bluej
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# 91
That's over 10K! Hope it's GBP, or EUR, or plain old $, and not like rupees or something 
Riddles and Puzzles: Extend the following to a valid equation
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As you did with the link, I guess.bluej said:
But we are still looking for the digits of the calculation.
bluej
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Sorry! I had to deduce it's punctuation since 5-somethingx10^(17k+) seems likely to exceed the available equivalent in world circulation in any currency and denomination. But, didn't take into account summing up over parallel universes, my badfresh_42 said:
bluej
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#91:
10k + 652
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Don't be so lazy!bluej said:
\begin{array}{lr}
&9567\\
+&1085\\
\hline \\
&10652
\end{array}
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92. Eight rooks are placed on the main diagonal of a chessboard. Then each of them makes a knight move. At the beginning none of the rooks threatened another rook. Is this possible even after their moves?
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fresh_42 said:
A1 -> B3, D4->C2, B2->A4, C3->D1. No rook is threatening another and the four rooks in that half stayed in their 1/4 of the board (they form the other diagonal now). Mirror it for the other four.
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93. Each of the seven dwarves has his own bed. One evening, the smallest dwarf decides to confuse the process a bit. He does not lie down in his own bed, but in another, randomly picked.
The second smallest dwarf, who goes to sleep next, goes to his bed - if this is free. If this is occupied, he randomly chooses another bed. The following dwarves do the same.
What is the probability that the biggest dwarf will sleep in his own bed this evening?
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The second smallest dwarf, who goes to sleep next, goes to his bed - if this is free. If this is occupied, he randomly chooses another bed. The following dwarves do the same.
What is the probability that the biggest dwarf will sleep in his own bed this evening?
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94. If ##(F_n)_n## is the ordinary Fibonacci sequence (##F_1=F_2=1##). What is
$$
\sum_{n=1}^\infty F_n \,10^{-(n+1)}
$$
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$$
\sum_{n=1}^\infty F_n \,10^{-(n+1)}
$$
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Not anonymous
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##9\;9\;9\;=\;6##
If cube-root and square-root operations are acceptable, then here is one set of solutions:
##(9\; /\; \sqrt 9) + \sqrt 9=\;6##
##\sqrt[3]8 + \sqrt[3]8 + \sqrt[3]8 =\;6##
##7 - (7\; /\; 7) =\;6##
##(6\; / \; 6) \times 6 =\;6##
##(5\; /\; 5) + 5 =\;6##
If cube-root and square-root operations are acceptable, then here is one set of solutions:
##(9\; /\; \sqrt 9) + \sqrt 9=\;6##
##\sqrt[3]8 + \sqrt[3]8 + \sqrt[3]8 =\;6##
##7 - (7\; /\; 7) =\;6##
##(6\; / \; 6) \times 6 =\;6##
##(5\; /\; 5) + 5 =\;6##
KnotTheorist
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We know that the sum converges by the ratio test. If we take the limit as n goes to infinity of |an+1/an|, we get Fn+110-(n+2) / Fn10-(n+1) = Fn+1/10Fn = φ/10 ≈.1618 < 1. Thus the summation converges.
From here, I am lost. I have done some calculations and written out the first hundred terms and they get close to .011235955056.
From here, I am lost. I have done some calculations and written out the first hundred terms and they get close to .011235955056.
