Math Journeys, Volume 1
A Rich Little Problem
Uncle Bob

 




 





We’ll stay mostly in the arithmetic sphere, and although blanket solutions, called generalizations, are what math people always look for, and they often require some algebra, our path will stay within the realm of numbers.


If you care to travel along and do part of the work, you may experience the solver’s exhilaration a few times. Stay curious, ask questions, keep searching, and you may experience the discoverer’s exhilaration, an even bigger high.

 

Let’s begin to solve our little alphanumeric in which each letter represents a digit in base ten. Let’s call the 4 a multiplier, RSTU the factor, and UTRS the product. Take in the big picture – it often saves work. The product has the same digits as the factor in reverse order. What values of R in the factor would produce a four-digit product? Estimate – if R in the factor is 3 or more, we would have a 5-digit product.

 

Think about parity, the odd-even property – is four times a number ever odd? Without using any pencil graphite, we’ve established that R is 2, and that means U in the product is 8 or possibly 9, but 4 x U is “blankety-2,” so we know that U is 8.

 

Solving for S and T involves a bit of trial and error. Go to it.

 

After you’ve solved it, think about related avenues to explore. There are many. Here is one of the questions that occurred to me right away.
 

1. Do multipliers other than 4 reverse a four-digit factor?
 

Here’s a leg up for you. The only other single-digit multiplier that reverses a 4-digit factor is 9. Can you find the factor and the reversed product?
 

Take time to finish solving the original problem and considering Question 1. Then scroll down.

You may have noticed this problem in "Alphanumerics" on the Math/Logic Puzzles menu, …

... but solving a math problem does not a journey make. I took a journey after solving, and I invite you to come along and even participate as I retrace my steps. This math journey will show just how rich a simple problem can be.

J1 RSTU.png
iLatt888fwPT.jpg

As we left it, we knew that R = 2 and U = 8.

 

Here is the full solution:

 

 

 

Then I proposed a question 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.


Solving ABCD x 9 = DCBA, we estimate first and see that ABCD must be less than 2000. Much less. So A is one (1) in both places, and 9 x D is “blankety-one,” so D has to be 9. Since 1200 x 9 is a 5-digit product, B must be 0 or 1, but 1 is A and is taken.

 

The answer is found quickly to be ...

J1 2178.png
J1 1089.png

Here is another path to explore:


2. Are there any 3-digit numbers that reverse when multiplied by a single digit?


There is some checking involved in looking for 3-digit examples, but not as much as you think.

 

Let’s use h, t, and u for the digits of the factor. Then our problems come in the form ...

where m is a single-digit multiplier 2, 3, 4, … up to 9.


 

J1 htu.png

Multipliers 5 and up are easy to deal with because if h were 2 or more, we would have a 4-digit product. So in all those cases h must equal one (1). It must be checked to see if the ones place in the product can be a one and if that leads to a reversal. For example, for m=9 and h=1 in both places, 9 x u must end in a one (1). We conclude that u must be 9, but are any reversals possible? Check it out.


Multipliers 2, 3, and 4 take a little more checking, but looking at h and u working in both factor and product should reduce the work. For m=3, h in the factor could be as large as 1, 2, or 3. If h in the units place is 1, the 3 times h is blankety-one, and that makes h = 7: impossible.


Have a fun ten minutes checking what I’ve left strewn about, and to think of some questions of your own.
 

I’ll drop a little “teaser” on you. Would the fact that 1.5 x 4356 = 6534 have any connection with our journey?
 

After you’ve spent some time on these, scroll down for more.

Question 2 asked you whether there are any 3-digit numbers that reverse when multiplied by a single digit.


There are no 3-digit examples involving single-digit multipliers. Keeping in mind estimation and parity, the odd/even property, we expedited the search but it proved fruitless. I must mention that any reversal like 257 x m = 752, does have a solution, but the multiplier must be a non-integer, in this case 752/257.


Making Connections: The result 1089, which happens to be 11 x 99, sent two new sparks through my brain, and they formed two new questions.


