1.5.1 Subtracting Reverses

A common first column problem from the early 2000s involves subtracting two numbers whose digits are reverses of each other (like $715 - 517$ or $6002 - 2006$). Let the first number $n_1 = abc = 100a + 10b + c$ so the second number with the digits reversed would be $n_2 = cba = 100c + 10b + a$ so:

$$n_1 - n_2 = (100a + 10b + c) - (100c + 10b + a)$$ $$= 100(a - c) + (c - a)$$ $$= 100(a - c) - (a - c)$$

So the gist of the trick is:

1. Take the difference between the most significant and the least significant digit and multiply it by 100 if it is a three-digit number, or if it is a four digit number multiply by 1000 (however, it only works for 4-digit numbers and above if the middle digits are 0’s; for example, $7002 - 2007$ the method works but $7012 - 2107$ it doesn’t work).
2. Then subtract from that result the difference between the digits.

Let’s look at an example:

812 - 218 =

• Step 1: $(8 - 2) \times 100 = 600$
• Step 2: $600 - 6 = 594$
• Answer: $594$

It also works for when the subtraction is a negative number, but you need to be careful:

105 - 501 =

• Step 1: $(1 - 5) \times 100 = -400$
• Step 2: $-400 - (1 - 5) = -396$
• Answer: $-396$

Like I said, you have to be careful with negative signs, a better (and highly recommended approach outlined in the next section) is to say: $105 - 501 = -(501 - 105) = -396$. By negating and reversing the numbers, you deal with positive numbers which are naturally more manageable. After you find the solution, you negate the result because of the sign switch.

Problem Set 1.5.1