I was experimenting with radicals on my calculator when I noticed that the units digit and the hundredths digit of $\sqrt{2}, \, 2\sqrt{2}, \, ... \, 11\sqrt{2} \,$ are the same. The next one does not work since in $12\sqrt{2} = 16.97...$ ($6 \neq 7$). On doing this up to $50\sqrt{2}$, I found that the numbers $\alpha\sqrt{2}$ which don't work are when $\alpha = 12, 19, 24, 31, 36, 38, 41, 43, 48, 50$.
My conjecture: The units digit and the hundredths digit of a number of the form $k\sqrt{2} \,$ ($k \in \mathbb{N}$) are the same, except when $k$ is of the form $12a+19b+41c$ where $a, b, c \in \mathbb{Z^+} \cup \left \{0 \right \}$.
Does anyone know how to prove this (or give a counterexample)?
Also, if this is true, are there similar conjectures for numbers of the form $k\sqrt{3}, \, k\sqrt{5}, \, ...$?
Thanks!
Other counterexamples are 111 and 140. I doubt that this is true.
Since the Frobenius number of $\{12, 19, 41, 70\}$ is $128$, the story ends. Thus you've proven the following result:
The units digit and the hundredths digit of a number of the form $k\sqrt{2}$, $k\in\mathbb{N}$ are the same, except when k is of the form $12a+19b+41c+70d$ where $a,b,c,d∈\mathbb{N}\cup{0}$.
The proof goes like this: You check all numbers up to 128. For these numbers your statement is true. For numbers greater 128 there is nothing left to proof, because all numbers can be written in the desired form.
In general, a statement like this is very unlikely to be true. Because: Assume it is true at least for all numbers $10^i$, $i\in\mathbb{N}$. This implies that all digits of $\sqrt{2}$ are the same.
The second reason is, there are really lots of numbers. If you find some pattern, test the pattern for many numbers before making a conjecture. Using the PC one can disprove false claims very fast.