The sum of all the elements in the set { n ∈ { 1 , 2 , . . . . . . , 1 0 0 } | H . C . F . o f n a n d 2 0 4 0 i s 1 } is equal to………

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 + − − − − + 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) = 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 = 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 = 3799 Again 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

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………

3 Views|Posted 7 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Alok Kumar Singh | Contributor-Level 10

7 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.

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

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

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

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

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

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

S(34) = 34 + 68 = 102

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

S (51) = 51

S (85) = 85, S (30) = 3

Upvote

Similar Questions for you

If |z + 1| = αz + β (i + 1) and  $z = \frac{1}{2}$ – 2i, find α + β.

Alok Kumar Singh

$| \frac{1}{2} - 2 i + 1 | = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\sqrt[]{\frac{9}{4} + 4} = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\frac{5}{2} = α (\frac{1}{2}) + β + i (- 2 α + β)$

$\frac{α}{2} + β = \frac{5}{2}$ ...(1)

–2α + β = 0 …(2)

Solving (1) and (2)

$\frac{α}{2} + 2 α = \frac{5}{2}$

$\frac{5}{2} α = \frac{5}{2}$

a =

Alok Kumar Singh

Variance = $\frac{\sum x^{2}}{n} - {(\bar{x})}^{2}$

$\frac{6 0^{2} + 6 0^{2} + 4 4^{2} + 5 8^{2} + 6 8^{2} + α^{2} + β^{2} + 5 6^{2}}{8} = {(58)}^{2} = 66.2$

$\frac{7200 + 1936 + 3364 + 4624 + 3136 + α^{2} + β^{2}}{8} = 3364 = 66.2$

$2532.5 + \frac{α^{2} + β^{2}}{8} - 3364 = 66.2$

α² + β² = 897.7 × 8

= 7181.6

Alok Kumar Singh

Start with

(1) $\bar{E} : \frac{6!}{2!} = 360$

(2) $\bar{G E} : \frac{5!}{2!},$ $\bar{G N} : \frac{5!}{2!}$

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

Alok Kumar Singh

${(1 + x)}^{11} =^{11} C_{0} +^{11} C_{1} x +^{11} C_{2} x^{2} + . . . . . +^{11} C_{11} x^{11}$

= \frac{2^{12} - 2 - 24}{12}

= \frac{2^{12} - 26}{12} = \frac{4070}{12} = \frac{2035}{6} = \frac{m}{n}

m + n = 2035 + 6 = 2041

Let $f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$ 

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ 

The range of fog(x) is

Alok Kumar Singh

$f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ ->g(x) = |x|, x Î (–3, 1)

$f (g (x)) = {\begin{matrix} 2 + 2 | x |, & | x | \in (- 1, 0) \Rightarrow x \in ? \\ 1 - \frac{| x |}{3}, & | x | \in [0, 3) \Rightarrow x \in (- 3, 1) \end{matrix}$