14
221
11 and 18
11 and 16.
11 is a tricky one, but this is how:if the sum of every other digit minus the sum of all the other digits equals a multiple of 11example:Q: Is 164725 divisible by 11?A: Yes. and this is why... first you take every other digit 164725, then add them 6+7+5=18. Now take the rest of the numbers 164725 and add them, 1+4+2=7. After that subtract them, 18-7=11, and 11 is a multiple of 11 (11x1=11).
21
There are 199 palindromic numbers between 0 and 1000. These include single-digit numbers (0-9), which total 10, and two-digit numbers (11, 22, ..., 99), which add up to 9. Additionally, there are 90 three-digit palindromic numbers, ranging from 101 to 999, that follow the format aba (where a and b are digits). Thus, the total is 10 (single-digit) + 9 (two-digit) + 90 (three-digit) = 109 palindromic numbers.
You can select 12 numbers for the first digit, 11 numbers for the second digit, and 10 numbers for the third digit; so 12*11*10 = 1320 sets of 3 numbers can be made out of 12 different numbers.
221
14
99990 Note that if you add 11 to my answer you get a 6-digit number therefore it must be the largest 5-digit whole multiple of 11.
11
11
Eleven digit numbers are those in the range between 9,999,999,999 and 100,000,000,000 exclusive. There are 90,000,000,000 of these.
There are more 12-digit palindromic numbers than 11-digit palindromic numbers. This is because the number of possible 12-digit palindromic numbers is greater than the number of possible 11-digit palindromic numbers. In general, the number of palindromic numbers of length n is 9 * 10^((n-1)/2), so for 11-digit palindromic numbers, there are 9 * 10^5 = 900,000 possibilities, while for 12-digit palindromic numbers, there are 9 * 10^6 = 9,000,000 possibilities.
The two 2-digit numbers that multiply to make 176 are 11 and 16. When you multiply 11 by 16, you get 176 (11 × 16 = 176). These are the only two-digit integers that satisfy this condition.
11 and 18
11 and 13.