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So I came across this neat pattern that I noticed, and so far I've only checked it for 5, and some two digit numbers that end with 5 and 3 digit numbers that end with 5.

The rule seems to be, if a number ends with 5, aside from 5 itself then the square will be a composition of 25 being the right most digits and then all digits after are equal to the square of the original number excluding the 5, plus that number again excluding the five. (below $ba$ is not a product)

So say $a$ will be the last two digits, and $b$ will be all of the other digits.

$$15^2=$$ $$a=5^2$$ $$b = 1^2 + 1$$ $$15^2 =ba= 225$$

and

$$2125^2$$ $$a=5^2$$ $$b=212^2 + 212$$ $$2125^2 = 4515625$$

Is this called anything in mathematics? Is there a way to prove this will be the behavior for all N digits numbers that end with 5.

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    $\begingroup$ I still use the rule that a number ending with 5gives a square that's an oblong number, $n(n+1)$, with 25 tacked on. $\endgroup$ Commented Sep 10, 2016 at 23:30
  • $\begingroup$ This trick is listed here: en.wikipedia.org/wiki/… $\endgroup$ Commented Sep 14, 2016 at 12:53

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If $n$ ends in $5$, we can write $n$ in the form $10k+5$ for some integer $k$. Then

$$n^2=(10k+5)^2=100k^2+100k+25=100(k^2+k)+25\;.$$

Clearly $100(k^2+k)$ ends in $00$, so $n^2$ ends in $25$.

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  • $\begingroup$ To answer the "is this called anything in mathematics" question: It's a special case of a square of a binomial, $$(a+b)^2 = a^2+2ab+b^2$$ $\endgroup$ Commented Sep 11, 2016 at 5:17
  • $\begingroup$ Yet another way to look at it is that (=5 mod 10) * (=5 mod 10) = (=25 mod 100). Is there a standard notation for modular arithmetic with mixed moduli? $\endgroup$ Commented Sep 11, 2016 at 10:35
  • $\begingroup$ It's definitely not true in general that $(a\mod m)*(b\mod n) = (ab\mod mn)$. Indeed, in general the answer is well-defined only $\mod\gcd(m,n)$, not $\mod mn$. Even if $a=b$ and $m=n$, this is not true in general. And any attempt to figure out an algebraic-seeming rule will rely on this simple argument anyway to determine validity, so it's best to just stick with the simple argument. $\endgroup$ Commented Sep 11, 2016 at 20:50
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$$(10n + 5)^2 = 100 n^2 + 100n + 25 = 100n(n + 1) + 25.$$

$\text{ }$

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  • $\begingroup$ And this n(n+1) shows how to easy get the other digits. $\endgroup$ Commented Sep 11, 2016 at 5:10
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Square of Every number ending in 5, say x5, is equal to x*(x+1) followed by 25. Number ending in 5 can be represented as 10n +5

(10n +5)^2 = 100n^2 + 100n+25= 10n * 10 (n +1) + 25.

A Vedic mathematics trick

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