Tag
From the set of {2, 3, 4, 5, 6, 7, 8, 9},
how many pairs of co-primes can
be formed?
Select one:
a. 19
b. 18
c. 20
d. 21
Answers
Answer:
We have to find possible pairs of coprimes from the set {2, 3, 4, 5, 6, 7, 8, 9}.
Remember, sets should be indicated with braces only and not with parentheses or square brackets!
Let A = {2, 3, 4, 5, 6, 7, 8, 9}.
What I'm going to do is given below:
→ First an element from A is considered.
→ Prime factorization of the element taken will be taken.
→ We know that prime factorization is to write a number in terms of its prime factors. After prime factorization, the elements which are multiples of prime factors seen in the prime factorization will be removed from A.
→ By removing such elements, we get another set which is a subset of A.
→ If this set obtained contains at least one element which was paired with the element considered earlier, then that element will be removed from the set.
→ Finally we make pairs with each element in the set obtained, with the considered element.
Consider the first element of the set A - 2.
The prime factorization of 2 is 2. So multiples of 2 in A should be removed from A to get a set which contain those, each are relatively prime to 2.
Multiples of 2 in A are 2, 4, 6 and 8. So,
A - {2, 4, 6, 8} = {3, 5, 7, 9}.
Thus 4 pairs can be obtained with 2 - (2, 3), (2, 5), (2, 7) and (2, 9).
Consider the next element of A - 3.
Prime factorization of 3 is 3, so multiples of 3 in A should be removed.
A - {3, 6, 9} = {2, 4, 5, 7, 8}
But since 3 is paired with 2 earlier, we deduct 2. Hence the set will be {4, 5, 7, 8}.
Thus 4 pairs can be made with 3 - (3, 4), (3, 5), (3, 7) and (3, 8).
Consider the next element - 4.
Prime factorization of 4 is 2². So multiples of 2 should be removed from A.
That set was found earlier, and it is none other than {3, 5, 7, 9}. But since 4 is paired with 3 earlier, we deduce 3. Hence the set becomes {5, 7, 9}.
Thus 3 pairs can be formed - (4, 5), (4, 7) and (4, 9).
Consider the next element - 5.
Prime factorization of 5 is 5. So multiples of 5 in A should be removed. There's only 5 as multiple of 5 in A.
A - {5} = {2, 3, 4, 6, 7, 8, 9}
But since 5 was paired with 2, 3 and 4 each earlier, we deduce these. Hence the set becomes {6, 7, 8, 9}.
Thus 4 pairs can be formed - (5, 6), (5, 7), (5, 8) and (5, 9).
Consider the next element - 6.
Prime factorization of 6 is 2 × 3. So multiples of 2 and 3 should be removed from A.
A - {2, 3, 4, 6, 8, 9} = {5, 7}
[Set containing multiples of 2 and 3 in A is removed from A here.]
But since 6 was paired with 5 earlier, we deduce 5. So the set becomes {7}.
Thus only one pair can be formed - (6, 7).
Consider the next element - 7.
Prime factorization of 7 is 7. So multiples of 7 in A should be removed, i.e., 7 only.
A - {7} = {2, 3, 4, 5, 6, 8, 9}
7 was paired with 2, 3, 4, 5 and 6 earlier each, so these are deducted. Hence the set becomes {8, 9}.
Thus 2 pairs can be formed - (7, 8) and (7, 9).
Consider the next element - 8.
Prime factorization is 2³. So multiples of 2 should be removed from A.
That set was found earlier - {3, 5, 7, 9}. But since 8 was paired with 3, 5 and 7 earlier each, these are deducted. Hence the set becomes {9}.
So one pair can be formed - (8, 9).
Consider the last element - 9.
Prime factorization is 3². So multiples of 3 should be removed.
This set was found earlier - {2, 4, 5, 7, 8}.
But 9 was paired with each of these elements earlier. So no new pairs are obtained.
Now we take the total no. of pairs obtained, which is the answer.
4 + 4 + 3 + 4 + 1 + 2 + 1 = 19.
Hence 19 pairs can be formed.
Let me include all these 19 pairs in a set, say X.
X = {(2, 3), (2, 5), (2, 7), (2, 9), (3, 4), (3, 5), (3, 7), (3, 8), (4, 5), (4, 7), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9), (6, 7), (7, 8), (7, 9), (8, 9)}
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀-ThesnowyPrince
Step-by-step explanation:
a.19
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