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Sia ? 6 years, 6 months ago
Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.
By Euclid’s division lemma, we have
a = bq + r; 0 ≤ r < b
For a = n and b = 3, we have
n = 3q + r ...(i)
Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.
Putting r = 0 in (i), we get
{tex}n = 3q{/tex}
∴ n is divisible by 3.
{tex}n + 1 = 3q + 1{/tex}
∴ n + 1 is not divisible by 3.
{tex}n + 2 = 3q + 2{/tex}
∴ n + 2 is not divisible by 3.
Putting r = 1 in (i), we get
{tex}n = 3q + 1{/tex}
∴ n is not divisible by 3.
{tex}n + 1 = 3q + 2{/tex}
∴ n + 1 is not divisible by 3.
{tex}n + 2 = 3q + 3 = 3(q + 1){/tex}
∴ n + 2 is divisible by 3.
Putting r = 2 in (i), we get
{tex}n = 3q + 2{/tex}
∴ n is not divisible by 3.
{tex}n + 1 = 3q + 3 = 3(q + 1){/tex}
∴ n + 1 is divisible by 3.
{tex}n + 2 = 3q + 4{/tex}
∴ n + 2 is not divisible by 3.
Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3.
Posted by Lakshay Sharma 6 years, 6 months ago
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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.
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Sia ? 6 years, 6 months ago
The given system of linear equation is
(a – b)x + (a + b)y = a2 – 2ab – b2 ....(1)
(a + b)(x + y) = a2 + b2 ....(2)
Equation (2) can be written as
(a + b)x + (a + b)y = a2 + b2 ....(3)
Subtracting equation (3) from equation (1), we get
– 2bx = – 2ab – 2b2
{tex}\Rightarrow{/tex} – 2bx = – 2b(a + b)
{tex}\Rightarrow{/tex} x = a + b
Substituting this value of x in equation (1), we get
(a – b)(a + b)(a + b)y = a2 – 2ab – b2
{tex}\Rightarrow{/tex} a2 – b2 + (a + b)y = a2 – 2ab – b2
{tex}\Rightarrow{/tex} (a + b)y = – 2ab
{tex}\Rightarrow \;y = - \frac{{2ab}}{{a + b}}{/tex}
Verification: Substituting x = a + b, {tex}y = - \frac{{2ab}}{{a + b}}{/tex},
We find that both the equations (1) and (2) are satisfied as shown below:
(a – b)x + (a + b)y = (a – b)(a + b) + (a + b) {tex}\left\{ { - \frac{{2ab}}{{a + b}}} \right\}{/tex}
= a2 – 2ab – b2
(a + b)(x + y) = (a + b){tex}\left\{ {(a + b) + (- \frac{{2ab}}{{a + b}})} \right\}{/tex}
= (a + b)2 – 2ab
= a2 + b2
hence, the solution we have got is correct.
Posted by Atharva Gosavi 7 years, 6 months ago
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Sia ? 6 years, 6 months ago
P(x, y), A(6,2), B(- 2, 6)
PA = PB or, PA2 = PB2
(x-6)2 + (y -2)2 = (x + 2)2 + (y - 6)2
or, x2 - 12x + 36 + y2- 4y + 4 = x2+ 4x + 4 + y2- 12y + 36
or, -12x - 4y = 4 x - 12y
or,12y -4y = 4x + 12x
or,8y = 16x
or,y = 2x
Hence Proved.
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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)
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