Define R(x) as the reverse of an integer x. For example, R(435) = 534.
Find the value of N, where:
 N is a 4digit positive integer which is divisible by each of 11 and 13, and:
 R(N) is also divisible by each of 11 and 13, where N < R(N), and:
 None of N and R(N) can contain any leading zero.
Prove that there are no others.
Clearly if a 4 digit number is divisible by 11, its reversal will be.
So we really only need to look for multiples of 11*13=143 whose reversals are divisible by 13.
So when is a 4 digit number divisible by 13.
One divisibility test for abcd (concatenation)
is 4a+3b1c4d must be divisible by 13.
But then for its reversal dcba to be divisible by 13
4d+3c1b4a must be divisible by 13
The sum of these is 2b+2c which would also be divisible by 13 so
b+c must be divisible by 13
This made it very easy to look for a solution by hand
1001 and its multiples are solutions disallowed by N<R(n)
between 1001 and 2002 the multiples of 143 (along with b+c) are
1144 (5)
1287 (10)
1430 (7)
1573 (12)
1716 (8)
1859 (13)
We see here the 1859 solution. Its reversal is 1573+8*1001.
You can also see a leading zero 'solution' 1573+7*1001=8580

Posted by Jer
on 20140404 13:10:20 