Return to Part 1 | Continue to Part 3

**Math Journeys, Volume 1**

**A Rich, Little Problem, Part 2**

**“Uncle Bob”**

In Part 1 we began an extended look into the solution and other questions related to the alphanumeric

**Part 1. Recap and Results.** Stop here if you’d
like more time to solve it. We’ll wait for you.... As we left it, we
knew that R = 2 and U = 8. The full solution is

2178 x 4 = 8712.

I proposed two ways to extend the journey.

1. Find another 4-digit factor that is reversed by a single-digit
multiplier. I mentioned that there was an example for a multiplier of 9.

2. Search for a 3-digit example. We checked for and eliminated 5 and 9 as
possible multipliers.

Oh, I also dropped a “teaser” fact and question – does 1.5 x 4356 = 6534 have a connection with our investigation?

1. Solving 9 x ABCD = DCBA, we estimate first and see that ABCD must be less than 2000. Much less. So A is one in both places, and D has be be 9. Since 1200 x 9 is a 5-digit product, B must be 0 or 1, but 1 is A. The answer is found quickly to be

1089 x 9 = 9801.

2. There are no 3-digit examples involving single-digit multipliers. Keeping in mind estimation and parity, the odd/even property, the search is expedited but proves fruitless.

**Part 2. Making Connections.**

The result 1089 sent two new sparks through my brain, and they formed two new questions.

3. Is there a connection between 1089 and 2178? We’ll get to that one shortly.

4. The number 1089 is a result of a number trick familiar to
many grade schoolers. It involves, not multiplication, but subtraction and
addition. Take any 3-digit number and reverse its digits and subtract the
lesser from the greater. Reverse the digits in the difference and add those
two numbers. The result is **always** 1089. Try it.

752 – 257 = 495; 495 + 594 = 1089.

“Always” is an uncompromising modifier in mathematics. It’s a generalization that requires a proof. Since it’s not part of our main investigation, I won’t use the space for it. But there are strong hints that this is connected to reversing multipliers. Generalizations and connections are mathematicians’ manna! In making the proof I found another enticing property of 1089. It factors into 99 x 11.

Back to Question 3. Have you tied 1089 with our first solution 2178? I did perform a search for other four-digit solutions, and came up dry. I factored 2178 and got

2 x 11 x 99.

Yikes! It’s the second multiple of 1089. I made a list including the next seven multiples.

1089, 2178, 3267, 4356, 5445, 6534, 7623, 8712, and 9801

Well now. Clouds are parting a bit. We’ve seen six of
these multiples already (recall the “teaser” fact).

In closing, I leave you some work. Find the multiplier that turns 3267 into
7623. Determine a commonality among that multiplier and 4, 9, and 1.5, and
apply that to 5445.

In Part 3 we look for more solutions and find – an infinite number of them.

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