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Sia ? 6 years, 6 months ago
Any positive integer is of the form 2q or 2q + 1, for some integer q.
{tex}\therefore{/tex} When n = 2q
{tex}\style{font-family:Arial}{n^2\;-\;n\;=\;n(n\;-\;1)\;=\;2q(2q\;-\;1)=\;2m,}{/tex}
where m = q(2q - 1) ( m is any integer)
This is divisible by 2
When n = 2q + 1
{tex}\style{font-family:Arial}{\begin{array}{l}n^2\;-\;n\;=\;n(n\;-\;1)\;=\;(2q\;+\;1)(2q+1-1)\\=2q(2q+1)\end{array}}{/tex}
= 2m, when m = q(2q + 1) ( m is any integer)
which is divisible by 2.
Hence, n2 - n is divisible by 2 for every positive integer n.
Posted by Gaurav Gupta 7 years, 6 months ago
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Sia ? 6 years, 6 months ago
the decimal expansion of given rational number is:
{tex}\frac { 15 } { 1600 } = \frac { 15 } { 2 ^ { 4 } \times 10 ^ { 2 } } = \frac { 15 \times 5 ^ { 4 } } { 2 ^ { 4 } \times 5 ^ { 4 } \times 10 ^ { 2 } } = \frac { 9375 } { 10 ^ { 6 } } = .009375{/tex}
Posted by Durga Kumari 7 years, 6 months ago
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Sia ? 6 years, 6 months ago
First, find the HCF of 210 and 55 by Euclid's Division Algorithm
210 = 55 {tex}\times{/tex} 3 + 45
55 = 45 {tex}\times{/tex} 1 + 10
45 = 10 {tex}\times{/tex} 4 + 5
10 = 5 {tex}\times{/tex} 2 + 0 (zero remainder)
therefore, HCF (210 , 55) = 5
Now,
{tex}\therefore {/tex} {tex}5 = 210 \times 5 + 55y{/tex}
{tex} \Rightarrow {/tex} {tex}5 - 1050 = 55y{/tex}
{tex} \Rightarrow {/tex} {tex} - 1045 = 55y{/tex}
{tex} \Rightarrow {/tex} {tex}y = - 19{/tex}
Posted by Hariom Dhakar 7 years, 6 months ago
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Sia ? 6 years, 6 months ago
n3 - n = n (n2 - 1) = n (n - 1) (n + 1)
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.
⇒ n (n – 1) (n + 1) is divisible by 3.
Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n is divisible by 2.
If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
⇒ n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.
∴ n (n-1) (n+1) = n3 - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)
Posted by Amandeep Saini 7 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let, {tex}α{/tex} = a - d, {tex}β{/tex} = a and {tex}\gamma {/tex} = a + d be the zeroes of the polynomial.
f(x) = 2x3 - 15x2 + 37x - 30
{tex}\alpha + \beta + \gamma = - \left( {\frac{{ - 15}}{2}} \right) = \frac{{15}}{2}{/tex} ....... (i)
{tex}\alpha \beta \gamma = - \left( {\frac{{ - 30}}{2}} \right) = 15{/tex} ......... (ii)
From (i)
a - d + a + a + d = {tex}\frac{{15}}{2}{/tex}
So, 3a = {tex}\frac{{15}}{2}{/tex}
a = {tex}\frac{{5}}{2}{/tex}
and From (ii)
a(a - d)(a + d) = 15
So, a(a2 - d2) = 15
{tex}⇒ \frac{{5}}{2}{/tex} {tex}\left[\left(\frac52\right)^2\;-d^2\right]{/tex} = 15
{tex}⇒ \frac{25}4\;-d^2{/tex}= 6
{tex}⇒ \;d^2\;=\;\frac{25}4-6\;{/tex}
{tex}⇒ {d^2} = \frac{1}{4}{/tex}
{tex}⇒ d = \frac{1}{2}{/tex}
Therefore, {tex}\alpha = \frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2{/tex}
{tex}\beta = \frac{5}{2}{/tex}
{tex}\gamma = \frac{5}{2} + \frac{1}{2} = 3{/tex}.
Posted by Girish Wadhwani 7 years, 6 months ago
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Sonali Aggarwal 7 years, 6 months ago
3x + y = -1 *2
-2x + 3y = 19 *3
Using elimination method,
6x + 2y = -2
-6x + 9y = 57
11y=55
y=5
and x= -2
Now since the point of their intersection lies on y= Mx + 3
Substituting values of x and y
5= -2M + 3
M= -1
Posted by Vinit Bajaj 7 years, 6 months ago
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Sonali Aggarwal 7 years, 6 months ago
LCM of 2 numbers is also a multiple of their HCF but in this case 380 is not a multiple of 18 so it is not possible to have 18 HCF and 380 as LCM of 2 numbers.
Posted by Singh Kalpana 7 years, 6 months ago
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Radhika Sharma 7 years, 6 months ago
1Thank You