bluej
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##\sum_{n=1}^\infty F_n \,10^{-(n+1)} =
\frac {1} {10} \sum_{n=1}^\infty \frac { F_n } {10^n}=\frac {1} {10} \frac {10} {89}=\frac {1} {89}=0.011235955##
\frac {1} {10} \sum_{n=1}^\infty \frac { F_n } {10^n}=\frac {1} {10} \frac {10} {89}=\frac {1} {89}=0.011235955##
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Solution of #94
The Fibonacci series has closed-form solution
$$ F_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right] $$
The sum that we want to calculate I will write more generally as
$$ S(a) = \sum_{n=0}^\infty F_n a^{n+1} $$
Inserting the Fibonacci series gives
$$ S(a) = \frac{a}{\sqrt{5}} \sum_{n=0}^\infty \left[ \left( a \frac{1+\sqrt{5}}{2} \right)^n - \left( a \frac{1-\sqrt{5}}{2} \right)^n \right] $$
We have geometric series here, so we get sums
$$ S(a) = \frac{a}{\sqrt{5}} \left[ \frac{1}{1 - a \frac{1+\sqrt{5}}{2}} - \frac{1}{1 - a \frac{1-\sqrt{5}}{2}} \right] = \frac{a^2}{1 - a - a^2} $$
Substituting in a = 1/10 gives
$$ S(1/10) = \sum_{n=1}^\infty F_n 10^{-n-1} = \frac{1}{89} $$
$$ F_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right] $$
The sum that we want to calculate I will write more generally as
$$ S(a) = \sum_{n=0}^\infty F_n a^{n+1} $$
Inserting the Fibonacci series gives
$$ S(a) = \frac{a}{\sqrt{5}} \sum_{n=0}^\infty \left[ \left( a \frac{1+\sqrt{5}}{2} \right)^n - \left( a \frac{1-\sqrt{5}}{2} \right)^n \right] $$
We have geometric series here, so we get sums
$$ S(a) = \frac{a}{\sqrt{5}} \left[ \frac{1}{1 - a \frac{1+\sqrt{5}}{2}} - \frac{1}{1 - a \frac{1-\sqrt{5}}{2}} \right] = \frac{a^2}{1 - a - a^2} $$
Substituting in a = 1/10 gives
$$ S(1/10) = \sum_{n=1}^\infty F_n 10^{-n-1} = \frac{1}{89} $$
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95. What are the smallest emirp and permutable three digit primes in the decimal system?
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KnotTheorist
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113
131
311
131
311
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Do you know the emirp, too?KnotTheorist said:
bluej
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fresh_42 said:
smallest: 13; smallest 3-digit: 107
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Three decimal digits!bluej said:
bluej
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Knew you'd say datfresh_42 said:
added above in spoilerQuestion not stated very precisely Mr 42..
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bluej said:
Question not read very precisely Mr. bluej..fresh_42 said:
bluej
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Can be read in 2 parts as 1) smallest emirp and 2) permutable three digit primes (i.e. all?)fresh_42 said:
Smallest 3-digit emirp in 10base system is 107.
Its more interesting in base 16!
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Here's one I might regret ... for having to check the solutions ...
96. The rule is simple: choose a pin, jump over another horizontally or vertically into and if there is a hole, and remove the pin you jumped over. Is there a strategy which leads to a single pin in the middle of the board?
In order that we have a common notation, the first moves could (not have to) be:
1. 5-10-H / 10
2. 12-11-10 / 11
3 ...
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96. The rule is simple: choose a pin, jump over another horizontally or vertically into and if there is a hole, and remove the pin you jumped over. Is there a strategy which leads to a single pin in the middle of the board?
In order that we have a common notation, the first moves could (not have to) be:
1. 5-10-H / 10
2. 12-11-10 / 11
3 ...
D104
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The Fibonacci sum without magic and without using the explicit formula:
Take out the first two elements:
$$\sum_{n=1}^\infty F_n 10^{-n-1} = 0.011 + \sum_{n=3}^\infty F_n 10^{-n-1}$$
Now use the recursive definition:
$$\sum_{n=3}^\infty F_n 10^{-n-1} = \sum_{n=3}^\infty (F_{n-2}+F_{n-1}) 10^{-n-1}$$
Some index shifting:$$= \sum_{n=1}^\infty F_{n} 10^{-n-3} + \sum_{n=2}^\infty F_{n} 10^{-n-2}$$
Let the second sum start at n=1 and subtract its first term, in addition unify the exponents:
$$= -0.001 + 0.01\sum_{n=1}^\infty F_{n} 10^{-n-1} + 0.1\sum_{n=1}^\infty F_{n} 10^{-n-1}$$
If we call the original sum X, then we can summarize this as ##X = 0.011 - 0.001 + 0.01 X + 0.1 X## or simplified ##0.89 X = 0.01##, which is solved by ##X=1/89##.
