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4
10
2014
Problem of the Week
Problem C and Solution
Happy Palindrome Day!
Problem
The date, April 10, 2014, is palindromic when the date is written in the form
month-day-year, 4102014. The date, October 4, 2014, is also a palindrome
when the date is written in the form day-month-year, 4102014. A palindrome
is a word, phrase, sentence, or number that reads the same forwards and
backwards. In honour of the first Palindrome Day of 2014 we offer the following
problem: The number 14 741 is a five-digit palindromic number. Determine the
largest five-digit palindromic number which can be divided exactly by 15.
Solution
We are looking for a five-digit number of the form abcba.
For a number to be divisible by 15, it must be divisible by both 3 and 5.
To be divisible by 5, a number must end in 0 or 5. If the required number ends
in 0, it must also begin with 0 in order to be a palindrome. But the number
0bcb0 is not a five-digit number. Therefore, the number cannot end in a 0 and
hence must start and end with a 5. The required number looks like 5bcb5.
For a number to be divisible by 3, the sum of its digits must be divisible by 3.
Since we want the largest possible number, let b = 9 in 5bcb5. We must find
the largest value of c so that 59c95 is divisible by 3. The sum of the digits is
5 + 9 + c + 9 + 5 = c + 28 and c can take on any integer value from 0 to 9. It
follows that c + 28 can take on integer values from 28 to 37. The largest
number in this range divisible by 3 is 36 so c + 28 = 36 and c = 8.
The largest five-digit palindromic number exactly divisible by 15 is 59 895.
(As an aside, the smallest five-digit palindromic number exactly divisible by 15
is 50 205. And there are only 33 five-digit palindromic numbers exactly divisible
by 15. The verification of this is left as a possible extension for the solver.)