**If the tens digit is $a$ and the unit digit is $5$, the square can be found by computing $a(a+1)$ and appending $25$ to the result.**

For instance $75^2$ is computed by multiplying $7$ and $8$ ($= 56$)and adding $25$ at the end to get $5625$. This also work with $3$-digit numbers (and longer numbers). For instance, to compute $115^2$, we compute $11\times 12 = 132$ and get $13225$.

To show why this is true is a simple exercise. A number with tens digit $a$ and unit digit $5$ is equal to $10a+5$. Squaring this number results in $(10a+5)^2 = 100a^2 + 100a + 25 = 100a(a+1) + 25$ which corresponds to the rule above.

This same argument is true when multiplying two 2-digit numbers where the tens digits are equal and the unit digits add up to 10. In this case we are multiplying $10a+b$ and $10a+c$ where $b+c =10$. The product $(10a+b)(10a+c)$ is equal to $100a^2+100a+bc = 100a(a+1)+bc$. So the corresponding rule is:

**Multiply $a$ with $a+1$ and append $bc$ at the end.**

The only caveat is that when $b=1$ and $c = 9$, you append $09$ (rather than just $9$) at the end. For instance to compute $83\times 87$, we get $8\times 9 = 72$ and append $3\times 7$ to get $7221$. Similarly for $124\times 126$, we compute $12\times 13 = 156$ and append $24$ to get $15624$.

You can use a similar trick to multiply two 2-digit number where the tens digits add up to 10 and the unit digits are the same, i.e. multiplying $10b+a$ with $10c+a$ where $b+c = 10$. The product is

$$(10b+a)(10c+a) = 100bc + 100a + a^2 = 100(bc+a) + a^2$$

Thus the corresponding rule is:

**Multiply $b$ with $c$, add $a$ and then append $a^2$ to the result.**

**Again, if $a^2$ has only one digit, prefix it with a $0$ before appending it to $bc+a$. For example, to compute $43\times 63$, we compute $4\times 6 + 3 = 27$ and append $3^2 = 09$ to obtain $2709$.**

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