Redundancy in divisibility problems

[B_Naber]

How do I know when prime factors are redundant?

Example: To find out whether x is divisible by 120, it's given that x is divisible by 12 and by 30.

We need three 2's, one 3 and one 5 for x to be divisible by 120, since x is divisible by 12, it's divisible by two 2's and a 3 and since x is divisible by 30, it's given that it has a 2, 5 and 3 in it's prime factorization. However, one of the 2's could be redundant.

Now I've found another example, which doesn't seem to use this concept:

Questions: "Is the integer x divisible by 36?"

1) x is divisible by 12
2) x is divisible by 9

So x needs two 2's and three 3's. Statement 1 gives it two 2's and a three and statement 2 gives it three 3's. Which is according to the answer sufficient for x to be divisible by 36. Why isn't one of the 3's redundant in this problem?

Is there any general rule when prime factors are redundant?

[tim]

One of the threes IS redundant in your second example. You’ve accounted for four threes, only three of which you need. The best way to think of this is that if you are told two different things about the same number (which we have in both of your examples) there could well be overlap in the information you are given..

[krishnan.anju1987]

I believe that for the first question, 120 has factors 2,2,2,3,5
12 has factors 2,2,3
30 has factors 3,2,5

confirmed factors- 2,2,3,5. one 3 and 2 could be redundant

[jnelson0612]

I believe that for the first question, 120 has factors 2,2,2,3,5
12 has factors 2,2,3
30 has factors 3,2,5

confirmed factors- 2,2,3,5. one 3 and 2 could be redundant

Agreed.