# Puzzle of the Week for 4 January 1999: Solution

1. 1999, 1234, 765, 469, 296, 173, 123, ...
2. 1999, 11990, 21901, 32813, 64636, 128282, 411103, 712217,...
3. 1999, 2, 3, 2, 2003, 2, 5, 2, 3, 2, 7, ...

Now, a few words of explanation:

Each term after the first two in sequence a is given by the difference of the previous terms: 1999 - 1234 = 765, 1234 - 765 = 469, etc. The term required by the puzzle is given by 296 - 173, or 123. The successive difference method (see the hints for this puzzle) reveals immediately how the sequence was constructed.

To construct sequence b, reverse the digits of the first term (to get 9991) and add them to get the next term (1999 + 9991 = 11990). Repeat this procedure to get each succeeding term. The term required by the puzzle (712217) is a palindromic number (reversing its digits does not change it). Interestingly, this procedure will yield a palindromic number starting with almost any first term. The smallest number for which the procedure apparently does not yield a palindromic number is 196; for this reason, the procedure is known as the 196-algorithm. There is (apparently) an infinite number of such numbers, although they are rare; you can determine as many such numbers as you wish by applying the 196-algorithm to 196 itself, for example, yielding 196 + 691 = 887, 887 + 788 = 1675, etc.

Sequence b yields its secret readily to the successive difference method. By comparing the original sequence:

1999, 11990, 21901, 32813, 64636, 128282, 411103, ...
with the sequence of its first differences:
9991, 9911, 10912, 31823, 63646, 282821, ...
we can see that the sequence of first differences contains the members of the original sequence with their digits reversed.

Sequence c consists entirely of primes (integers evenly divisible only by themselves and 1). Although the successive difference method doesn't help in this case, the appearance of 2003 in the fourth term after 1999 suggests that each term in sequence c:

1999, 2, 3, 2, 2003, 2, 5, 2, 3, 2, ...
is related to one in the simpler sequence:
1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, ...
In fact, each term in sequence c is the smallest prime factor of the corresponding term in this simpler sequence. The term in the simpler sequence that corresponds to the one required by the puzzle is 2009, which has prime factors of 7, 7, and 41; so the required term is 7. This year, 1999, is the last of three prime-numbered years in this decade; the last decade with three prime-numbered years was the 1870s, and the next will be the 2080s (with four prime-numbered years, the maximum possible).