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.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

I re-phrased the question as follows:

Let n be an integer in base 10 such that n = abc_{7}and n = cba_{9}. What is n?

3. 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, 666_{7}= 342_{10}is our upper bound. But, 600_{9}= 486_{10}. So, c cannot be 6. By the same argument c cannot be 5.

From the problem stating that both abc_{7}and cba_{9}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 100_{7}, 200_{7}, 300_{7}, 400_{7}, 500_{7}, and 600_{7}. Then I checked 100_{9}, 200_{9}, 300_{9}, and 400_{9}. I was looking for base 10 values that appeared to converge. 500_{7}and 300_{9}were very close; 245 and 243 in base 10, respectively. Since the base 7 and base 9 numbers have to be inverted, I tried 503_{7}and 305_{9}and was successful: 248_{10}.

ii.

I kept the same restrictions in mind from approach (i).

1 < a < 6

0 < b < 6

1 < c < 4

Let n = abc_{7}and n = cba_{9}. Then,

n = a7^{2}+ b7^{1}+ c7^{0}= 49a + 7b + c and

n = c9^{2}+b9^{1}+ c9^{0}= 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?

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# Homework Help: MAA Problem 9/22/10

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