14. Subsets
a. A subset of the set of integers from 1 to 100 has the property that the absolute value of
the difference between an element and twice another element is always greater than 1.
What is the maximum number of elements in the subset?
If I pick 1, I can’t pick 2 or 3.
If I pick 2, I can’t pick 3, 4, or 5 (or 1)
3: 5, 6, or 7 (or 1 or 2)
4: 7, 8, or 9 (or 2)
5: 9, 10, or 11 (or 2 or 3)
6: 11, 12, or 13 (or 3)
Starting at the bottom, I can choose 1, 4-6, 14-26, 54-100: 1 + 3 + 13 + 47 = 64.
Starting at the top, I can choose 100-51, 24-13, 5-4, 1: 50 + 12 + 2 + 1 = 65!
b. A subset of the set of integers from 1 to 100 has the property that any pair of numbers in
the subset has a greatest common factor that is a prime number. What is the maximum
number of elements in the subset?
If you have any prime, you’ll only get a few of its multiples. Best case is 2, which lets us have
some of the evens, but not all. For example, once we’ve got 4, we can’t have any other multiple
of 4. Same for 6, 8, 10, 12, 14, I guess for all. Anything of the form 2*p, you can’t have
anything of the form 2*p*q. Anything of the form 2*p*q, you can’t have anything of the form
2*p*r or 2*q*r. So if we have 2, we could also have 4, 6, 10, 14, 22, 26, 34, 38, 46, 58, 62, 74,
82, 86, 94, for a total of 16. Not including a prime number resulted in a set with many fewer
elements.
c. A subset of the set of integers from 1 to 100 has the property that the product of any pair
of numbers in the subset is not a factor of 1080. What is the maximum number of elements
in the subset?
1080 = 23 × 33 × 5. We need to make sure that every pair of numbers either includes a factor
that isn’t a 2, 3, or 5, or that every pair of numbers includes more than three 2’s, more than three
3’s, or more than one 5 (or some combination of these).
Every prime greater than 5 should be in, as should their multiples. 7 gives 14 elements, 11 gives
8 _additional_ elements, 13 gives 6 _more_, 17 & 19 give 5 each, 23 gives 4, 29 & 31 give 3
each, 37-47 give 2 each for a subtotal of 8, and 53-97 give 1 each for a subtotal of 10 and a total
of 66 so far.
Multiples of 24 = 16 that haven’t already been counted are 16, 32, 48, 64, 80, and 96, for 6
additional elements. 34 = 81 is another element, as are 52 = 25, 50, 75, and 100, for another 4
elements. This makes the current count 77.
Now we can add some factors of 1080. If they are multiplied by one of the numbers we’ve
already selected, there will be a prime factor other than 2, 3, & 5, which is good. If they multiply
one another, they need to generate too many 2’s, 3’s, or 5’s. If I only add multiples of 4, they’ll
combine to give multiples of 16, which is fine. I could use the same reasoning for multiples of 9
or 5, but there are more unused 4’s, I think. That means I can have my 66 prime-based elements,
16 _new_ multiples of 4, 81, and the four 25’s, but I double-counted 100, for a total of 86.
d. A subset of the set of integers from 1 to 1000 has the property that the positive
difference between any pair in the subset is not between 4 and 7 inclusive. What is the
maximum number of elements in the subset?
If 1 is chosen, 5-8 can’t be. If 2 is chosen, 6-9 can’t be. This pattern continues indefinitely, so
4
we can choose 1-4, 12-15, 23-26, …, which is 11 of the 1000 numbers, or 360 of the first 990
numbers, then the next four, for an answer of 364.
e. A subset of the set of integers from 1 to 10000 has the property that the positive
difference between any pair in the subset is not a multiple of seven or eleven. What is the
maximum number of elements in the subset?
I can only pick seven numbers that aren’t equivalent in mod 7, and it’s easy to have them also
have no equivalence in mod 11, so the answer is 7.
f. A subset of the set of integers from 1 to 10000 has the property that the product of any
pair of numbers in the subset is not a multiple of 12. What is the maximum number of
elements in the subset?
So long as we don’t include any multiple of 3, we can’t possibly have a product that is a multiple
of 12, so we can include at least 6667 elements. If we remove the multiples of 2 that are not
multiples of 4, we can add the odd multiples of 3. There are 2500 of the former and only 1667 of
the latter, so it’s not worth it, and the answer is 6667.