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Apr 22 2016 11:59am
was wondering if anyone could give me some guidance with these three questions:

3. Prove that there are no solutions in positive integers to the equation x3 + y3 = 100



11 Devise a recursive algorithm to find the nth term of the sequence defined by a0 = 1,
a1 = 2, and an = an-1 ∙ an-2 for n = 2, 3, 4 …



12 DNA sequences are made up of 4 bases: A (adenine), C (cytosine), G (guanine), or T
(thymine). How many 8-element DNA sequences
a) Do not end with C or T?
b ) Start with A and ends with T?
c) Start with G?
d) Contain only G or C?

This post was edited by Trig on Apr 22 2016 11:59am
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Apr 22 2016 12:58pm
3) The easiest way is simply to check any possible positive x, y. There are not many of them, given that x^3 and y^3 are both inferior to 100 : x and y must be chosen among 1,2,3 or 4 because 5^3 is already too large.

11)
1 -> A
2 -> B
For i = 1 to (n-1) :
A*B -> C
B -> A
C -> B
End
Display B

Notice that a_n = 2^f_n, where f_n is the nth term of the Fibonacci sequence starting with 0 and 1. Hence, you can produce an explicit formulae for a_n.


12)
a) 2 * 4^7 : 2 choices for the last base, 4 choices for every 7 other bases.
b ) 4^6 : 4 choices for every 6 bases in the middle.
c) 4^7 : 4 choices for every 7 finishing bases.
d) 2^8 : 2 choices for every 8 bases.

... unless I miss an important property of DNA sequences.

This post was edited by feanur on Apr 22 2016 12:58pm
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Apr 24 2016 10:16pm
Quote (feanur @ Apr 22 2016 01:58pm)
3) The easiest way is simply to check any possible positive x, y. There are not many of them, given that x^3 and y^3 are both inferior to 100 : x and y must be chosen among 1,2,3 or 4 because 5^3 is already too large.

11)
1 -> A
2 -> B
For i = 1 to (n-1) :
A*B -> C
B -> A
C -> B
End
Display B

Notice that a_n = 2^f_n, where f_n is the nth term of the Fibonacci sequence starting with 0 and 1. Hence, you can produce an explicit formulae for a_n.


12)
a) 2 * 4^7 : 2 choices for the last base, 4 choices for every 7 other bases.
b ) 4^6 : 4 choices for every 6 bases in the middle.
c) 4^7 : 4 choices for every 7 finishing bases.
d) 2^8 : 2 choices for every 8 bases.

... unless I miss an important property of DNA sequences.


Only thing I can think of on # 12 is that AT and CG would be paired in normal DNA. Not sure they care about that for this problem though.

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