3. Have you tied 1089 with our first solution 2178?

 

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. In Part 1, I dropped a “teaser” fact and question – does 1.5 x 4356 = 6534 have a connection with our investigation? So we have seen now six of these multiples already: 1089, 2178, and 4356 have reversals in the list. I grant you that 1.5 is not a single-digit multiplier.


4. The number 1089 is the result in 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 a mathematician’s manna.


For now, I leave you some homework: Find the multiplier that turns 3267 into 7623 in simplest terms. Determine a commonality among that multiplier and 4, 9, and 1.5, and apply that idea to the partner-less 5445. Hint. Think about fraction form as in 9 = 9/1.


When you're ready for the final leg of this journey, scroll down. We'll look for more whole number solutions.

Crossl5staind.jpg
p21panel3fourths.jpg

 In 4 digits, we discovered that 1089 x 9 = 9801, and that 2178 and 8712 were also multiples of 1089. The full list of the first nine multiples reveals other reversed numbers:


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


By simply dividing reversals we can find the other multipliers, but they, unlike 4 and 9, are not whole. Is there a connection regardless?


6534 / 4356 = 1.5 and 7623 / 3267 = 2.33333....

We look for a pattern. The first multiple 1089 is reversed in the ninth; the second in the 8th. It helps here to know that 2.33333... is the fraction 7/3. Hmm.


One and 9 ... 2 and 8... 3 and 7.


The pattern would continue with 4 and 6 and finally 5 and 5. Ah, now we see that the multipliers are the ratios 9/1, 8/2, 7/3 and finally 6/4 gives 1.5. Our original problem and solution hid the fact that


2178 x (8/2) = 8712.


For homework you were asked to apply this relationship to 5445, the fifth multiple. Well, the terms of each ratio seem to add to ten; and 5 and 5 add to ten.


5445 x 5/5 = 5445


Two New Directions:. Yes, we can make much more of this. Every good solution leads to at least two more good questions. Factors of 1089 suggest that we look back to smaller numbers (now that we know the multipliers are not necessarily whole), and ahead to larger ones.


1089 = 99 x 11 = 9 x 11 x 11


Five-digit Solutions: By a combination of luck, experience, and multiplying the old fashioned way – by hand – I discovered that I could stick a 9 in the middle of the four-digit factors and not spoil the reversing process. 10989 x 9 and 21978 x 4 have reversing products.

 

 

 

Perhaps that maneuver would have been suggested if I thought of 99 elevens as just 11 short of 1100, or


1089 = 1000 + 100 - 10 - 1.


What combination would make 10989?


10000 + 1000 - 10 - 1


What factors? (10989 = 99 x 111) Is this a pattern? What about 99 x 1111 and 99 x 11,111? Check them out.


Heady with my discovery, I slipped additional 9s in the middle, and sure enough 109989, 1099989, 10999989, and so on, all have reversing families. It can be shown algebraically that 1) this will work for any number of nines in the middle, and 2) the needed multipliers are ratios of those same numbers that add to ten, 9/1 and 8/2 and so on. Try expanding 21978 in a similar way. You will have found a second infinite set of solutions.


Smaller Solutions. If 99 x 11 and 99 x 111 generate reversing families, we should backtrack and see why we didn’t find 3-digits solutions. That’s your final assignment in this journey. Throw these patterns into reverse and look for 3-digit reversers, and find the needed ratios. Are they generated by multiples of 99 x 1? Are any of the multipliers whole? Do the multiplier ratios follow the same pattern?


I’ll end here with a 2-digit family: multiples of 9. What is the reversed product and the multiplier for the factor 18? We’re not really at an end – just a stop while we think of more avenues to investigate or other journeys to share. I’m looking forward to them. Ta ta!

We are taking an extended look into the solution and other questions related to the alphanumeric at the right.

The Solution is 2178 x 4 = 8712. We wondered if there were other multiplications that reversed a factor’s digits to form the product. We stepped down and looked for 3-digit examples but found no “whole” multipliers that did so.
 

J1 RSTU.png