Since there are no palindromes, the question cannot be answered.
Since there are no palindromes, the question cannot be answered.
Since there are no palindromes, the question cannot be answered.
Since there are no palindromes, the question cannot be answered.
2/6, 3/6, 4/6, 5/6
2/8, 3/8, 4/8, 5/8
Multiply it by successive counting numbers.
Multiply only the numerator by whole numbers. The resulting fraction may then be simplified.So, for example, the first few multiples of the unit fraction 1/6 are:1/6, 2/6, 3/6, 4/6, 5/6, 6/6 and 7/6. Some of these can be simplified to give:1/6, 1/3, 1/2, 2/3, 5/6, 1 and 7/6.
No multiples of 2 are factors of 3.
-4
8/25.
32% * * * * * * * * 8/25
3/5,6/5,9/5
There are 10 3-digit odd palindromes that are divisible by five.
For there to be palindromes, each digit must be replicated. Therefore there are at most three distinct digits.If there are 3 pairs of different digits, then there are 6 palindromes. If there can be more duplicate digits, then there are 27 palindromes.
90
90 of them.
The answer depends on what the fraction is to be multiplied by!
Nine. The sum of the digits must be a multiple of 9; because of the repeated digits, this is only possible if the first two digits add up to 9.
The smallest digit palindrome that is the sum of two 3-digit palindromes is 121. This is achieved by adding the two 3-digit palindromes 101 and 20, both of which are palindromic. Therefore, 101 + 101 = 202, but if we consider a valid case with two different palindromes, we can use 111 and 110, which gives us 221, the next smallest palindrome. However, the smallest individual palindrome formed by the sum of any two 3-digit palindromes remains 121.
2/8, 3/8, 4/8, 5/8