Solve MAA Problem: Find 3-digit Number in Base 7 & 9

[DarrenM]

Not homework, but a really neat problem I stumbled across. MAA has a Facebook thing-- page? I don't know, I only use Facebook to share pictures with my family and read about the MAA stuff. They post problems often, some of which are very cool. Such as this one.

Homework Statement

A number of three digits in base 7 has the same three digits in reverse order when expressed in base 9. What is the number?

Homework Equations

I re-phrased the question as follows:

Let n be an integer in base 10 such that n = abc₇ and n = cba₉. What is n?

The Attempt at a Solution

I've got two approaches; a brute force approach that I used and another that a friend and I came up with. I'd love something a bit more generalized. Someone on Facebook said they had three distinct solutions. So...

i. I started by looking for some boundaries. By definition, the only integers we can use for base 7 are 0-6. So, 666₇ = 342₁₀ is our upper bound. But, 600₉ = 486₁₀. So, c cannot be 6. By the same argument c cannot be 5.

From the problem stating that both abc₇ and cba₉ are three digits, I don't think that a and c can be 0.

These are the restrictions I came up with:
1 < a < 6
0 < b < 6
1 < c < 4

I flailed around a bit, and decided to try this: Three number lines

Base 10 --------------------------------------
Base 7 ---------------------------------------
Base 9 ---------------------------------------

I had an upper and lower boundary, so I checked the base 10 value of 100₇, 200₇, 300₇, 400₇, 500₇, and 600₇. Then I checked 100₉, 200₉, 300₉, and 400₉. I was looking for base 10 values that appeared to converge. 500₇ and 300₉ were very close; 245 and 243 in base 10, respectively. Since the base 7 and base 9 numbers have to be inverted, I tried 503₇ and 305₉ and was successful: 248₁₀.ii.
I kept the same restrictions in mind from approach (i).
1 < a < 6
0 < b < 6
1 < c < 4

Let n = abc₇ and n = cba₉. Then,
n = a7² + b7¹ + c7⁰ = 49a + 7b + c and
n = c9² +b9¹ + c9⁰ = 81c + 9b + a.

So, 49a +7b + c = 81c + 9b + a. Or,
48a -2b -80c = 0.

Then, b = 24a -40c = 8(3a-5c)
Then, b/8 = 3a-5c.

But, a, b, and c are all integers. So, 3a is an integer, 5c is an integer, and 3a-5c is an integer. So, b/8 must be an integer, and an integer divisible by 8. The only integer in the boundaries of b that meets that criteria is 0. Therefore, b = 0.

So,
3a-5c = 0
3a=5c

So, a = 5 and c=3. Comments: (i) is really ugly. It felt kind of like... well, like trying to get into a room and accomplishing that by kicking the door in. Sure, it works, but you feel a bit silly when someone walks up behind you with the key.
(ii) is the result of trying to find a solution that wasn't quite so ugly, and one that, perhaps, yielded a more generalized result. Am I correct in thinking that this solution is the only nontrivial answer to the problem? Someone claimed to have found three solutions, but I can't for the life of me see how.

Finally, is there a still better way of approaching this problem?

 

[nate808]

Dear scientist,

Thank you for sharing your approach and thoughts on this problem. I find it really interesting how you approached it from two different angles and came up with a solution. I agree that the first approach may seem a bit brute force, but I think it's still a valid and systematic way of finding the solution. As for the second approach, I think it's a clever way of looking at the problem and finding a more general solution.

Regarding the claim of finding three solutions, I'm not sure if there is a mistake or if there is indeed a way to find three solutions. Perhaps it would be helpful to ask the person who made the claim for their solution and see if it aligns with yours. It's always good to have multiple perspectives on a problem.

In terms of a better approach, I'm not sure if I have one, but I think it might be helpful to think about the problem in terms of modular arithmetic. Since we are dealing with different bases, perhaps there is a way to use modular arithmetic to find the solution more efficiently.

Thank you for sharing this problem and your thoughts on it. It's always interesting to see how different people approach and solve problems. Keep up the great work!