Take out the first two elements:
$$\sum_{n=1}^\infty F_n 10^{-n-1} = 0.011 + \sum_{n=3}^\infty F_n 10^{-n-1}$$
Now use the recursive definition:
$$\sum_{n=3}^\infty F_n 10^{-n-1} = \sum_{n=3}^\infty (F_{n-2}+F_{n-1}) 10^{-n-1}$$
Some index shifting:$$= \sum_{n=1}^\infty F_{n} 10^{-n-3} + \sum_{n=2}^\infty F_{n} 10^{-n-2}$$
Let the second sum start at n=1 and subtract its first term, in addition unify the exponents:
$$= -0.001 + 0.01\sum_{n=1}^\infty F_{n} 10^{-n-1} + 0.1\sum_{n=1}^\infty F_{n} 10^{-n-1}$$
If we call the original sum X, then we can summarize this as ##X = 0.011 - 0.001 + 0.01 X + 0.1 X## or simplified ##0.89 X = 0.01##, which is solved by ##X=1/89##.
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97. What is the smallest natural number ##n## such that ##\varphi(n) \neq \varphi(p)## for all primes ##p\,?##
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#97 - presumably the Euler phi function, the count of all positive integers less than the arg that do not evenly divide the arg.
For prime number p, ## \varphi(p) = p - 1 ##. Thus, for some n where there is no p that makes ## \varphi(n) ## and ## \varphi(p) ## equal, ## \varphi(n) + 1 ## must be a composite number. The smallest n that makes that happen is n = 15, and ## \varphi(15) = 8 ##.
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98. How many options are there to represent the number ##1,000## as sum of two or more consecutive natural numbers?
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#98
Let us first add the consecutive natural numbers between any two of them, ##n_1## and ##n_2##. That sum is
$$ N = \sum_{n=n_1}^{n_2} n = \frac12 n_2(n_2 + 1) - \frac12 n_1(n_1 - 1) = \frac12 (n_2 - n_1 + 1) (n_2 + n_1) $$
or ## 2N = (n_2 - n_1 + 1) (n_2 + n_1) ##. This expression means that a solution can be obtained by factoring 2N. Doing so gives ## 2N = f_1 f_2 ## with the two f's being natural numbers, and one sets each factor equal to a factor of the sum expression: ## f_1 = n_2 - n_1 + 1 ## and ## f_2 = n_1 + n_2 ##. Solving gives
$$ n_1 = \frac12 (f_2 - f_1 +1) ,\ n_2 = \frac12 (f_1 + f_2 - 1) $$
The pairs of lowest and highest natural numbers are thus
$$ N = \sum_{n=n_1}^{n_2} n = \frac12 n_2(n_2 + 1) - \frac12 n_1(n_1 - 1) = \frac12 (n_2 - n_1 + 1) (n_2 + n_1) $$
or ## 2N = (n_2 - n_1 + 1) (n_2 + n_1) ##. This expression means that a solution can be obtained by factoring 2N. Doing so gives ## 2N = f_1 f_2 ## with the two f's being natural numbers, and one sets each factor equal to a factor of the sum expression: ## f_1 = n_2 - n_1 + 1 ## and ## f_2 = n_1 + n_2 ##. Solving gives
$$ n_1 = \frac12 (f_2 - f_1 +1) ,\ n_2 = \frac12 (f_1 + f_2 - 1) $$
The pairs of lowest and highest natural numbers are thus
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99. The regular ##4,294,967,295-##gon can be constructed with compass and ruler. How?
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