Question

The sum of all the elements of the set {α ∈ {1, 2, …, 100} : HCF(α, 24) = 1} is

Answer 1

Correct Answer: 1633

To find the sum of all elements α in the set {1, 2, …, 100} such that the highest common factor (HCF) of α and 24 is 1, we are essentially identifying numbers that are coprime with 24. To achieve this, identify numbers in the range that are not divisible by any prime factors of 24.

The prime factorization of 24 is 2³×3. Numbers coprime with 24 are not divisible by 2 or 3. We can use the principle of inclusion-exclusion to solve this problem.

1. Count numbers divisible by 2: ⌊100/2⌋=50.
2. Count numbers divisible by 3: ⌊100/3⌋=33.
3. Count numbers divisible by 6 (2×3): ⌊100/6⌋=16.

Using inclusion-exclusion, find numbers divisible by 2 or 3:

N(2∪3)=N(2)+N(3)−N(6)=50+33−16=67.

The count of numbers coprime with 24 is:

100−67=33.

Now to sum all α coprime with 24, use the principle of generating numbers not divisible by 2 or 3 and calculate their sum. The arithmetic sum of numbers 1 to 100 is n(n+1)/2=100×101/2=5050. Calculate sums of numbers divisible by 2 and 3, and subtract them.

1. Sum of multiples of 2: 2+4+...+100=2(1+2+...+50)=2×50×51/2=2550.
2. Sum of multiples of 3: 3+6+...+99=3(1+2+...+33)=3×33×34/2=1683.
3. Sum of multiples of 6: 6+12+...+96=6(1+2+...+16)=6×16×17/2=816.

Applying inclusion-exclusion, the sum of numbers divisible by 2 or 3 is:

S(2∪3)=S(2)+S(3)−S(6)=2550+1683−816=3417.

The sum of numbers coprime with 24 is:

5050−3417=1633.

The computed sum is 1633, which matches the expected range, thus confirming the answer.