The sum of all the elements of the set $\{\alpha \in\{1,2, \ldots, 100\}: \operatorname{HCF}(\alpha, 24)=1\}$ is ______.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Permutations and Combinations

The sum of all the elements of the set

{α \in {1, 2, \dots, 100} : HCF (α, 24) = 1}

is ______.

Answer: 1633

We need

α

coprime to

24 = 2^{3} \cdot 3

, i.e.,

α

not divisible by 2 or 3. Use inclusion-exclusion: Sum =

S (1

100) - S

(multiples of 2)

- S

(multiples of 3)

+ S

(multiples of 6).

S (1

100) = \frac{100 \cdot 101}{2} = 5050

S

(multiples of 2)

= 2 (1 + 2 + \dots + 50) = 2 \cdot \frac{50 \cdot 51}{2} = 2550

S

(multiples of 3)

= 3 (1 + 2 + \dots + 33) = 3 \cdot \frac{33 \cdot 34}{2} = 1683

S

(multiples of 6)

= 6 (1 + 2 + \dots + 16) = 6 \cdot \frac{16 \cdot 17}{2} = 816

. Sum

= 5050 - 2550 - 1683 + 816 = 1633

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Integer answer