The sum of all the elements in the set <math> <mrow> <mrow> <mo>{</mo> <mrow> <mi>n</mi> <mo>∈</mo> <mrow> <mo>{</mo> <mrow> <mn>1</mn> <mo>,</mo> <mn>2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> </mrow> <mo>}</mo> </mrow> <mrow> <mo>|</mo> <mrow> <mi>H</mi> <mo>.</mo> <mi>C</mi> <mo>.</mo> <mi>F</mi> </mrow> </mrow> <mo>.</mo> <mtext> </mtext> <mtext> </mtext> <mi>o</mi> <mi>f</mi> <mtext> </mtext> <mtext> </mtext> <mi>n</mi> <mtext> </mtext> <mtext> </mtext> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <mn>2</mn> <mn>0</mn> <mn>4</mn> <mn>0</mn> <mtext> </mtext> <mtext> </mtext> <mi>i</mi> <mi>s</mi> <mtext> </mtext> <mtext> </mtext> <mn>1</mn> </mrow> <mo>}</mo> </mrow> </mrow> </math> is equal to………

2040 = 23 × 3 × 5 × 17If HCF between {n & 2040} = 1 ∴  n can not be multiple of 2, 3, 5, 17.Let S(n) denote sum of numbers divisible by n. S ( 2 ) = 2 + 4 + − − − − + 1 0 0 = 2 × 5 0 × 5 1 2 = 5 0 × 5 1  S ( 3 ) = 3 + 6 + − − − + 9 9 = 3 × 3 3 × 3 4 2 = 3 3 × 5 1  S ( 5 ) = 5 + 1 0 − − − + 1 0 0 = 5 × 2 0 × 2 1 2 = 5 0 × 2 1  S ( 1 7 ) = 1 7 + 3 4 + − − − + 8 5 = 1 7 × 5 × 6 2 = 5 × 5 1  S ( 6 ) = 6 + 1 2 + − − − = 6 × 1 6 × 1 7 2 = 1 6 × 1 5  S ( 1 0 ) = 1 0 + 2 0 + − − − = 1 0 × 1 0 × 1 1 2 = 5 0 × 1 1 S(34) = 34 + 68 = 102 S ( 1 5 ) = 1 5 + 3 0 + − − − + 9 0 = 1 5 × 6 × 7 2 = 4 5 × 7 S (51) = 51S (85) = 85, S (30) = 30 + 60 + 90 = 180 S (all other combinations) = 0 ∴ Sum of all numbers which are either divisible by 2, 3, 5, or 17 is = 1 5 × 5 1 + 3 3 × 5 1 + 5 0 × 2 1 + 5 × 5 1 -[16 × 51 + 50 × 11 + 102 + 45 × 7 × 51 + 85] + 180 = 3799Again sum of all number of { 1 , 2 , − − − − 1 0 0 } = 1 0 0 × 1 0 1 2 = 5 0 5 0  ∴ Sum of required number = 5050 – 3799 = 1251

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Maths Class 11th Sets

The sum of all the elements in the set ${n \in {1, 2, . . . . . ., 100} | H . C . F . o f n a n d 2040 i s 1}$ is equal to………

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alok kumar singh | Contributor-Level 10

2 months ago

2040 = 2³ × 3 × 5 × 17

If HCF between {n & 2040} = 1

$∴$ n can not be multiple of 2, 3, 5, 17.

Let S(n) denote sum of numbers divisible by n.

