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Sets and Relations question

2022 · 24 Jun · Shift 2 · Q40
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Sets and Relations question

2022 · 24 Jun · Shift 2 · Q40

JEE MainMathematicsSets and RelationsNumerical+4 / −1
The sum of all the elements of the set {α∈{1,2,.....,100}:HCF(α,24)=1}\{ \alpha \in \{ 1,2,.....,100\} :HCF(\alpha ,24) = 1\}{α∈{1,2,.....,100}:HCF(α,24)=1} is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 1633

  1. We need the sum of all integers α∈{1,2,…,100}\alpha \in \{1,2,\dots,100\}α∈{1,2,…,100} such that gcd⁡(α,24)=1.\gcd(\alpha,24)=1.gcd(α,24)=1.

  2. First factorize 242424: 24=23⋅3.24=2^3\cdot 3.24=23⋅3. So, for gcd⁡(α,24)=1\gcd(\alpha,24)=1gcd(α,24)=1, the number α\alphaα must be divisible by neither 222 nor 333.

  3. Thus we need numbers from 111 to 100100100 which are not divisible by 222 and not divisible by 333.

  4. Let us find their sum using inclusion-exclusion.

  • Sum of all numbers from 111 to 100100100: S=100⋅1012=5050.S=\frac{100\cdot 101}{2}=5050.S=2100⋅101​=5050.

  • Sum of multiples of 222 up to 100100100: 2(1+2+⋯+50)=2⋅50⋅512=2550.2(1+2+\cdots+50)=2\cdot \frac{50\cdot 51}{2}=2550.2(1+2+⋯+50)=2⋅250⋅51​=2550.

  • Sum of multiples of 333 up to 100100100: There are ⌊100/3⌋=33\lfloor 100/3\rfloor=33⌊100/3⌋=33 such multiples, so 3(1+2+⋯+33)=3⋅33⋅342=1683.3(1+2+\cdots+33)=3\cdot \frac{33\cdot 34}{2}=1683.3(1+2+⋯+33)=3⋅233⋅34​=1683.

  • Multiples of both 222 and 333, i.e. multiples of 666: There are ⌊100/6⌋=16\lfloor 100/6\rfloor=16⌊100/6⌋=16 such multiples, so 6(1+2+⋯+16)=6⋅16⋅172=816.6(1+2+\cdots+16)=6\cdot \frac{16\cdot 17}{2}=816.6(1+2+⋯+16)=6⋅216⋅17​=816.

  1. By inclusion-exclusion, sum of numbers divisible by 222 or 333 is 2550+1683−816=3417.2550+1683-816=3417.2550+1683−816=3417.

  2. Therefore, the required sum is 5050−3417=1633.5050-3417=1633.5050−3417=1633.

  3. Hence, 1633\boxed{1633}1633​

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