of $\triangle ABN = A_1$} \\ \text{Let the area of $\triangle BNC = A_2=42$} \\ \text{Let the area of $\triangle MNC = A_3=\ ?$} \\ \text{Let $h=h_1+h_2$ or $h_2=h-h_1$} \)

\(\begin{array}{|rcll|} \hline A_3 &=& \dfrac{xh_1}{2} \\ \mathbf{2A_3} &=& \mathbf{xh_1} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline A_1+A_2 &=& \dfrac{2xh}{2} \\ \mathbf{A_1+A_2} &=& \mathbf{xh} \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A_1 &=& \dfrac{2xh_2}{2} \\ A_1 &=& xh_2 \quad | \quad h_2=h-h_1 \\ A_1 &=& x(h-h_1) \\ A_1 &=& xh-xh_1 \quad | \quad xh=A_1+A_2,\ xh_1=2A_3 \\ A_1 &=& A_1+A_2-2A_3 \\ 2A_3+A_1 &=& A_1+A_2 \\ 2A_3 &=& A_2 \\ A_3 &=& \dfrac{A_2}{2} \quad | \quad A_2 = 42 \\ A_3 &=& \dfrac{42}{2} \\ \mathbf{A_3} &=& \mathbf{21} \\ \hline \end{array}\)

The area of  \(\triangle MNC=\mathbf{21}\)

a=⟨3, −1⟩ and b=⟨−1,−4⟩ .

 complete the head to tail method, draw a +b on a graph without drawing unnecessary vectors.

Can someone help me?

\(\vec{a}=<5,-1>\), and \(\vec{b}=<-3,-4>\).
Represent \(\vec{a+b}\) by using the head to tail method.

Do not include any unnecessary vectors.

a and b are positive integers such that \(ab + 5a + b = 2000\).
Find b.

\(\begin{array}{|rcll|} \hline ab + 5a + b &=& 2000 \\ (a+1)(b+5)-5 &=& 2000 \\ \mathbf{(a+1)*(b+5)} &=& \mathbf{2005} \\ && \text{find all integer products }\ (a+1)*(b+5) = 2005 \\ \hline \end{array} \)

All integer products:

\(\begin{array}{|ccl|r|r|r|r|l|} \hline && (a+1)*(b+5) & (a+1) & (b+5) & a & b & \text{$a$ and $b$ are positive integers} \\ \hline 2005 &=& 1\times 2005 & 1 & 2005 & 0 & 2000 \\ 2005 &=& 5\times 401 & 5 & 401 & \mathbf{4} & \mathbf{396} & \checkmark \\ 2005 &=& 401\times 5 & 401 & 5 & 400 & 0 \\ 2005 &=& 2005\times 1 & 2005 & 1 & 2004 & -4 \\ \hline \end{array} \)

\(\text{$b+5 = 401$, so $b=401-5$ hence $\mathbf{b = 396}$ }\)

The sum of the first n terms of a sequence is \(n^2 + 5n\).  
What is the \(1000th\) term of the sequence?

Formula:  \(\mathbf{\boxed{s_n =\dfrac{(a_1+a_n)}{2}\cdot n }}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_1} &=& \mathbf{s_1} \quad | \quad s_n =n^2 + 5n \\ &=& 1^2 + 5*1 \\ \mathbf{a_1} &=& \mathbf{6} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_n &=& \dfrac{(a_1+a_n)}{2}\cdot n \quad | \quad a_1=6,\ s_n = n^2 + 5n \\\\ n^2 + 5n &=& \dfrac{(6+a_n)}{2}\cdot n \\\\ 2n^2 + 10n &=& (6+a_n)n \quad | \quad :n \\ 2n + 10 &=& 6+a_n \\ 2n + 4 &=& a_n \\ \mathbf{a_n} &=& \mathbf{2n+4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{2n+4} \quad | \quad n=1000 \\ a_{1000} &=& 2*1000+4 \\ \mathbf{a_{1000}}&=& \mathbf{2004} \\ \hline \end{array}\)

Please help with this number theory problem: Find the largest integer n such that \(2^n \)divides \(17^9 - 9^9\).

My attempt:

Formula:

\(\begin{array}{|rcll|} \hline a^9-b^9 &=& (a^3-b^3)(a^6+a^3b^3+b^6) \quad | \quad (a^3-b^3) = (a-b)(a^2+ab+b^2) \\ \mathbf{a^9-b^9} &=& \mathbf{(a-b)(a^2+ab+b^2)(a^6+a^3b^3+b^6)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a^9-b^9} &=& \mathbf{(a-b)(a^2+ab+b^2)(a^6+a^3b^3+b^6)} \\\\ 17^9-9^9 &=& (17-9)(17^2+17*9+9^2)(17^6+17^39^3+9^6) \\ 17^9-9^9 &=& 8*523*28250587 \\ && \boxed{523 ~\text{ is odd and }~ 28250587~\text{ is odd}\\ \text{and } 8=2^3 } \\ 17^9-9^9 &=& 2^3*523*28250587 \\ \hline \end{array} \)

The largest integer n is 3

Find:

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}} \\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\ x-1 &=& \cfrac{1}{2 + x-1} \\ (x-1)(2 + x-1) &=& 1 \\ x+x^2-x-2-x+1 &=& 1 \\ x^2-2 &=& 0 \\ x^2 &=& 2 \\ \mathbf{ x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array}\)

Find the distance between \(Q = (3, -7, -1)\) and the line through \(A = (1, 1, 2)\) and \(B = (2, 3, 4)\).
This distance is equal to \(\dfrac{\sqrt{d}}{3}\) for some integer \(d\).
What is \(d\)?

