heureka

What is the value of the sum\( \dfrac {1}{1\cdot 3} + \dfrac {1}{3\cdot 5} + \dfrac {1}{5\cdot 7} + \dfrac {1}{7\cdot 9} + \cdots + \dfrac {1}{199\cdot 201}\)?

Express your answer as a fraction in simplest form.

\(\begin{array}{rcll} && \dfrac{1}{1*3} + \dfrac{1}{3*5} + \dfrac{1}{5*7}+ \dfrac{1}{7*9}+\ldots+\dfrac{1}{199*201} \\ &=& \dfrac{1}{1*3} + \dfrac{1}{3*5} + \dfrac{1}{5*7}+ \dfrac{1}{7*9}+\ldots+\dfrac{1}{(2n-1)(2n+1)} \\ \hline && \dfrac{1}{(2n-1)(2n+1)} = \dfrac12\left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right) \\ && \dfrac{1}{1*3} = \dfrac12\left( \dfrac{1}{1} - \dfrac{1}{3} \right) \\ && \dfrac{1}{3*5} = \dfrac12\left( \dfrac{1}{3} - \dfrac{1}{5} \right) \\ && \dfrac{1}{5*7} = \dfrac12\left( \dfrac{1}{5} - \dfrac{1}{7} \right) \\ && \dfrac{1}{7*9} = \dfrac12\left( \dfrac{1}{7} - \dfrac{1}{9} \right) \\ && \ldots \\ && \dfrac{1}{199*201} = \dfrac12\left( \dfrac{1}{199} - \dfrac{1}{201} \right) \\ \hline &=& \dfrac12\left( \dfrac{1}{1} - \dfrac{1}{3} \right) + \dfrac12\left( \dfrac{1}{3} - \dfrac{1}{5} \right) + \dfrac12\left( \dfrac{1}{5} - \dfrac{1}{7} \right) + \dfrac12\left( \dfrac{1}{7} - \dfrac{1}{9} \right)+\ldots+\dfrac12\left( \dfrac{1}{199} - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( \dfrac{1}{1} - \underbrace{\dfrac{1}{3} + \dfrac{1}{3}}_{=0} - \underbrace{\dfrac{1}{5}+\dfrac{1}{5}}_{=0} - \underbrace{\dfrac{1}{7} + \dfrac{1}{7}}_{=0} - \underbrace{\dfrac{1}{9}+ \dfrac{1}{9}}_{=0} +\ldots- \underbrace{\dfrac{1}{199}+\dfrac{1}{199}}_{=0} - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( \dfrac{1}{1} - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( 1 - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( \dfrac{201-1}{201} \right) \\ &=& \dfrac12\left( \dfrac{200}{201} \right) \\ &=& \dfrac{100}{201} \\ \end{array}\)

What is the sum of the following sequence:
\(1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000\).

\(\begin{array}{|rcll|} \hline &&\mathbf{ 1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000 } \\ &=& \sum \limits_{k=1}^{996} k(k+1)(k+2)(k+3)(k+4) \\ &=& \sum \limits_{k=1}^{996} k^5 + 10 k^4 + 35 k^3 + 50 k^2 + 24 k \\ &=& \sum \limits_{k=1}^{996} k^5 + 10\sum \limits_{k=1}^{996} k^4 + 35 \sum \limits_{k=1}^{996}k^3 + 50\sum \limits_{k=1}^{996} k^2 + 24\sum \limits_{k=1}^{996} k \\ && \boxed{ \sum \limits_{k=1}^{n} k^1= \dfrac12 n^2 + \dfrac12 n^1 \\ \sum \limits_{k=1}^{n} k^2= \dfrac13 n^3 + \dfrac12 n^2 + \dfrac16 n^1 \\ \sum \limits_{k=1}^{n} k^3= \dfrac14 n^4 + \dfrac12 n^3 + \dfrac14 n^2 \\ \sum \limits_{k=1}^{n} k^4= \dfrac15 n^5 + \dfrac12 n^4 + \dfrac13 n^3 -\dfrac1{30} n^1 \\ \sum \limits_{k=1}^{n} k^5= \dfrac16 n^6 + \dfrac12 n^5 + \dfrac{5}{12} n^4 -\dfrac1{12} n^2 } \\ &=& \dfrac16 996^6 + \dfrac12 996^5 + \dfrac{5}{12} 996^4 -\dfrac1{12} 996^2 + 10\left(\dfrac15 996^5 + \dfrac12 996^4 + \dfrac13 996^3 -\dfrac1{30} 996 \right) \\ &&+ 35\left( \dfrac14 996^4 + \dfrac12 996^3 + \dfrac14 996^2 \right) + 50\left( \dfrac13 996^3 + \dfrac12 996^2 + \dfrac16 996 \right) \\ &&+ 24\left( \dfrac12 996^2 + \dfrac12 996 \right) \\ &=& \dfrac16 996^6 \\ && + \dfrac12 996^5 + \dfrac{10}5 996^5 \\ && + \dfrac{50}{3} 996^4+ \dfrac{10}2 996^4+ \dfrac{35}4 996^4 \\ && + \dfrac{10}3 996^3+ \dfrac{35}2 996^3+ \dfrac{50}3 996^3 \\ && -\dfrac1{12} 996^2 + \dfrac{35}4 996^2 + \dfrac{50}2 996^2 + \dfrac{24}2 996^2 \\ && -\dfrac{10}{30} 996 + \dfrac{50}6 996 + \dfrac{24}2 996 \\ &=& \dfrac16 996^6 + \dfrac{5}{2}\cdot 996^5 + \dfrac{85}{6} 996^4 + \dfrac{75}2 996^3 + \dfrac{274}6 996^2 +20\cdot 996 \\ &=& \mathbf{165170830829004000} \\ \hline \end{array}\)