$S (2) = 2 + 4 + - - - - + 100 = 2 \times \frac{50 \times 51}{2} = 50 \times 51$

$S (3) = 3 + 6 + - - - + 99 = 3 \times \frac{33 \times 34}{2} = 33 \times 51$

$S (5) = 5 + 10 - - - + 100 = 5 \times \frac{20 \times 21}{2} = 50 \times 21$

$S (17) = 17 + 34 + - - - + 85 = 17 \times \frac{5 \times 6}{2} = 5 \times 51$

$S (6) = 6 + 12 + - - - = 6 \times \frac{16 \times 17}{2} = 16 \times 15$

$S (10) = 10 + 20 + - - - = 10 \times \frac{10 \times 11}{2} = 50 \times 11$

S(34) = 34 + 68 = 102

$S (15) = 15 + 30 + - - - + 90 = 15 \times \frac{6 \times 7}{2} = 45 \times 7$

S (51) = 51

S (85) = 85, S (30) = 30 + 60 + 90 = 180 S (all other combinations) = 0

$∴$ Sum of all numbers which are either divisible by 2, 3, 5, or 17 is

$= 15 \times 51 + 33 \times 51 + 50 \times 21 + 5 \times 51$

...more

2040 = 2³ × 3 × 5 × 17

If HCF between {n & 2040} = 1

$∴$ n can not be multiple of 2, 3, 5, 17.

Let S(n) denote sum of numbers divisible by n.

$S (2) = 2 + 4 + - - - - + 100 = 2 \times \frac{50 \times 51}{2} = 50 \times 51$

$S (3) = 3 + 6 + - - - + 99 = 3 \times \frac{33 \times 34}{2} = 33 \times 51$

$S (5) = 5 + 10 - - - + 100 = 5 \times \frac{20 \times 21}{2} = 50 \times 21$

$S (17) = 17 + 34 + - - - + 85 = 17 \times \frac{5 \times 6}{2} = 5 \times 51$

$S (6) = 6 + 12 + - - - = 6 \times \frac{16 \times 17}{2} = 16 \times 15$

$S (10) = 10 + 20 + - - - = 10 \times \frac{10 \times 11}{2} = 50 \times 11$

S(34) = 34 + 68 = 102

$S (15) = 15 + 30 + - - - + 90 = 15 \times \frac{6 \times 7}{2} = 45 \times 7$

S (51) = 51

S (85) = 85, S (30) = 30 + 60 + 90 = 180 S (all other combinations) = 0

$∴$ Sum of all numbers which are either divisible by 2, 3, 5, or 17 is

$= 15 \times 51 + 33 \times 51 + 50 \times 21 + 5 \times 51$

-[16 × 51 + 50 × 11 + 102 + 45 × 7 × 51 + 85] + 180 = 3799

Again sum of all number of ${1, 2, - - - - 100} = \frac{100 \times 101}{2} = 5050$

