Patratele numerelor terminate in 5

[mihai miculita]

Observatii:
1). Ultima cifra a unui numar coincide cu restul impartirii acelui numar la $10.$
Asa ca notand cu $u(n),$ ultima cifra a numarului $n\in\mathbb{N},$ avem: $\boxed{n=10k+r; 0\le r\le 9\Rightarrow u(n)=r}.$
2). In cazul in care
$n=\overline{a_ma_{m-1}\dots a_2a_1a_0}=$$\overline{a_ma_{m-1}\dots a_2a_10}+a_0=10.\overline{a_ma_{m-1}\dots a_2a_1}+a_0\Rightarrow$$\left\{\begin{array}{c} k=\overline{a_ma_{m-1}\dots a_2a_1}\\ r=a_0 \end{array}\right$
3). Orice numar terminat in $5$ (adica: $u(n)=5$) este de forma: $\boxed{n=10k+5}.$
4). $n=10k+5\Rightarrow n^2=(10k+5)^2=100k^2+100k+25=$$\boxed{100k(k+1)+25}.$
Asa ca, pentru a gasi patratul unui numar terminat in $5,$ facem produsul $k(k+1)=k^2+k$ si scriem la capatul acestui produs $25.$
Exemple:
$n=15\Rightarrow k=1\Rightarrow k(k+1)=1.2=2\Rightarrow n^2=15^2=225;$
$n=25\Rightarrow k=2\Rightarrow k(k+1)=2.3=6\Rightarrow n^2=25^2=625;$
$n=35\Rightarrow k=3\Rightarrow k(k+1)=3.4=12\Rightarrow n^2=35^2=1225;$
$n=45\Rightarrow k=4\Rightarrow k(k+1)=4.5=20\Rightarrow n^2=45^2=2025;$
$n=55\Rightarrow k=5\Rightarrow k(k+1)=5.6=30\Rightarrow n^2=55^2=3025;$
$n=65\Rightarrow k=6\Rightarrow k(k+1)=6.7=42\Rightarrow n^2=65^2=4225;$
$n=75\Rightarrow k=7\Rightarrow k(k+1)=7.8=56\Rightarrow n^2=75^2=5625;$
$n=85\Rightarrow k=8\Rightarrow k(k+1)=8.9=72\Rightarrow n^2=85^2=7225;$
$n=95\Rightarrow k=9\Rightarrow k(k+1)=9.10=90\Rightarrow n^2=95^2=9025;$
$n=105\Rightarrow k=10\Rightarrow k(k+1)=10.11=110\Rightarrow n^2=105^2=11025;$
$n=115\Rightarrow k=11\Rightarrow k(k+1)=k^2+k=11^2+11$$=121+11=132\Rightarrow n^2=25^2=13225;$
$n=125\Rightarrow k=12\Rightarrow k(k+1)=k^2+k=12^2+12=$$144+12=156\Rightarrow n^2=125^2=15625;$
$n=135\Rightarrow k=10\Rightarrow k(k+1)=k^2+k=13^2+13=$$169+13=182\Rightarrow n^2=135^2=18225;\dots$