Do the division, if there is no remainder, it is divisible. Seriously, many of the "divisibility rules" that have been discovered become more complicated than doing the actual division. For practical purposes, just learn the divisibility rules for a few simple cases (divisibility rules by 2, 4, 8, 5, 10, 3, 9, 7, 11, and 13), and for all other cases, just do the division.
It is divisible by any of its factors which are: 1, 3, 31 and 93
26
1 3 9 19 57 95 171 285 855
Those for 1, 2, 4, 5 and 8.
It is divisible by any of its factors which are: 1, 2, 3, 6, 7, 14, 21 and 42
The divisibility rules for a prime number is if it is ONLY divisible by 1, and itself.
The rule for 1 and the rule for 3.
The divisibility rules will show that 53 is not divisible by anything other than 1 and itself. Since it is already prime, it doesn't have a factorization.
It is divisible by any of its factors which are: 1, 3, 31 and 93
26
You only have to test the numbers 1 through 5. If you know the rules of divisibility, you know that 3, 4 and 5 aren't factors.
1 3 9 19 57 95 171 285 855
Those for 1, 2, 4, 5 and 8.
Factors are divisors. If you know the divisibility rules, you know that 80 is divisible by 1, 2, 4, 5 and 8. If you divide 80 by those numbers, you find the other half of the factor pairs.
It is divisible by any of its factors which are: 1, 2, 3, 6, 7, 14, 21 and 42
21
I am not sure what you are asking but 141 is divisible by the number 1, itself (141) and certainly by the number 3 so it is not a prime number.