3-Digit Numbers & Their Reverses Divisible By 7

[JSITD]

Pairs of Numbers Divisible By Seven (Unusual Question)

First of all, I don't go to school in the states so I don't know what the difference between "pre-algebra," and "intermediate/advanced algebra" is. If I've posted this in the wrong place, then I apologise as I'm honestly not sure where it should go. It's a bit of an odd question.

The question: n is a three digit number and m is the number obtained by reversing the order of the digits in n. Suppose that m does not equal n, and that n+m and n-m are both divisible by seven. Find all possible pairs of n and m.

My work so far:
n = 100a + 10b + c
m = 100c + 10b + a
n+m = 101a + 20b + 101c
n-m = 99c - 99a

There are 128 three digit multiples of 7, including 64 even and 64 odd. For each pair of m and n, either both numbers are even or both are odd. Therefore, for each value of n the first and last digit are both either even or odd. 546 & 553 are the 64th & 65th three digits multiples of 7 (those in the middle.) n and m​ are always both multiples of seven.
I'd greatly appreciate any help that could be offered.

 

[Dr.Peterson]

I'd use modular arithmetic to solve this. That doesn't really fit under any category except perhaps "advanced" (number theory, in particular). For example, 101 is congruent to 3 (mod 7), so you can work with smaller numbers.

The important question is, what math do you know that might apply? Did the question come from some sort of algebra class, or some other source? And do you know any modular arithmetic?

Also, I assume you meant this:

My work so far:
n = 100a + 10b + c
m = 100c + 10b + a
n+m = 101a + 20b + 101c
n-m = 99c - 99a

 

[JSITD]

I know some basic modular arithmetic, but not a lot. The question's actually from an external source, so it's not aimed specifically at my class. It is, however, aimed at my level, but it could include topics that I personally haven't covered yet. Where I go to school (in Scotland), maths is just one subject until your final year of school, when it splits into Pure Maths, Mechanics, and Statistics. And it's simply called 'Maths' from the ages of 4 to 16. (Yeah, Scotland's weird.) I'll try looking at some modular arithmetic to solve it, though. Is there a particular branch or technique that I should research?

 

[Dr.Peterson]

I know some basic modular arithmetic, but not a lot. The question's actually from an external source, so it's not aimed specifically at my class. It is, however, aimed at my level, but it could include topics that I personally haven't covered yet. Where I go to school (in Scotland), maths is just one subject until your final year of school, when it splits into Pure Maths, Mechanics, and Statistics. And it's simply called 'Maths' from the ages of 4 to 16. (Yeah, Scotland's weird.) I'll try looking at some modular arithmetic to solve it, though. Is there a particular branch or technique that I should research?

There may be something I'm not aware of, but I just used the basic concept: do all the work "mod 7", so that the fact that 101a + 20b + 101c is divisible by 7 can be restated as

\(\displaystyle 101a + 20b + 101c \equiv 0 \left(mod\; 7\right)\)
​

and then as

\(\displaystyle 3a + 6b + 3c \equiv 0 \left(mod\; 7\right)\)
​

A little more work with the equations, and I could just list possible values for a, and for each of those list possibilities for b and c. That list amounts to the answer. I get something like 11 or 12 distinct pairs of numbers (which I haven't actually written out).

You will also benefit from some basic number theory facts, such as that if two numbers are divisible by 7, then so are their sum and difference, as you evidently know. But your initial statements about digits being odd or even doesn't agree with my work.

Having communicated with many "foreigners" over the years, my sense is that American school math is weird! And as for categories, pre-algebra would not use variables, and algebra does. There's not much more to it than that. The fact that our categories don't provide a place for elementary number theory is one of the problems in our system!

 

[JSITD]

A little more work with the equations, and I could just list possible values for a, and for each of those list possibilities for b and c.

What do you mean by this? I've been trying to rearrange them so that I can solve simultaneously, but I can't manage to do so.

 

[Dr.Peterson]

What do you mean by this? I've been trying to rearrange them so that I can solve simultaneously, but I can't manage to do so.

Can you show me what you did with the equations, so I can see if you are going in a helpful direction?

 

[JSITD]

This is what I've got so far (Sorry about the poor image quality

 

[Dr.Peterson]

It looks like you're struggling to solve the system of equations, because you are forgetting the goal of eliminating one variable at a time. Once you've eliminated a variable between two equations, you should be eliminating the SAME variable between another pair.

Or, you can use substitution. Note that one equation you are starting with is 2c - 2a == 0. This tells you that c == a! (I'm using "==" for congruence.) So replace c in the other equation with a, and you'll have one equation in a and b.

Solve that for, say, b; then for each of the ten possibilities for a, you can write what b has to be (and of course c as well). Keep in mind, though, that this is congruence, not equality; so if a is 2, x could be either 2 or 2 + 7 = 9.

I hope these hints will be enough. Let me know.

 

[JSITD]

I can get to the point where a == (-b). Following this, I find that n could be only 861 & 952. (Bearing in mind that, per the terms of the question, palindromes are not allowed.)
Thank you for all your help so far, by the way. It's been really useful.

 

[Dr.Peterson]

I can get to the point where a == (-b). Following this, I find that n could be only 861 & 952. (Bearing in mind that, per the terms of the question, palindromes are not allowed.)
Thank you for all your help so far, by the way. It's been really useful.

Looks like you've got it (with additional help from my own site, I see).

I had missed the fact that m and n must be different, and happened to have only checked the two cases in my list that aren't palindromes!