heureka

can someone help me solve numbers 25 and 27? Thanks you :)

without mistake:

25.
\(\begin{array}{rcll} \text{Isosceles triangle: } & \angle{OAD} &=& \angle{ODA} \\\\ \text{Exterior Angle Theorem: } & \angle{AOP} &=& \angle{OAD} + \angle{ODA} \\ & \angle{AOP} &=& \angle{OAD} + \angle{OAD} \\ & \angle{AOP} &=& 2\cdot \angle{OAD} \qquad | \qquad : 2\\ & \frac{\angle{AOP} } {2} &=& \angle{OAD} \\ & \frac{ 41^{\circ} } {2} &=& \angle{OAD} \\ & 20.5^{\circ} &=& \angle{OAD} \\ \end{array}\)

can someone help me solve numbers 25 and 27? Thanks you :)

25.

\(\begin{array}{rcll} \text{Isosceles triangle: } & \angle{OAD} &=& \angle{ODA} \\\\ \text{Exterior Angle Theorem: } & \angle{OAP} &=& \angle{OAD} + \angle{ODA} \\ & \angle{OAP} &=& \angle{OAD} + \angle{OAD} \\ & \angle{OAP} &=& 2\cdot \angle{OAD} \qquad | \qquad : 2\\ & \frac{\angle{OAP} } {2} &=& \angle{OAD} \\ & \frac{ 41^{\circ} } {2} &=& \angle{OAD} \\ & 20.5^{\circ} &=& \angle{OAD} \\ \end{array} \)

How do you calculate the coefficients of a power series such as this one:
[ 1/4x^4 + 1/4x^5 + 1/4x^6]^10. Just the first 3 coefficients, with steps, would be great.
I would greatly appreciate any help. Thanks a million and have a great day.

\(\begin{array}{rcll} && [~ \frac14 \cdot x^4 + \frac14 \cdot x^5 + \frac14 \cdot x^6 ~]^{10} \\ &=& [~ \frac{1}{4} \cdot ( x^4 + x^5 + x^6 ) ~]^{10} \\ &=& \left( \frac{1}{4} \right)^{10} \cdot ( x^4 + x^5 + x^6 )^{10} \\ &=& \left( \frac{1}{4^{10} } \right) \cdot ( x^4 + x^5 + x^6 )^{10} \\ &=& \left( \frac{1}{1048576} \right) \cdot ( x^4 + x^5 + x^6 )^{10} \\\\ && ( a + b + c )^{n} \\ &=& \sum \limits_{k=0}^{n} \sum \limits_{i=0}^{k} \dbinom{n}{k} \dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \qquad | \qquad (n-k)+(k-i)+(i) = n \\\\ && ( x^4 + x^5 + x^6 )^{10} \\ &=& \sum \limits_{k=0}^{10} \sum \limits_{i=0}^{k} \dbinom{10}{k} \dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \qquad | \qquad (10-k)+(k-i)+(i) = 10 \\\\ \end{array} \)

\(\begin{array}{lcll} (k=0,i=0): & \dbinom{10}{0} \dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\ \hline (k=1,i=0): & \dbinom{10}{1} \dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \cdot x^{41} \\ (k=1,i=1): & \dbinom{10}{1} \dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \cdot x^{42} \\ \hline (k=2,i=0): & \dbinom{10}{2} \dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \cdot x^{42} \\ (k=2,i=1): & \dbinom{10}{2} \dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \cdot x^{43} \\ (k=2,i=2): & \dbinom{10}{2} \dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \cdot x^{44} \\ \hline (k=3,i=0): & \dbinom{10}{3} \dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \cdot x^{43} \\ (k=3,i=1): & \dbinom{10}{3} \dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \cdot x^{44} \\ (k=3,i=2): & \dbinom{10}{3} \dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \cdot x^{45} \\ (k=3,i=3): & \dbinom{10}{3} \dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \cdot x^{46} \\ \cdots \end{array}\)

\(\small{ \begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\ & & 10 \cdot x^{41} & 10 \cdot x^{42} &\\ & & & 45 \cdot x^{42} & 90 \cdot x^{43} & 45 \cdot x^{44} & \\ & & & & 120 \cdot x^{43} & 360 \cdot x^{44} & 360 \cdot x^{45} & 120 \cdot x^{46}\\ & & & & & \cdots & \cdots & \cdots\\ \hline sum & x^{40} & 10\cdot x^{41} & 55\cdot x^{42} & 210 \cdot x^{43} & \cdots & \cdots & \cdots \end{array} } \)

