3n+1_conjecture 11, When 101(binary number) continues 6 times, find the remainder of 7, 2進数101が6回続くとき、これの7の剰余を求めてみましょう。

投稿者: | 01/18/2022

Now, Let’s preparatory exercise for the solution of 3n + 1 conjecture today too.

By the way, we often see questions about binary numbers,
perhaps due to the development of artificial intelligence.
Also, for some reason it is a normal distribution.

今日も準備運動です。
ところで、人工知能開発の影響なのでしょうか、二進数に関する問題をよくみかけます。
また、何らかの理由で正規分布です。

Q: When 101(2) continues six times, can you find the remainder of 7(10)?


First, 101101(2) is 45(10).
Threrfore, 101101101101101101(2) can be transformed into the following.

101101101101101101(2) = 45(212 + 26 + 20) (decimal) = 45(64 x 65 + 1) (decimal) = 187245(10) — (1)

From (1), get below.

187245(10) % 7(10) = 2(10)

Therefore, the answer is “2(10)“.

This question definitely wants you to use “8n ≡ 1n mod 7″.
Therefore, it is intentionally written as 101(2).
But if you can solve it standard, you should solve it normally.

この問題は、 “8n ≡ 1n mod 7″ を積極的に使うことを想定しています。
そのため、わざわざ 101(2) と書いてあるのです。
しかし、普通に解けるのなら、普通に解くべきです。
その合同式に気が付く前に、四則演算が終わります。