In this post, I will discuss some simple tricks I use to square numbers that begin or end with the digit \( 5 \). We will first see a few illustrations of each trick. Then we will generalise the tricks for any number that begins or ends with the digit \( 5 \).

## Squaring a 2-Digit Number That Ends With 5

I learnt this from an arithmetic book during my childhood days. If the first digit of a 2-digit number is \( a \) and the second digit is \( 5 \) in decimal representation, then its square can then be written as the result of \( a \times (a + 1) \) followed by \( 25 \) in decimal representation, i.e. the first few digits of the square is given by \( a \times (a + 1) \) and the last two digits are \( 25 \). Here are some examples:

- \( 25^2 = 625 \). (since \( 2 \times 3 = 6 \))
- \( 85^2 = 7225 \). (since \( 8 \times 9 = 72 \))

## Squaring a 2-Digit Number That Begins With 5

After learning the previous trick, I wondered if I could make more such tricks for myself. This is the first one I could come up with. If the first digit of a 2-digit number is \( 5 \) and the second digit is \( a \), then its square can be written as the result of \( 25 + a \) followed by \( a^2 \). In other words, the first two digits of the square are obtained from the result of \( 25 + a \) and the last two digits are obtained from the result of \( a^2 \). Here are some examples:

- \( 52^2 = 2704 \). (since \( 25 + 2 = 27 \) and \( 2^2 = 4 \))
- \( 57^2 = 3249 \). (since \( 25 + 7 = 32 \) and \( 7^2 = 49 \))

## Squaring Any Number That Ends with 5

Let us represent all digits except the last one as \( a \), e.g. if we are given the number \( 115 \). We say, \( a = 11 \). Then we can express the given number algebraically as \( 10a + 5 \). Note that the square of this number is \[ (10a + 5)^2 = 100a(a + 1) + 25. \] In decimal representation, this amounts to writing the result of \( a(a + 1) \) followed by \( 25 \). Here are some examples:

- \( 115^2 = 13225 \). (since \( 11 \times 12 = 132 \))
- \( 9995^2 = 99900025 \). (since \( 999 \times 1000 = 999000 \))

## Squaring Any Number That Begins With 5

Let us represent all digits except the first one as \( a \), e.g. if we are given the number \( 512 \), we say, \( a = 12 \). Then we can express the given number algebraically as \( 5 \times 10^n + a \) where \( n \) is the number of digits in \( a \). Note that \[ (5 \times 10^n + a)^2 = 25 \times 10^{2n} + 10^{n + 1} a + a^2. \] In decimal reprensetation, this amounts to performing the following steps:

1. Write \( 25 \) as the first two digits.
2. Then write \( a^2 \) as a \( 2n \)-digit number immediately after \( 25 \). Prefix \( a^2 \) with appropriate number of zeros so that \( a^2 \) is written with \( 2n \) digits.
3. Write the \( + \)-sign directly below the first digit, that is, write the \( + \)-sign directly before the first \( 2 \).
4. Write every digit of \( a \) including any preceding zeros immediately after the \( + \)-sign.
5. Finally add the numbers in both rows column by column performing the carrying operation whenever necessary.

Here are some examples: \[ 502^2 = \\ \left\{ \begin{array}{cccccc} 2 & 5 & 0 & 0 & 0 & 4 \\ + & 0 & 2 \\ \hline 2 & 5 & 2 & 0 & 0 & 4 \end{array} \right\} = 252004. \] \[ 512^2 = \\ \left\{ \begin{array}{cccccc} 2 & 5 & 0 & 1 & 1 & 4 \\ + & 1 & 2 \\ \hline 2 & 6 & 2 & 1 & 1 & 4 \end{array} \right\} = 262114. \] \[ 564^2 = \\ \left\{ \begin{array}{cccccc} 2 & 5 & 4 & 0 & 9 & 6 \\ + & 6 & 4 \\ \hline 3 & 1 & 8 & 0 & 9 & 6 \\ \end{array} \right\} = 318096. \]

## Applying Both Tricks Together

Let us now see an example where we use both the tricks together. Let us find \( 5195^2 \). This is a number that begins with the digit \( 5 \) as well as ends with the digit \( 5 \). We need to use the second trick to find \( 5195^2 \). But the second trick begins with writing \( 25 \) immediately followed by the result of \( 195^2 \). So we use the first trick to calculate \( 195^2 \).

To write the result of \( 195^2 \), we first write \( 380 \) which we obtain as the result of \( 19 \times 20 \) and then we write \( 25 \) immediately after it. Thus \( 195^2 = 38025 \). Now we perform the second trick as follows: \begin{align*} 5195^2 = \left\{ \begin{array}{cccccccc} 2 & 5 & 0 & 3 & 8 & 0 & 2 & 5 \\ + & 1 & 9 & 5 \\ \hline 2 & 6 & 9 & 8 & 8 & 0 & 2 & 5 \end{array} \right\} = 26988025. \end{align*}