How can we quickly square numbers divisible by 5?

Lets go into an example: \( 145^2 \).

We split it into 2 parts : 14 and 5.

The last two digits are 25.

Then, the next succesive digits are \(14 \cdot 15 \), or \( 14 \cdot (14+1) \).

Therefore, \( 145^2=21025 \).

Why does this work?

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TopNewestSimple. Take an \(n\) digit number, where \(n\ge2\), as \(10x+5\). Thus, the number squared is \((10x+5)^{2}\) \(=\) \(100x^{2} + 100x + 25\). Since the coefficients of \(x^{2}\) and \(x\) are multiplied by \(100\), the last two digits are \(25\). Now the next digits are given by \(100(x)(x+1)\), thus proving the property

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@David Lee, I knew that!

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