A student firstly wrote $x=3$ on the board. For each procces, the stutent deletes the number x and replaces it with either $(2x+4)$ or $(3x+8)$ or $(x^2+5x)$. Is this possible to make the number $(20^{17}+2016)$ on the board? (Explain your answer)
HIDE: Note This type of the question is well known but I am going to make a collection so,Problem
Source: 2017 Azerbaijan Junior\Senior National Olympiad
Tags: combinatorics, Sequence
Telman
03.05.2022 10:00
Looking mod 7 we see that the number written to the board is always 3 mod 7 20¹⁷+2016 gives -1 mod 7 so 20¹⁷+2016 cant be written on the board
KhayalAliyev
03.05.2022 10:20
Good job @telman
lian_the_noob12
05.03.2023 22:33
It was so easy! There is an invariant in mod 7!!
mathmax12
31.07.2023 18:46
Going in mod 7, this must be 3 mod 7, but 20^17+2016=-1 mod 7 so no solutions
zaidova
14.11.2023 20:25
For x=3 (at the first time) 2x+4=10 then 3x+8=17 and x^2+5x=9+15=24 The numbers can be 10, 17 or 24. And such questions we must look the numbers on the same mod. In this problem it always gives 3 for mod 7. 20^17+2016==20^17(mod 7) Fermat says that for a prime number p, we can say; and also (a,p)=1 a^p-1==1(mod p) let's use it: We get 20^6==1(mod 7) 20^5==-6==1(mod 7) So it can't be the written number.
Sadigly
31.08.2024 03:02
mod 7