The integers 1 through 9 are arranged

Question

The integers 1 through 9 are arranged as follows on a rectangular keypad:

1 2 3
4 5 6
7 8 9

Consider the 6-digit palindromes (numbers that have the same sequence of digits from right to left as left to right) formed by going back and forth across each row, column or diagonal; for example, going back and forth across the top row we form 123321, as well as 321123. One diagonal yields 159951 and 951159, and the second column gives us 258852 and 852258.

a) Explain why each of the numbers obtained as described above is divisible by 111. Your explanation should not involve going through each case; instead, your goal is to find a property that these numbers share, that ensures divisibility of 111.

b) Can you generalize your result above? Can you have keypads with different combinations of numbers that retain this property? What happens if you consider numbers in a different base?

Answer 1

For the first question note that your numbers are of the form $abccba=$ where $b=a+1$ and $c=a+2$ then we have it is equal to:

$a×100000+b×10000+c×1000+c×100+b×10+a×1=$

$a(100000+1)+b(10000+10)+c(1000+100)=$

$a (100001)+b (10010)+c (1100)=$

$a×100001+a×10010+10010+a×1100+2×1100=$

$a(100001+10010+1100)+(10010+2200)=$

$a (1001×111)+(110×111)= $

$111 (a×1001+110)$

and then $111$ divides the numbers.

I think you can try the second part analyzing how this was solved, hope it helps.

Edit: I meant to say the first case you want to solve and I wanted you to try the column and diagonal case based in this approach.

Answer 2

Hint: $10^{2k+1}+1$ (k>=0) is always divisible by 11.