数学の質問なんですが、nを2乗するとき、n=1から1ずつ増やしていくと、n^2の1ケタ目が1,4,9,6,5,6,9,4...|質問・相談が会員登録不要のQ&AサイトSooda!(ソーダ)

此、

レディットで　聞いてみました、
This is true - nice observation! This follows because if we write n = a_0 + a_1 * 10 + a_2 * 100 + ... + a_k * 10^k (here, a_0 is the units digit, a_1 is the tens digit, and so on. Concretely, this is like writing 123 = 1 * 100 + 2 * 10 + 3 * 1, just done in general), and then square that expression and expand, every term is divisible by 10, except for the term a_0 * a_0. That term is the only term that can contribute to the units digit (since everything else is divisible by 10, so the units digit of everything else will be 0). Therefore, the units digit of n^2 is the same as the units digit of a_0^2. This tells us "the units digit of n^2 is the units digit of (the units digit of n)^2". Because of this, the units digit of n^2 will cycle based on the units digit of n, and so this pattern will repeat forever.

What you've discovered here is also a special case of modular arithmetic. https://en.wikipedia.org/wiki/Modular_arithmetic modular arithmetic is best thought of as arithmetic with remainders. In this case, the last digit is the remainder when we divide the number by 10. What you've done is a very specific case of the fact that, if x and y are numbers, the remainder when we divide (xy) by n is the product of the individual remainders when we divide x by n and y by n. You've found this for n = 10, and x = y. If this kind of problem interests you, I highly recommend learning some number theory! You'll enjoy it.
との　事です。

===補足===
まあ、

整数同士の　掛け算では、
自身以下には　影響を、
与えません。

詰まり、

10の桁は　1の桁に、
影響を　与えません。

ですから、

全ての　整数は、
I×10+jで　表せますよね、

で　I×10は、
一の位に　影響を、
示さないのだから、

j|j=0→9　此が、
循環しているだけな訳ですよ。

更には、

1、9、
は　m±1、
2、8、
は　m±2、
3、7、
は　m±3、
4、6、
は　m±4、
ですから、
同じに　なるのでしょうね？

回答者：Nouble (質問から1日後)
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