The math behind Lenny's cheap tie.

http://www.bbc.co.uk/bbcone/magicians/video/lennys-cheap-tie/

Dear readers,

Choose a 3 digit number whose all digits are different. Flip it around to get another 3 digit number. Subtract the smaller from the larger.

Let us assume that our number looks like $xyz$, which means it is actually $100x + 10y + z$. The flipped number would be $100z + 10y + z$. Assume that $x > z$, i.e. the first number is larger - you may choose otherwise. Their subtraction will give

$[100x + 10y + z] - [100z + 10y + x] = 100(x - z - 1) + 90 + (10 + z - x)$

Please note that the final form is like that because $x > z$ and a borrow/carry-over is performed. If we flip this number, we get $100(10 + z - x) + 90 + (x - z - 1)$. The sum of $100(x - z - 1) + 90 + (10 + z - x)$ and $100(10 + z - x) + 90 + (x - z - 1)$ is 1089.

The price of Lenny's tie was revealed to be GBP 10.89, a finite number no matter what your choice was. They did mention that the digits must be different. For this, it is strictly required that the hundreds and units must be different, i.e. $x \neq z$. A condition on the tens digit is not needed.

Regards.

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