# Thread: R(n)=4n+3 Complicated Word Problem

1. ## R(n)=4n+3 Complicated Word Problem

Hello, I'm having trouble with a question frmo the British mathematical Olympiad (1997)

N is a four-digit integer which does not end in a zero, and R(n) is the four digit integer obtained by reversing the digits of N. E.g. R(3275)=5723

Determine all such integers N for which R(N)=4N+3

I used the year (1997) as a reference and I found out it does work for the above statement so that's one but I don't know how many others there are.

I've worked out that according to the above statement, R(N) would have to start with either 1,3,5,7 or 9

I've tried laying it out so N=1000a +100b+10c+d and because R(N) must be odd due to +3 it would =4000d+400c+40b+4a and therefore a must be odd, narrowing down the first and last integer of both N and R(N) to be odd.

2. Originally Posted by Mukilab
I've tried laying it out so N=1000a +100b+10c+d and because R(N) must be odd due to +3 it would =4000d+400c+40b+4a ....
That'll simplify to d = (332a + 20b - 130c -1) / 1333

1997 : 7991 is ONLY integer solution.

3. Explanation?

I've gotten that sum... it doesn't prove anything...

4. Originally Posted by Mukilab
Explanation?
I've gotten that sum... it doesn't prove anything...
Not every "puzzle" has a proof...

5. Since d is an integer (between 0 and 9), 332a+ 20b- 130c- 1
must be one of

1: 0
2: 1333
3: 2(1333)= 2666
4: 3(1333)= 3999
5: 4(1333)= 5332
6: 5(1333)= 6665
7: 6(1333)= 7998
8: 7(1333)= 9331
9: 8(1333)= 10664
10: 9(1333)= 11997

That should help.