Question

Let there be a set of different natural numbers, each of which is less than 11, where there are no numbers that are exactly twice as large as another number in the set. How many numbers are in the largest possible such set?

a. 5
b. 6
c. 7
d. 8

Answer 1

To construct the largest set of different natural numbers less than 11 without any number being double another, we sequentially include numbers while excluding their multiples and divisors by 2. The set {1, 2, 3, 5, 6, 9} meets the criteria, containing six numbers.

Step-by-step explanation:

The question asks for the largest set of natural numbers less than 11 where no number is twice another number in the set. We can build such a set by starting with the smallest natural number, which is 1, and then adding each subsequent number that is not double or half of any number already in the set.

• Starting with 1, the next natural number is 2, which can be included since it is not double another number yet.
• 3 can be included as well, but 4 cannot since it is double of 2.
• We can include 5 and 6, but not 7, because it's over the allowed range.
• 8, as double of 4, is not allowed, but 9 can be included.
• Finally, 10 can't be included because it is double of 5.

Therefore, the largest set is {1, 2, 3, 5, 6, 9}. There are six numbers in this set, making the answer (b) 6.