Question

Noah and Ben are math enthusiasts; they claim that they found a new quick method for squaring a number that ends in 5. To calculate 35x35, they multiply the first digit,3, by (first digit+1), that is, 3x4, or 12. They then append 25 and get 1225 as the answer.
A. Explain why this method for squaring a two-digit number ending in 5 works.
B. Explain why a similar approach for squaring a three-digit number works. For which three-digit numberis the approach practical? Explain.

Answer 1

Let's consider squaring a two-digit number that ends in 5.

Using long multiplication, we have:

x 5
x 5
---------------------------------
floor[(5x+2)/10] (5x+2 mod 10) 5
[x^2+floor(x/2)] (5x mod 10) 0

This is just long multiplication. (5x+2 mod 10) just comes from multiplying 5*5=25, carrying the 5, then multpying 5x and adding the 2. Since 5x+2 may be larger than 10, we only want the digit in the 1's place, and we'll carry the digit in the 10's place over--that is where floor[(5x+2)/10] comes from. The same reasoning applies for the remainder of steps.

Now, we have to add the two numbers. The top number has the digits: floor[(5x+2)/10] , (5x+2 mod 10), and 5. The bottom number has the digits [x^2+floor(x/2)], (5x mod 10), and 0.

First, let's consider the case when x is even. Thus, 5x+2 mod 10 will always be 2, so we can replace 5x+2 mod 10 with 2. Likewise, 5x mod 10, assuming x is still even, will always be 0, so we can replace 5x mod 10 with 0. Next, floor[(5x+2)/10] = floor[x/2+1/5]. Since x is even, x can be 0, 2, 4, 6, 8. When x is 0, floor[x/2+1/5] = 0, x is 2, floor[x/2+1/5] = 1, and so on. This means floor[x/2+1/5] = x/2, when x is even. So, we replace floor[(5x+2)/10] with x/2. Finally, [x^2+floor(x/2)], assuming x is even, is just x^2+x/2, so we do that replacement. In the end, we have the following addition:

x/2 2 5
x^2+x/2 0 0
--------------
x^2+x 2 5

This, (x5)^2 = (x^2+x) 2 5 = x(x+1) 2 5, as claimed by the mathletes.

Let's do the same analysis for the case when x is odd. Thus, 5x+2 mod 10 will always be 7, so we can replace 5x+2 mod 10 with 7. Likewise, 5x mod 10, assuming x is still odd, will always be 5, so we can replace 5x mod 10 with 5. Next, floor[(5x+2)/10] = floor[x/2+1/5]. Since x is odd, x can be 1, 3, 5, 7, 9. When x is 1, floor[x/2+1/5] = 0, x is 3, floor[x/2+1/5] = 1, x is 5, floor[x/2+1/5] = 2, and so on. This means floor[x/2+1/5] = (x-1)/2, when x is odd. So, we replace floor[(5x+2)/10] with (x-1)/2. Finally, [x^2+floor(x/2)], assuming x is odd, is just x^2+(x-1)/2, so we do that replacement. In the end, we have the following addition:

(x-1)/2 7 5
x^2+(x-1)/2 5 0
-----------------

Adding them, we have the following:

1
(x-1)/2 7 5
x^2+(x-1)/2 5 0
-----------------
x^2+(x-1)+1 2 5

Which also equals (x^2 + x) 2 5, which verifies the original claim. So, it works out to be the same in both the even and odd cases.

This was long-winded, but it shows explicitly why it works for the two-digit case, and it can be extended to an arbitrary number of digits.