If positive two-digit integers x and y have the same digits, but in reverse order, which of the following must be a factor of \(x + y\)?
I call these kinds of problems Theorem Problems. What I mean by that is that there is some deep mathematical rule that forms the basis of the problem. In this case, the rule is not that deep, and I’ll discuss it in just a moment – after we solve the problem without worrying about the rule. Remember, we’re playing a matching game, not taking a math test.
STEP 1: Pick a two digit number.
Let’s go with 24 for illustrative purposes.
STEP 2: Reverse the two digits of your pick.
That gives us 42.
STEP 3: Add.
24 + 42 = 66.
STEP 4: Use your example number to eliminate answers. Process of elimination is always a good idea on must be problems.
Sixty-six is not divisible by 9, 10, or 14, so the only possible answers are A and D.
STEP 5: Rinse and repeat.
12 + 21 = 33. Thirty-three is not divisible by 6.
The correct answer is D.
THE RULE: There are actually two rules, both of which have shown up in multiple editions of the Official Guide, so it’s worth your while to take a look:
RULE 1: Let’s say the two digits of x are a and b. Thus, \(x = ab = 10a + b\) because a is in the tens place and b is in the units (or ones) place. Now we reverse the digits to get y: \(y = ba = 10b + a\). If we take the sum of x and y we get \(x + y = (10a + b) + (10b + a) = 11a + 11b\). Clearly, the sum is divisible by 11.
RULE 2: If we were to take the difference between x and y we would get \(x – y = (10a + b) – (10b + a) =9a – 9b\). Clearly, the difference is divisible by 9.
Remember, we solved the problem without using or deriving these rules, but having these rules in your toolbox may come in handy.