\(\begin{array}{rcll} && [~ \frac14 \cdot x^4 + \frac14 \cdot x^5 + \frac14 \cdot x^6 ~]^{10} \\ &=& \frac{1}{1048576}\cdot x^{40} + \frac{10}{1048576}\cdot x^{41} + \frac{55}{1048576}\cdot x^{42} + \frac{210}{1048576}\cdot x^{43} + \frac{615}{1048576}\cdot x^{44}\\ &+& \frac{1452}{1048576}\cdot x^{45} + \frac{2850}{1048576}\cdot x^{46} + \frac{4740}{1048576}\cdot x^{47} + \frac{6765}{1048576}\cdot x^{48} + \frac{8350}{1048576}\cdot x^{49}\\ &+& \frac{8953}{1048576}\cdot x^{50} + \frac{8350}{1048576}\cdot x^{51} + \frac{6765}{1048576}\cdot x^{52} + \frac{4740}{1048576}\cdot x^{53} + \frac{2850}{1048576}\cdot x^{54}\\ &+& \frac{1452}{1048576}\cdot x^{55} + \frac{615}{1048576}\cdot x^{56} + \frac{210}{1048576}\cdot x^{57} + \frac{55}{1048576}\cdot x^{58} + \frac{10}{1048576}\cdot x^{59}\\ &+& \frac{1}{1048576}\cdot x^{60} \end{array} \)

Find the smallest positive integer that satisfies the system of congruences

a)

n congruency sign 2 (mod ll)

n congruency sign 3 (mod 17)

\(\begin{array}{rcll} n &\equiv& {\color{red}2} \pmod {{\color{green}11}} \\ n &\equiv& {\color{red}3} \pmod {{\color{green}17}} \\ \text{Let } m &=& 11\cdot 17 = 187 \\\\ \end{array}\)

Because 11 and 17 are relatively prim ( gcd(11,17) = 1! ) we can go on:

\(\begin{array}{rcll} n &=& {\color{red}2} \cdot {\color{green}17} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ {\color{green}17}^{\varphi({\color{green}11})-1} \pmod {{\color{green}11}} ] }_{=\text{modulo inverse 17 mod 11} } }_{=17^{10-1} \mod {11} }}_{=17^{9} \mod {11}}}_{=2} + {\color{red}3} \cdot {\color{green}11} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ {\color{green}11}^{\varphi({\color{green}17})-1} \pmod {{\color{green}17}} ] }_{=\text{modulo inverse 11 mod 17} } }_{=11^{16-1} \mod {17} }}_{=11^{15} \mod {17}}}_{=14}\\\\ n &=& {\color{red}2} \cdot {\color{green}17} \cdot [ 2] + {\color{red}3} \cdot {\color{green}11} \cdot [14] \\ n &=& 68 + 462 \\ n &=& 530 \\\\ && n \pmod {m}\\ &=& 530 \pmod {187} \\ &=& 156 \\\\ n &=& 156 + k\cdot 187 \qquad k \in Z\\\\ \mathbf{n_{min}} & \mathbf{=}& \mathbf{156} \end{array}\)

b)

n congruency sign 1 (mod 7)

n congruency sign 7 (mod 13)

n congruency sign 13 (mod 20)

\(\begin{array}{rcll} n &\equiv& {\color{red}1} \pmod {{\color{green}7}} \\ n &\equiv& {\color{red}7} \pmod {{\color{green}13}} \\ n &\equiv& {\color{red}13} \pmod {{\color{green}20}} \\ \text{Let } m &=& 7\cdot 13\cdot 20 = 1820 \\ \end{array} \)

Because 7 and 13 and 20 are relatively prim  we can go on:

\(\small{ \begin{array}{l} n = {\color{red}1} \cdot {\color{green}13\cdot 20} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}13\cdot 20) }^{\varphi({\color{green}7}) -1 } \pmod {{\color{green}7}} ] }_{=\text{modulo inverse }(13\cdot 20) mod 7 } }_{=(13\cdot 20)^{6-1} \mod {7}} }_{=(13\cdot 20)^{5} \mod {7}} }_{=(260\pmod{7})^{5} \mod {7}} }_{=(1)^{5} \mod {7}} }_{=1} + {\color{red}7} \cdot {\color{green}7\cdot 20} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}7\cdot 20) }^{\varphi({\color{green}13}) -1} \pmod {{\color{green}13}} ] }_{=\text{modulo inverse } (7\cdot 20) mod 13 } }_{=(7\cdot 20)^{12-1} \mod {13}} }_{=(7\cdot 20)^{11} \mod {13}} }_{=(140\pmod{13})^{11} \mod {13}} }_{=(10)^{11} \mod {13}} }_{=4} + {\color{red}13} \cdot {\color{green}7\cdot 13} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}7\cdot 13) }^{\varphi({\color{green}20}) -1 } \pmod {{\color{green}20}} ] }_{=\text{modulo inverse } (7\cdot 13) mod 20 } }_{=(7\cdot 13)^{8-1} \mod {20}} }_{=(7\cdot 13)^{7} \mod {20}} }_{=(91\pmod{20})^{7} \mod {20}} }_{=(11)^{7} \mod {20}} }_{=11}\\\\ n = {\color{red}1} \cdot {\color{green}13\cdot 20} \cdot [1] + {\color{red}7} \cdot {\color{green}7\cdot 20} \cdot [4] + {\color{red}13} \cdot {\color{green}7\cdot 13} \cdot [11] \\ n = 260+ 3920 + 13013 \\ n = 17193 \\\\ n \pmod {m}\\ = 17193 \pmod {1820} \\ = 813 \\\\ n = 813 + k\cdot 1820 \qquad k \in Z\\\\ \mathbf{n_{min}} \mathbf{=} \mathbf{813} \end{array} }\)

\(\begin{array}{l} \text{In number theory, }\\ \text{Euler's theorem (also known as the Fermat–Euler theorem or Euler's totient theorem)}\\ \text{states that if } \mathbf{n} \text{ and } \mathbf{a} \text{ are coprime positive integers, then }\\ \hline a^{\varphi (n)} \equiv 1 \pmod{n} \end{array}\)

find x and y

\(\begin{array}{lrcl} (1) & 8\cdot 9 &=& x\cdot 2x \\\\ (2) & 4\cdot y &=& x\cdot 2x \\\\ (2) = (1) & 4\cdot y &=& 8\cdot 9 \\ & y &=& \frac{8\cdot 9}{4} \\ & y &=& 2\cdot 9 \\ & \mathbf{y} & \mathbf{=} & \mathbf{ 18 } \\\\ (2) & 4\cdot y &=& x\cdot 2x \\ & 4\cdot 18 &=& x\cdot 2x \\ & 72 &=& 2x^2 \\ & 36 &=& x^2 \\ & 6 &=& x \\ & \mathbf{x} & \mathbf{=} & \mathbf{ 6 } \\\\ \end{array}\)

how many roots does the following have

p(x)=(x^2-5)(x^2+4)(x^2+10)(2x+6)

\(\small{ \begin{array}{rcrcrcrcrcr} && (x^2-5) &\times& (x^2+4) &\times& (x^2+10) &\times& (2x+6) \\ &=& (x^2-5) &\times& (x^2+4) &\times& (x^2+10) &\times& (2\cdot(x+3)) \\ &=& \underbrace{(x^2-5)}_{\text{binomial} } &\times& (x^2+4) &\times& (x^2+10) &\times& (x+3) &\times& 2 \\ &=& (x-\sqrt{5})\cdot (x+\sqrt{5}) &\times& (x^2+4) &\times& (x^2+10) &\times& (x+3) &\times& 2 \\ && && \boxed{~ \begin{array}{rcll} i^2 = -1 \\ -4i^2 = 4 \\ \end{array} ~} && \boxed{~ \begin{array}{rcll} i^2 = -1 \\ -10i^2 = 10 \end{array} ~}\\ &=& (x-\sqrt{5})\cdot (x+\sqrt{5}) &\times& (x^2-4i^2) &\times& (x^2-10i^2) &\times& (x+3) &\times& 2 \\ &=& (x-\sqrt{5})\cdot (x+\sqrt{5}) &\times& \underbrace{(x^2-4i^2)}_{\text{binomial} } &\times& \underbrace{(x^2-10i^2)}_{\text{binomial} } &\times& (x+3) &\times& 2 \\ &=& (x-\sqrt{5})\cdot (x+\sqrt{5}) &\times& (x-2i)\cdot (x+2i)&\times&(x-\sqrt{10}i)\cdot (x+\sqrt{10}i)&\times& (x+3) &\times& 2 \\\\ &=& \underbrace{(x-\sqrt{5})}_{1.\text{ root: }\sqrt{5}} \cdot \underbrace{(x+\sqrt{5})}_{2.\text{ root: }-\sqrt{5}} &\times& \underbrace{(x-2i)}_{3.\text{ root: }2i} \cdot \underbrace{(x+2i)}_{4.\text{ root: }-2i} &\times& \underbrace{(x-\sqrt{10}i)}_{5.\text{ root: }\sqrt{10}i} \cdot \underbrace{(x+\sqrt{10}i)}_{6.\text{ root: }-\sqrt{10}i} &\times& \underbrace{(x+3)}_{7.\text{ root: }-3} &\times& 2 \\ \end{array} }\)

r = 2cos5θ

und

r=2-2cosθ

Hier die Graphen:

Hallo zusammen,

hier kommt meine nächste kleine Knobelei:

Sei Q(n) die Quersumme einer natürlichen Zahl (Summe ihrer Ziffern).