Given that m and n are positive integers such that m is congruent to 6 (mod 9) and n is congruent to 0 (mod 9), 

what is the largest integer that mn is necessarily divisible by?

I assume,

\(\begin{array}{|rclcl|} \hline m &\equiv& 6 \pmod {9} & \text{or} & m=6+9u,\quad u \in \mathbb{Z} \\ n &\equiv& 0 \pmod {9} & \text{or} & n=9v,\quad v \in \mathbb{Z} \\\\ mn &=& 9v(6+9u) \\ &=& 54v + 81uv \quad &|\quad gcd(54,81)=27 \\ \hline \end{array} \)

If m and n are positive integers, then the largest integer that mn is necessarily divisible by is 27

James has a collection of 20 miniature cars and trucks.
Cars make up 40% of the collection.
If James adds only cars to his collection,
how many cars must he add to make the collection 75% cars?

\(\begin{array}{|rcll|} \hline \mathbf{\text{cars }} &=& \mathbf{20\cdot40\ \%} \\\\ &=& \dfrac{20\cdot 40}{100} \\ &=& \mathbf{8} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{(20+x)\cdot75\ \%} &=& \mathbf{8+x} \\\\ (20+x)\cdot 0.75 &=& 8+x \\ 15+ 0.75x &=& 8+x \\ 7+ 0.75x &=& x \\ x- 0.75x &=& 7 \\ 0.25x &=& 7 \\ x &=& \dfrac{7}{0.25} \\ \mathbf{ x } &=& \mathbf{28} \\ \hline \end{array}\)

He must add 28 cars to make the collection 75% cars.

What is the sum of the tens digit and the units digit in the decimal representation of \(9^{2004}\)?

\(\begin{array}{|rcll|} \hline && 9^{2004} \pmod{100} \\ &\equiv & \left(3^2\right)^{2004} \pmod{100} \\ &\equiv & 3^{2\cdot 2004} \pmod{100} \\ &\equiv & 3^{4008} \pmod{100} \quad &| \quad 3^{\phi(100)} \equiv 1 \pmod{100} \qquad \phi(100) = 40 \\ & & \quad &| \quad 3^{40} \equiv 1 \pmod{100} \\ &\equiv & 3^{40\cdot 100 + 8} \pmod{100} \\ &\equiv & \left(3^{40}\right)^{100} 3^8 \pmod{100} \\ &\equiv & 1^{100} 3^8 \pmod{100} \\ &\equiv & 3^8 \pmod{100} \\ &\equiv & 6561 \pmod{100} \\ &\equiv & \mathbf{61\pmod{100}} \\ \hline \end{array} \)

The sum of the tens digit and the units digit in the decimal representation of \(9^{2004}\) is \(6+1 =\mathbf{ 7}\)

For a positive integer \(n,\ \phi(n)\) denotes the number of positive integers less than or equal to \(n\) that are relatively prime to \(n\).
What is \(\phi(2^{100}) \)?

