# Quick multiplication

How can we explain this with equations ?

Quick Multiplication of any two numbers whose last two digits add up to 10 and all other numbers are the same

$32\times 38 = 3\times (3+1)|(2\times 8)\implies (3\times 4)|16\implies 12|16 \implies 1216$

$81\times 89 \implies 8\times(8+1)|(1\times9) \implies(8\times9)|09\implies72 | 09 \implies 7209$

$124 \times 126 \implies 12\times(12+1) | (4\times6) \implies(12\times13)|24\implies156|24 =15624$

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$(10 a + b)(10 a + d) = 100 a^2 + 10 a (b + d) + b d$

Since $b + d = 10$, that's the same as $100(a^2 + a) + bd = 100a(a+1) + b d$.

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Express the numbers as 10*x + y and 10*x + (10-y).

Here x is the part of the number other than last digit, y and 10-y are the last digits.

So in case of 32 * 38, x=3, y=2 in case of 124 * 126, x=12, y=4 etc.

Now multiply the expressions for the 2 numbers i.e. (10*x + y) * (10*x + (10-y))

You will get 100*x*x+100*x+a*(10-a) which after simplification becomes 100*x*(x+1)+a*(10-a)

The first part is a number ending in 2 zeros as it has a factor of 100. the last part is nothing but the multiplication of last digits of the original pair of numbers.

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So the two numbers are $10d + a$ and $10d + 10 - a$, where is the last digit, necessarily non-zero, and $d$ is the number formed by the rest of the digits. Hence $$(10d + a)(10d + 10 - a) = 100(d^2 + d) + a(10 -a)$$ The number $a(10 -a)$ is either one or two digits, so $100(d^2 + d)$ appears to the left of $a(10 -a)$ (including a zero in the case of $1 \cdot 9$.

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