Math Journey Archive

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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.

Return to Part 1 | Continue to Part 3

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