\(\phi(n)\) is Euler's totient function

Formula, value for a prime power argument:

\(\text{If $p$ is prime and $k \ge 1$, then $\\ \phi(p^k) = p^{k-1}(p-1)$ }\)

\(\begin{array}{|rcll|} \hline \mathbf{\phi(p^k)} &=& \mathbf{ p^{k-1}(p-1) } \\\\ \phi(2^{100}) &=& 2^{100-1}(2-1) \\ \mathbf{\phi(2^{100})} &=& \mathbf{2^{99}} \\ \hline \end{array}\)

Let A be a matrix, and let x and y be linearly independent vectors such that
\(Ax=y,\ Ay=x+2y\)
Then we have that
\(A^5x=ax+by\)  for some scalars of a and b. Find the ordered pair (a,b).

\(\begin{array}{|rcll|} \hline \mathbf{Ax} &=&y \quad & | \quad \cdot A \\ A^2x &=& Ay \quad & | \quad \mathbf{Ay =x+2y} \\ &=& x+2y \quad & | \quad \cdot A \\\\ A^3x &=& Ax+2 Ay \\ &=& y+2(x+2y) \\ &=& 2x+5y \quad & | \quad \cdot A \\\\ A^4x &=& 2Ax+5 Ay \\ &=& 2y+5(x+2y) \\ &=& 5x+12y \quad & | \quad \cdot A \\\\ A^5x &=& 5Ax+12 Ay \\ &=& 5y+12(x+2y) \\ &=& \mathbf{12x+29y} \\ \hline \end{array}\)

1) 

Compute the product

\(\dfrac{(1998^2 - 1996^2)(1998^2 - 1995^2) \dotsm (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \dotsm (1997^2 - 0^2)}.\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{(1998^2 - 1996^2)(1998^2 - 1995^2) \dotsm (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \dotsm (1997^2 - 0^2)}} \\\\ &=& \dfrac{(1998 - 1996)(1998 + 1996)(1998 - 1995)(1998 + 1995) \dotsm (1998 - 0)(1998 + 0)} {(1997 - 1996)(1997 + 1996)(1997 - 1995)(1997 + 1995) \dotsm (1997 - 0)(1997 + 0)} \\\\ &=& \dfrac{2\cdot 3994 \cdot 3 \cdot 3993 \dotsm 1998\cdot 1998 } {1\cdot 3993 \cdot 2 \cdot 3992 \dotsm 1997\cdot 1997 } \\\\ &=& \dfrac{2\cdot 3 \dotsm 1998 } {1 \cdot 2 \dotsm 1997 } \times \dfrac{3994 \cdot 3993 \dotsm 1998 } {3993 \cdot 3992 \dotsm 1997 } \\\\ &=& 1998 \times \dfrac{3994 } {1997 } \\\\ &=& 1998 \times \dfrac{2\cdot 1997} {1997 } \\\\ &=& 2\cdot 1998 \\ &=& \mathbf{3996 } \\ \hline \end{array}\)

3)
Find \(f(k)\) such that: \(\sum \limits_{k=1}^n f(k) = n^3\).

\(\begin{array}{|rcll|} \hline \sum \limits_{k=1}^n f(k) &=& n^3 \\ \underbrace{\sum \limits_{k=1}^{n-1} f(k) }_{=(n-1)^3} + f(n) &=& n^3 \\\\ (n-1)^3 + f(n) &=& n^3 \\ f(n) &=& n^3-(n-1)^3 \\\\ \mathbf{ f(k) } &=& \mathbf{k^3-(k-1)^3} \\ \hline \end{array}\)

Let u, v and w be vectors satisfying

\(u*v=3, u*w=4, v*w=5\).

Then what are 

\((u+2v)*w, (w-u)*v, (3v-2w)*u\)

equal to?

\(\begin{array}{|rcll|} \hline && \mathbf{ (u+2v)*w} \\ &=& u*w+2v*w \\ &=& 4+ 2*5 \\ &=& \mathbf{14} \\ \hline && \mathbf{(w-u)*v} \\ &=& w*v-u*v \\ &=& 5 -3 \\ &=& \mathbf{2} \\ \hline && \mathbf{(3v-2w)*u} \\ &=& 3v*u-2w*u \\ &=& 3*3-2*4 \\ &=& \mathbf{1} \\ \hline \end{array}\)