The are: 111, 141, 171, 222, 252, 282, 303, 333, 363, 393, 414, 444, 474, 525, 555, 585, 606, 636, 666, 696, 717, 747, 777, 828, 858, 888, 909, 939, 969 and 999.
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3, 6, 9
1221, 2112, 3003, 10401, 11211, 12021
Trick question all numbers are divisible by four ,but only 175 result in whole numbered answers
All palindromes between 10000 and 99999 are of the form abcba where a is a digit in {1, 2, ..., 9} and b and c are digits in {0, 1, ..., 9} All multiples of 25 end in 00, 25, 50 or 75. As a cannot be 0, this means the only multiple of 25 we are interested in are those which end in 25 and 75. This means we are looking for integers of the form 52c25 and 57c75. There are 10 possible values for the digit c in each case (one of {0, 1, ..., 9}), which means: There are 20 palindromes between 10,000 and 99,999 which are divisible by 25.
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