# If $a_{n+1}=\lfloor 1.05\times a_n\rfloor$, does there exist $N$ such that $a_N\equiv0 \$(mod$\ $$10)? I've known the following theorem. Theorem: Supposing that a_{n+1}=\lfloor 1.05\times a_n\rfloor for any natural number n, there exists N such that a_N\equiv0 \ (mod\$$10$) for any integer $20\le a_1\le100$.

Proof: $\{a_n\}$ is a monotonic increase sequence, so let's observe the minimum $n$ such that $a_n\ge100$ for any $a_1$. The observation shows you that you'll always get any one of $100, 101, 102, 103$. Then, you get $$101\to106\to111\to116\to121\to127\to133\to139\to145\to152\to159\to166\to174\to182\to191\to200$$

$$102\to107\to112\to117\to122\to128\to134\to140$$

$$103\to108\to113\to118\to123\to129\to135\to141\to148\to155\to162\to170,$$ so the proof is completed.

Then, here are my questions.

As far as I know, the next question still remains unsolved.

Question1: Supposing that $a_{n+1}=\lfloor 1.05\times a_n\rfloor$ for any natural number $n$, does there exist $N$ such that $a_N\equiv0 \$(mod$\ $$10) for any integer a_1\ge20\ ? It is likely that such N would exist, but I'm facing difficulty. I'm also interested in the following generalization. Question2: Supposing that \alpha\gt1 is a real number and that a_{n+1}=\lfloor \alpha\times a_n\rfloor for any natural number n, does there exist N such that a_N\equiv0 \ (mod\$$10$) for any integer $a_1\ge \frac{1}{\alpha-1} \$?

Any help would be appreciated.

• It may help to observe that $\lfloor1.05 a_n\rfloor=a_n+\lfloor0.05a_n\rfloor$. (Or it may not.) – alex.jordan Aug 8 '13 at 15:24
• @alex.jordan: Thank you. This sequence equals an operation to add the quotient, when you divide $a_n$ by 20, to $a_n$. – mathlove Aug 8 '13 at 15:37
• How the one's digit changes might be useful, or not. – mathlove Aug 9 '13 at 14:55

If I choose $$a_1=23$$ then $$a_2=24$$, $$a_3=25$$, $$a_4=26$$, $$a_5=27$$, $$a_6=28$$, $$a_7=29$$, $$a_8=30$$. Does this answer your question?

In fact I have found that there exit $$N$$ less than 30 for all $$a_1$$ between 20 and 100. Here is a python implementation

import math

for seed in range(20,100):
print('The seed is now: ',seed)
for i in range(30):
seed = math.floor(1.05*seed)
if seed%10 == 0:
print(i)
print(seed)
break


N.B: If $$a_1$$ is $$96$$ then $$a_2$$ is 100.