3, 6, 9
Trick question all numbers are divisible by four ,but only 175 result in whole numbered answers
1221, 2112, 3003, 10401, 11211, 12021
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.
-7
3, 6, 9
Trick question all numbers are divisible by four ,but only 175 result in whole numbered answers
There are 151 3-digit numbers that are divisible by 6.
1221, 2112, 3003, 10401, 11211, 12021
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.
-7
There are three such numbers: 12, 24 and 36.
All 4 digit numbers that are divisible by 9 are also divisible by 3. The first is 1008 and the last is 9999.
They all are.
Not all of them are.
All you need to do is multiply the three numbers together:5 * 6 = 3030 * 9 = 270
None. Extremely simple proof: To be divisible by 5 the units digit must be 5. But then it would not be divisible by 2.