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'Weekly Problem 5 - 2014' printed from http://nrich.maths.org/
In order to be a multiple of 15, a number must be a multiple both of 3 and of 5. So its units digit must be 0 or 5. However, since the number is a palindrome, the units digit must also equal the thousands digit and this cannot be 0, so the required number is of the form '5aa5'.
The largest such four-digit numbers are 5995, 5885, 5775. Their digit sums are 28, 26, 24 respectively. In order to be a multiple of 3, the digit sum of a number must also be a multiple of 3, so 5775 is the required number. The sum of its digits is 24.
This problem is taken from the UKMT Mathematical Challenges.View the previous week's solutionView the current weekly problem