The sequence $1,2,3,4,0,9,6,9,4,8,7,\ldots$ is formed so that each term, starting from the fifth, is the units digit of the sum of the previous four. (a) Do the digits $2,0,0,4$ occur in the sequence in this order? (b) Will the initial digits $1,2,3,4$ ever occur again in this order?
Problem
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Tags: algebra, Sequence
RedFireTruck
08.04.2021 19:09
WRONG Part (a): assume 2004 occurs. then 22004 must also occur. now lets find the x in x22004 such that it also occurs. we must have 0=0+2+2+x so x=-4. however, this is not a units digit, so 2004 will not ever occur in the sequence. QED sorry for bad proof. i can make nicer if needed.
donian9265
08.04.2021 19:12
@above. That is false. $x=6$ also gives a unit digit of $0$. I really hope 1st graders aren't solving this.
No.
Suppose at some point, we have finitely many odd numbers. Then, because we had an odd number at the beginning, at some point, there must be a final odd number, let it be $O$. If we let $E_1,E_2,E_3$ be the even numbers after $O$, we will have $$O,E_1,E_2,E_3$$Adding these up gives an odd number. Therefore, we have reached a contradiction because $O$ was not the last odd number. Therefore, we will have infinitely many odd numbers.
However, suppose we ever a sequence $E_1,E_2,E_3,E_4$ being $4$ even numbers, and the next number is odd. However, this obviously can't happen. Therefore, if we ever have $4$ consecutive even numbers, every number after those numbers must also be even. Therefore, there was finitely many odd numbers, a contradiction. Hence $2,0,0,4$ can never appear.
RedFireTruck
08.04.2021 19:16
Part (b): proof = [1, 2, 3, 4, 0]while proof[-4:] != [1, 2, 3, 4]: proof += [sum(proof[-4:]) % 10]print("yes", proof, "QED")proof = [1, 2, 3, 4, 0] while proof[-4:] != [1, 2, 3, 4]: proof += [sum(proof[-4:]) % 10] print("yes", proof, "QED")RunResetPop Out / woah imagine trying to bash this out rip croatian first graders
donian9265
08.04.2021 19:30
Yea, above beat me to bashing with a much better code as well. But here is an actual solution (without bash)
There are $10^4$ sequences of $4$ numbers. Therefore, after $10^4+c$ for some finite $c$ (it's like $4$ or something but I don't want to be off by $1$ lol), the sequence must have already repeated itself.
Now, suppose the first set of $4$ numbers, $a,b,c,d$ that repeats itself is not $1,2,3,4$ and suppose that the first time this sequence occurs, the number before is $e$ and the second time, it is $f$. It is clear that the number before this sequence is uniquely determined, for example, if the sequence is $1,3,4,5$, the number before must be $7$. (Working modulo $10$, makes this obvious because there are $10$ digits and each will correspond to a different sum modulo $10$). Therefore, if $a,b,c,d$ repeats itself, $e=f$ and thus $e,a,b,c$ must have repeated itself before. A contradiction, thus $1,2,3,4$ repeats itself.
Too lazy to make an actually good proof, sorry.
brmorgan82
08.04.2021 20:03
Dear Friends of AoPS,
it goes odd even odd even even so 2004 cant happen because its 4 evens in a row
we see it repeats every 5 if u take it mod 2 and repeats every 312 if u take it mod 5 so it must repeat for mod 10