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Q: Odd number of digits in a palindrome to be divisible by 11?

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Nope - To be divisible by nine, the sum of the digits must also be divisible by 9 (the digits total 21). Additionally, since it's an odd number - it cannot be divided by 6 !

odd digits such as 1,3,5,7, and 9.

No, 26871 is an odd number. No odd number is divisible by 2.

No odd numbers are divisible by even numbers.

If the sum of the digits in the odd positions starting from the ones digit less the sum of the digits in the even positions (starting with the 10s digit) is divisible by 11, then so is the original number. for 8003: odd position digits: 3 + 0 = 3 even position digits: 0 + 8 = 8 3 - 8 = -5 which is not divisible by 11, so 8003 is NOT divisible by 11.

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No. 101 is prime.

Every palindrome with an even number of digits is divisible by 11. The easiest way to see this is to recall the divisibility rule by 11: if a number X is written as ABCDEFG... (here A,B,C, ... are digits), then it's divisible by 11 if and only if the sum A-B+C-D+E-F+G-... is divisible by 11. In a palindrome with an even number of digits, each digit will appear in an odd position and in an even position, so when we calculate this sum, it will be added once and subtracted once, canceling. Since all the digits cancel, the sum A-B+C-D+... will be 0, which is divisible by 11. So the original number ABCD....DCBA was also divisible by 11.

4 and all of its multiples are divisible by 4. They end in even digits. No number that ends in an odd digit is divisible by 4.

ez.....9999

336

To find the smallest number divisible by 7 and 6 with two odd digits, we need to consider the least common multiple (LCM) of 7 and 6, which is 42. The smallest two-digit odd number is 11. Therefore, the smallest number that meets all these criteria is 11 x 42, which equals 462.

If the number formed by the last three digits is divisible by 8. This requires that: if the digit in the hundreds place is even, the last two digits must form a number divisible by 8 and if the digit in the hundreds place is odd, the last two digits must form a number divisible by 4 but not by 8.

There are 5760 such numbers.

Nope - To be divisible by nine, the sum of the digits must also be divisible by 9 (the digits total 21). Additionally, since it's an odd number - it cannot be divided by 6 !

The sum of the digits in odd position minus the sum of the digits in even position is divisible by 11.

odd digits such as 1,3,5,7, and 9.

If the sum of the digits in the odd positions starting from the ones digit less the sum of the digits in the even positions (starting with the 10s digit) is divisible by 11, then so is the original number. for 8003: odd position digits: 3 + 0 = 3 even position digits: 0 + 8 = 8 3 - 8 = -5 which is not divisible by 11, so 8003 is NOT divisible by 11.