$∴$ Sum of required number = 5050 – 3799 = 1251

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2040 = 23 × 3 × 5 × 17If HCF between {n & 2040} = 1 <math> <mrow> <mo>∴</mo> </mrow> </math>   n can not be multiple of 2, 3, 5, 17.Let S(n) denote sum of numbers divisible by n. <math> <mrow> <mi>S</mi> <mrow> <mo>(</mo> <mrow> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>2</mn> <mo>+</mo> <mn>4</mn> <mo>+</mo> <mo>−</mo> <mo>−</mo> <mo>−</mo> <mo>−</mo> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mo>=</mo> <mn>2</mn> <mo>×</mo> <mfrac> <mrow> <mn>5</mn> <mn>0</mn> <mo>×</mo> <mn>5</mn> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>5</mn> <mn>0</mn> <mo>×</mo> <mn>5</mn> <mn>1</mn> </mrow> </math>               <math> <mrow> <mi>S</mi> <mrow> <mo>(</mo> <mrow> <mn>3</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>3</mn> <mo>+</mo> <mn>6</mn> <mo>+</mo> <mo>−</mo> <mo>−</mo> <mo>−</mo> <mo>+</mo> <mn>9</mn> <mn>9</mn> <mo>=</mo> <mn>3</mn> <mo>×</mo> <mfrac> <mrow> <mn>3</mn> <mn>3</mn> <mo>×</mo> <mn>3</mn> <mn>4</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <mn>3</mn> <mo>×</mo> <mn>5</mn> <mn>1</mn> </mrow> </math>              <math> <mrow> <mi>S</mi> <mrow> <mo>(</mo> <mrow> <mn>5</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>5</mn> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mo>−</mo> <mo>−</mo> <mo>−</mo> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mo>=</mo> <mn>5</mn> <mo>×</mo> <mfrac> <mrow> <mn>2</mn> <mn>0</mn> <mo>×</mo> <mn>2</mn> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>5</mn> <mn>0</mn> <mo>×</mo> <mn>2</mn> <mn>1</mn> </mrow> </math>             <math> <mrow> <mi>S</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mn>7</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>1</mn> <mn>7</mn> <mo>+</mo> <mn>3</mn> <mn>4</mn> <mo>+</mo> <mo>−</mo> <mo>−</mo> <mo>−</mo> <mo>+</mo> <mn>8</mn> <mn>5</mn> <mo>=</mo> <mn>1</mn> <mn>7</mn> <mo>×</mo> <mfrac> <mrow> <mn>5</mn> <mo>×</mo> <mn>6</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>5</mn> <mo>×</mo> <mn>5</mn> <mn>1</mn> </mrow> </math>                <math> <mrow> <mi>S</mi> <mrow> <mo>(</mo> <mrow> <mn>6</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>6</mn> <mo>+</mo> <mn>1</mn> <mn>2</mn> <mo>+</mo> <mo>−</mo> <mo>−</mo> <mo>−</mo> <mo>=</mo> <mn>6</mn> <mo>×</mo> <mfrac> <mrow> <mn>1</mn> <mn>6</mn> <mo>×</mo> <mn>1</mn> <mn>7</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mn>6</mn> <mo>×</mo> <mn>1</mn> <mn>5</mn> </mrow> </math>               <math> <mrow> <mi>S</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>1</mn> <mn>0</mn> <mo>+</mo> <mn>2</mn> <mn>0</mn> <mo>+</mo> <mo>−</mo> <mo>−</mo> <mo>−</mo> <mo>=</mo> <mn>1</mn> <mn>0</mn> <mo>×</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mo>×</mo> <mn>1</mn> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>5</mn> <mn>0</mn> <mo>×</mo> <mn>1</mn> <mn>1</mn> </mrow> </math> S(34) = 34 + 68 = 102 <math> <mrow> <mi>S</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mn>5</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>1</mn> <mn>5</mn> <mo>+</mo> <mn>3</mn> <mn>0</mn> <mo>+</mo> <mo>−</mo> <mo>−</mo> <mo>−</mo> <mo>+</mo> <mn>9</mn> <mn>0</mn> <mo>=</mo> <mn>1</mn> <mn>5</mn> <mo>×</mo> <mfrac> <mrow> <mn>6</mn> <mo>×</mo> <mn>7</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>4</mn> <mn>5</mn> <mo>×</mo> <mn>7</mn> </mrow> </math>        S (51) = 51S (85) = 85, S (30) = 30 + 60 + 90 = 180 S (all other combinations) = 0 <math> <mrow> <mo>∴</mo> </mrow> </math> Sum of all numbers which are either divisible by 2, 3, 5, or 17 is <math> <mrow> <mo>=</mo> <mn>1</mn> <mn>5</mn> <mo>×</mo> <mn>5</mn> <mn>1</mn> <mo>+</mo> <mn>3</mn> <mn>3</mn> <mo>×</mo> <mn>5</mn> <mn>1</mn> <mo>+</mo> <mn>5</mn> <mn>0</mn> <mo>×</mo> <mn>2</mn> <mn>1</mn> <mo>+</mo> <mn>5</mn> <mo>×</mo> <mn>5</mn> <mn>1</mn> </mrow> </math>         -[16 × 51 + 50 × 11 + 102 + 45 × 7 × 51 + 85] + 180 = 3799Again sum of all number of <math> <mrow> <mrow> <mo>{</mo> <mrow> <mn>1</mn> <mo>,</mo> <mn>2</mn> <mo>,</mo> <mo>−</mo> <mo>−</mo> <mo>−</mo> <mo>−</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> </mrow> <mo>}</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>5</mn> <mn>0</mn> <mn>5</mn> <mn>0</mn> </mrow> </math>   <math> <mrow> <mo>∴</mo> </mrow> </math> Sum of required number = 5050 – 3799 = 1251

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