Let the thumb be \(1,\) the index finger \(2,\) the middle finger \(3,\) etc. Finger \(x,\) on the left hand, touches finger \(y\) on the right hand. That is, if \(x = 2\) and \(y = 3,\) then the index finger on the left hand is touching the middle finger on the right hand. There are \(x - 1\) fingers to the right of finger \(x\) on the left hand, and \(5 - x\) fingers to the left. Similarly, there are \(y - 1\) fingers to the left of finger \(y\) on the right hand, and \(5 - y\) fingers to the right. The number of ones in the final product, is thus

$$(5 - x)(5 - y) = xy - 5x - 5y + 25$$The number of tens in the final product, is equal to the fingers touching, plus the number of fingers below the touching fingers, which is

$$2 + (x - 1) + (y - 1) = x + y$$So the number of ones it contributes to the final product, is

$$10(x + y) = 10x + 10y$$If we want to multiply \(6\) by \(9,\) we would put fingers \(x = 1\) and \(y = 4\) together. In general, if we want to multiply \(a\) and \(b,\) we're putting together \(x = a - 5\) and \(y = b - 5.\) Thus \(a = x + 5\) and \(b = y + 5,\) are the two numbers we're multiplying. From all of this, we can see the equation we want to prove is

$$(x + 5)(y + 5) = (10x + 10y) + (xy - 5x - 5y + 25)$$Expanding the left-hand side gives us

$$xy + 5x + 5y + 25$$ But simplifying the right-hand side, by combining like terms, gives us the same thing. Thus, our equation is true, and the trick has been proven to always work.Conclude by giving your students these challenges: