Using the trick that for a number ending in $5$, $(\overline{a5})^{2} = a(a+1) \,|\, 25$, find $45^{2}$.

Using the trick that for a number ending in $5$, $(\overline{a5})^{2} = a(a+1) \,|\, 25$, find $45^{2}$.
  1. $2025$
  2. $2050$
  3. $2125$
  4. $1925$

Solution

Here $a = 4$, so $a(a+1) = 4 \cdot 5 = 20$, and the result is $20\,|\,25 = 2025$.

Asked in: IMO

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