Berechne: Q(Q(Q(\(2016^{777}\))))

\(\begin{array}{rcll} Q( 2016^{777} ) &=& 11700 \\ Q( 11700 ) &=& 9 \\ Q( 9 ) &=& 9 \\\\ \mathbf{ Q(Q(Q( 2016^{777} ))) } & \mathbf{=} & \mathbf{9} \end{array}\)

see: 

Solve for n:

1000= 50 x(1.00125)^n + 10 x(1.00125^n - 1/1.00125 - 1)

Please show your steps.

Let a = 1.00125

\(\small{ \begin{array}{rcll} 1000 &=& 50 \cdot (1.00125)^n + 10 \cdot (1.00125^n - \frac{1}{1.00125} - 1) \qquad & | \qquad a = 1.00125\\ 1000 &=& 50 \cdot a^n + 10 \cdot (a^n - \frac{1}{a} - 1) \qquad & | \qquad :10\\ 100 &=& 5 \cdot a^n + (a^n - \frac{1}{a} - 1)\\ 100 &=& 5 \cdot a^n + a^n - \frac{1}{a} - 1\\ 100 &=& 6 \cdot a^n - \frac{1}{a} - 1 \qquad & | \qquad +1\\ 101 &=& 6 \cdot a^n - \frac{1}{a} \\\\ 6 \cdot a^n - \frac{1}{a} &=& 101 \qquad & | \qquad + \frac{1}{a}\\\\ 6 \cdot a^n &=& 101 + \frac{1}{a} \qquad & | \qquad :6\\\\ a^n &=& \dfrac { 101 + \frac{1}{a} }{ 6 } \qquad & | \qquad \log_{10}()\\\\ \log_{10}(a^n) &=& \log_{10} \left( \dfrac { 101 + \frac{1}{a} }{ 6 } \right)\\\\ n\cdot \log_{10}(a) &=& \log_{10} \left( \dfrac { 101 + \frac{1}{a} }{ 6 } \right) \qquad & | \qquad \log_{10}(a)\\\\ n &=& \dfrac{\log_{10} \left( \dfrac { 101 + \frac{1}{a} }{ 6 } \right) }{ \log_{10}(a)} \qquad & | \qquad a = 1.00125\\\\ n &=& \dfrac{ \log_{10} \left( \dfrac { 101 + \frac{1}{1.00125} }{ 6 } \right) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ \log_{10} \left( \dfrac { 101 + 0.99875156055 }{ 6 } \right) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ \log_{10} \left( \dfrac { 101.998751561 }{ 6 } \right) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ \log_{10} ( 16.9997919268 ) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ 1.23044360575 }{ 0.00054252909} \\\\ \mathbf{n} & \mathbf{=} & \mathbf{2267.97718911} \end{array} }\)

One question, heureka.......did you you find the number of partitions by "trial and error"  or did you use some algorithm for this???

Hallo CPhill,

i did it by computing program. I wrote a algorithm for this in c++.

Here is the counting for all sums from 1 to 60:

   1 = 0
   2 = 0
   3 = 0
   4 = 0
   5 = 0
   6 = 0
   7 = 0
   8 = 0
   9 = 0
  10 = 1
  11 = 10
  12 = 55
  13 = 220
  14 = 715
  15 = 2002
  16 = 4995
  17 = 11340
  18 = 23760
  19 = 46420
  20 = 85228
  21 = 147940
  22 = 243925
  23 = 383470
  24 = 576565
  25 = 831204
  26 = 1151370
  27 = 1535040
  28 = 1972630
  29 = 2446300
  30 = 2930455
  31 = 3393610
  32 = 3801535
  33 = 4121260
  34 = 4325310
  35 = 4395456
  36 = 4325310
  37 = 4121260
  38 = 3801535
  39 = 3393610
  40 = 2930455
  41 = 2446300
  42 = 1972630
  43 = 1535040
  44 = 1151370
  45 = 831204
  46 = 576565
  47 = 383470
  48 = 243925
  49 = 147940
  50 = 85228
  51 = 46420
  52 = 23760
  53 = 11340
  54 = 4995
  55 = 2002
  56 = 715
  57 = 220
  58 = 55
  59 = 10
  60 = 1