Line through \(A = (1, 1, 2)\) and \(B = (2, 3, 4)\)

\(\small{ \begin{array}{|rclrcl|} \hline \vec{x} &=& \vec{A}+t(\vec{B}-\vec{A}) & \vec{r} &=& \vec{B}-\vec{A} \quad | \quad \vec{A} = (1, 1, 2)\quad \vec{B} = (2, 3, 4)\\ &&& \vec{r} &=& \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}-\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \\ &&&\vec{r} &=& \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \\ \mathbf{\vec{x}} &=& \mathbf{\vec{A}+t\vec{r}} \\ \hline (\vec{x}-\vec{Q})\cdot \vec{r} &=& 0 \quad | \quad \vec{PQ} \text{ is } \perp \text{ to } \vec{r} \\ (\mathbf{\vec{A}+t\vec{r}}-\vec{Q})\cdot \vec{r} &=& 0 \\ (\vec{A}-\vec{Q} + t\vec{r})\cdot \vec{r} &=& 0 & \vec{A}-\vec{Q}&=& \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}-\begin{pmatrix} 3 \\ -7 \\ -1 \end{pmatrix} \quad | \quad \vec{Q} = (3, -7, -1)\\ & & & \vec{A}-\vec{Q}&=& \begin{pmatrix} -2 \\ 8 \\ 3 \end{pmatrix} \\\\ \left[\begin{pmatrix} -2 \\ 8 \\ 3 \end{pmatrix} + t\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \right]\cdot\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} &=& 0 \\\\ \left[\begin{pmatrix} -2+ t \\ 8+2t \\ 3+2t \end{pmatrix} \right]\cdot\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} &=& 0 \\\\ (-2+ t)\cdot 1 + (8+2t)\cdot 2+ (3+2t)\cdot 2 &=& 0 \\ -2+t+16+4t+6+4t &=& 0 \\ 9t+20 &=& 0 \\ \mathbf{t} &=& \mathbf{ -\dfrac{20}{9} } \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline \mathbf{\vec{P}} &=& \vec{A}+t\vec{r} \\ &=& \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}-\mathbf{ \dfrac{20}{9} }\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \\\\ &=& \begin{pmatrix} 1-\dfrac{20}{9} \\ 1-\dfrac{40}{9} \\ 2-\dfrac{20}{9} \end{pmatrix} \\\\ &=& \begin{pmatrix} \mathbf{ -\dfrac{11}{9}} \\ \mathbf{-\dfrac{31}{9}} \\ \mathbf{ -\dfrac{22}{9}} \end{pmatrix} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\vec{PQ}} &=& \vec{Q}- \vec{P} \\ &=& \begin{pmatrix} 3 \\ -7 \\ -1 \end{pmatrix}- \begin{pmatrix} \mathbf{ -\dfrac{11}{9}} \\ \mathbf{-\dfrac{31}{9}} \\ \mathbf{ -\dfrac{22}{9}} \end{pmatrix} \\\\ &=& \begin{pmatrix} 3 \\ -7 \\ -1 \end{pmatrix}+ \begin{pmatrix} \dfrac{11}{9} \\ \dfrac{31}{9} \\ \dfrac{22}{9} \end{pmatrix} \\\\ &=& \begin{pmatrix} 3 +\dfrac{11}{9} \\ -7+\dfrac{31}{9} \\ -1+\dfrac{22}{9} \end{pmatrix} \\\\ &=& \begin{pmatrix} \mathbf{ \dfrac{38}{9}} \\ \mathbf{-\dfrac{32}{9}} \\ \mathbf{ \dfrac{13}{9}} \end{pmatrix} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline |\vec{PQ}| &=& \sqrt{ \mathbf{ \left( \dfrac{38}{9} \right)^2} +\mathbf{\left(-\dfrac{32}{9}\right)^2 } +\mathbf{\left( \dfrac{13}{9}\right)^2 } } \\ &=& \sqrt{ \dfrac{38^2}{9^2} + \dfrac{32^2}{9^2} + \dfrac{13^2}{9^2} } \\\\ &=& \sqrt{ \dfrac{1}{9}\left( \dfrac{38^2}{9} + \dfrac{32^2}{9} + \dfrac{13^2}{9} \right) } \\\\ &=& \dfrac{1}{3}\sqrt{ \dfrac{38^2}{9} + \dfrac{32^2}{9} + \dfrac{13^2}{9} } \\\\ &=& \dfrac{1}{3}\sqrt{ \dfrac{1}{9}\left( 38^2+32^2+13^2 \right) } \\\\ &=& \dfrac{1}{3}\sqrt{ \dfrac{2637}{9} } \\\\ &=& \dfrac{1}{3}\sqrt{293} \\\\ &=& \mathbf{ \dfrac{\sqrt{293}}{3} } \quad | \quad \dfrac{\sqrt{d}}{3} \\ \\ \mathbf{d} &=& \mathbf{293}\\ \hline \end{array}\)