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 #2
avatar+26367 
+9

In how many different ways can 2/15 be represented as 1/A + 1/B,

if A and B are positive integers with A less than or equal to B

 

\(\text{For odd $n>2$ there is always at least one decomposition into exactly two unit fractions: $\dfrac{2}{n} = \dfrac{1}{A} + \dfrac{1}{B}$ } \\ \text{Finding all possibilities.}\\ \text{The prime factorization of $n^2$ results in all possible decompositions into two unit fractions.} \)

 

\(n=15\ \text{is odd} \\ n^2 =225 \\ \text{All divisors of $n^2=225$ are: $1,\ 3,\ 5,\ 9,\ 15,\ 25,\ 45,\ 75,\ 225$}\\ \text{Let $n^2=p\times q$ }\)

 

\(\begin{array}{|r|r|r|c|c|c|c|c|c|c|c|c| } \hline & p & q & n^2 & s & t & r & k & A & B & A\le B & \\ & & & =p\cdot q & = \frac{p+q}{2} & = \frac{p-q}{2} & =\frac{t}{2} & = \frac{15+\sqrt{15^2+t^2} }{2} & =k-r & =k+r & \\ \hline 1. & 225 & 1 & 225=225 \cdot 1 & 113 & 112 & 56 & 64 & 8 & 120 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{8} + \dfrac{1}{120}} \\ \hline 2. & 75 & 3 & 225= 73 \cdot 3 & 39 & 36 & 18 & 27 & 9 & 45 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{9} + \dfrac{1}{45}} \\ \hline 3. & 45 & 5 & 225= 45 \cdot 5 & 25 & 20 & 10 & 20 & 10 & 30 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{10} + \dfrac{1}{30}} \\ \hline 4. & 25 & 9 & 225= 25 \cdot 9 & 17 & 8 & 4 & 16 & 12 & 20 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{12} + \dfrac{1}{20}} \\ \hline 5. & 15 & 15 & 225= 15 \cdot 15 & 15 & 0 & 0 & 15 & 15 & 15 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{15} + \dfrac{1}{15}} \\ \hline \end{array}\)

 

laugh

22 ene 2019
 #7
avatar+26367 
+7

In the sequence 1,2,2,4,8,32,256...

each term (starting from the third term) is the product of the two terms before it. For example, the seventh term is 256  , which is the product of the fifth term (8) and the sixth term (32).

This sequence can be continued forever, though the numbers very quickly grow enormous! (For example, the 14'th term is close to some estimates of the number of particles in the observable universe.)

What is the last digit of the  term of the sequence

 

\(\begin{array}{|lcll|} \hline a_1 &=& 1 \\ a_2 &=& 2\\ a_3 = a_2\cdot a_1 &=& a_2^1 \\ a_4 = a_3\cdot a_2 = a_2^1\cdot a_2 &=& a_2^2 \\ a_5 = a_4\cdot a_3 = a_2^2\cdot a_2^1 &=& a_2^3 \\ a_6 = a_5\cdot a_4 = a_2^3\cdot a_2^3 &=& a_2^5 \\ a_7 = a_6\cdot a_5 = a_2^5\cdot a_2^3 &=& a_2^8 \\ a_8 = a_7\cdot a_6 = a_2^8\cdot a_2^5 &=& a_2^{13}\\ a_9 = a_8\cdot a_7 = a_2^{13}\cdot a_2^8 &=& a_2^{21} \\ \ldots \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \text{Let } \mathcal{F} \text{ are the Fibonacci number }\\\\ \mathcal{F}_1 &=& 1\\ \mathcal{F}_2 &=& 1\\ \mathcal{F}_3 &=& 2\\ \mathcal{F}_4 &=& 3\\ \mathcal{F}_5 &=& 4\\ \mathcal{F}_6 &=& 5\\ \mathcal{F}_7 &=& 13\\ \mathcal{F}_8 &=& 21\\ \mathcal{F}_9 &=& 34\\ \mathcal{F}_{10} &=& 55\\ \mathcal{F}_{11} &=& 89\\ \mathcal{F}_{12} &=& 144\\ \mathcal{F}_{13} &=& 233\\ \mathcal{F}_{14} &=& 377\\ \ldots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_n &=& a_2^{\mathcal{F}_{n-1}} \quad & | \quad a_2 = 2 \\ a_n &=& 2^{\mathcal{F}_{n-1}} \\ \hline \end{array} \)

 

If n = 14:

\(\begin{array}{|rcll|} \hline a_{14} &=& 2^{\mathcal{F}_{14-1}} \\ a_{14} &=& 2^{\mathcal{F}_{13}} \quad & | \quad \mathcal{F}_{13} = 233 \\ a_{14} &=& 2^{233} \\\\ && \text{ The last digit of the term } a_{14} \\ && 2^{233} \pmod{10} \\ &\equiv & 2^{13\cdot 17+12} \pmod{10}\\ &\equiv & \left(2^{13}\right)^{17} 2^{12} \pmod{10} \quad & | \quad 2^{13} \equiv 2 \pmod{10} \\ &\equiv & 2^{17}\cdot 2^{12} \pmod{10} \\ &\equiv & 2^{29} \pmod{10} \\ &\equiv & 2^{13\cdot 2+3} \pmod{10} \\ &\equiv & \left(2^{13}\right)^{2} 2^{3} \pmod{10} \quad & | \quad 2^{13} \equiv 2 \pmod{10} \\ &\equiv & 2^2\cdot 2^3 \pmod{10} \\ &\equiv & 2^{5} \pmod{10} \\ &\equiv & 32 \pmod{10} \\ &\mathbf{\equiv} & \mathbf{ 2 \pmod{10}} \\ \hline \end{array}\)

 

\(\text{ The last digit of the term $a_{14}$ is $\mathbf{2}$}\)

 

laugh

21 ene 2019