Problem of the Week
Problem B and Solution
"Palindr"-homes
Problem
Jorge and Phyllis live on a street where houses are numbered between 10001 and 10997, with
10001 being the first house number, and 10997 the last. However, only every sixth number is
used, i.e., 10001 is used, but 10002, 10003, 10004, 10005, 10006 are not, then 10007 is used,
and so on.
a) How many homes are on this street?
b) How many of the house numbers on this street are palindromes (i.e., the digits occur in
the same order both forward and reversed, such as 909)?
Solution
a) Since 10 997 − 10 001 = 996, and only every sixth number is used, there are
966 ÷ 6 = 166 possible house numbers after 10001. Thus, including the first
house (numbered 10001), the number of homes on this street is
166 + 1 = 167 homes.
b) Palindromes are numbers which appear the same when the digits are
reversed. Thus, between 10001 and 10997, there are ten palindromes,
10001, 10101, 10201, 10301, 10401, 10501, 10601, 10701, 10801, and 10901.
(The next palindrome would be 11011, which is greater than 10997.)
However, the desired house numbers must also be such that the palindrome
minus 10001 is divisible by 6, since only every sixth number is used. The ten
differences are
0, 100, 200, 300, 400, 500, 600, 700, 800, and 900.
Of these differences, we see that only four of the numbers, namely 0, 300,
600, and 900 are divisible by 6. It follows that only the four house numbers
10001, 10301, 10601, and 10901 are palindromes.