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Anant Yadav 7 years, 7 months ago
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Sia ? 6 years, 6 months ago
Let the number be (3q + r)
{tex}n = 3 q + r \quad 0 \leq r < 3{/tex}
{tex}\text { or } 3 q , 3 q + 1,3 q + 2{/tex}
{tex}\text { If } n = 3 q \text { then, numbers are } 3 q , ( 3 q + 1 ) , ( 3 q + 2 ){/tex}
{tex}3 q \text { is divisible by } 3{/tex}.
{tex}\text { If } n = 3 q + 1 \text { then, numbers are } ( 3 q + 1 ) , ( 3 q + 3 ) , ( 3 q + 4 ){/tex}
{tex}( 3 q + 3 ) \text { is divisible by } 3{/tex}.
{tex}\text { If } n = 3 q + 2 \text { then, numbers are } ( 3 q + 2 ) , ( 3 q + 4 ) , ( 3 q + 6 ){/tex}
{tex}( 3 q + 6 ) \text { is divisible by } 3{/tex}.
{tex}\therefore \text { out of } n , ( n + 2 ) \text { and } ( n + 4 ) \text { only one is divisible by } 3{/tex}.
Posted by Gurkirat Singh 7 years, 7 months ago
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Posted by Gurkirat Singh 7 years, 7 months ago
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Posted by Abhishek Rai 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
According to the question, we have to show that every positive integer is either even or odd.
Let us assume that there exists a smallest positive integer that is neither odd nor even, say n. Since n is the least positive integer which is neither even nor odd, n - 1 must be either odd or even.
Case 1: If n - 1 is even, n - 1 = 2k for some k.
But this implies n = 2k + 1
This implies n is odd.
Case 2: If n - 1 is odd, n - 1 = 2k + 1 for some k.
But this implies n = 2k + 2 = 2(k + 1)
This implies n is even.
Therefore,In both cases , we arrive at a contradiction.
Thus, every positive integer is either even or odd
Posted by Ajay Kumar 7 years, 7 months ago
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Posted by Shalvi Goyal 7 years, 7 months ago
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Kunal Rajour 7 years, 7 months ago
Sonu Goyal 7 years, 7 months ago
Posted by Sandhu Sandhu 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.
Since, Speed = {tex}\frac{{{\text{Distance travelled}}}}{{{\text{Time taken to travel that distance}}}}{/tex} {tex} \Rightarrow x = \frac{d}{t} \Rightarrow{/tex} d = xt ....(1)
According to the question,
x + 10 = {tex}\frac{d}{{t - 2}} \Rightarrow{/tex} (x + 10)(t - 2) = d
{tex}\Rightarrow{/tex} xt + 10t - 2x - 20 = d
{tex}\Rightarrow{/tex} -2x + 10t = 20 .....(2) [Using eq. (1)]
Again, x - 10 = {tex}\frac{d}{{t + 3}} \Rightarrow{/tex} (x - 10)(t + 3) = d
{tex}\Rightarrow{/tex} xt - 10t + 3x - 30 = d
{tex}\Rightarrow{/tex} 3x - 10t = 30 .....(3) [Using eq. (1)]
Adding equations (2) and (3), we obtain:
x = 50
Substituting the value of x in equation (2), we obtain:
(-2) {tex}\times{/tex} (50) + 10t = 20 {tex}\Rightarrow{/tex}-100 + 10t = 20
{tex}\Rightarrow{/tex}10t = 120
t = 12
From equation (1), we obtain:
d = xt = 50 {tex}\times{/tex} 12 = 600
Thus, the distance covered by the train is 600 km.
Posted by Aastha Asati 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
The system of equation is given by :
bx + cy = a + b ......(i)
{tex}ax\left( {\frac{1}{{a - b}} - \frac{1}{{a + b}}} \right) + cy\left( {\frac{1}{{b - a}} + \frac{1}{{b + a}}} \right){/tex}{tex}= \frac{{2a}}{{a + b}}{/tex} ...(ii)
From equation (i)
bx + cy - (a + b) = 0 ............ (iii)
From equation (ii)
{tex} ax\left( {\frac{1}{{a - b}} - \frac{1}{{a + b}}} \right) + cy\left( {\frac{1}{{b - a}} + \frac{1}{{b + a}}} \right){/tex} {tex} - \frac{{2a}}{{a + b}} = 0{/tex}
{tex}⇒ x\left( {\frac{{2ab}}{{(a - b)(a + b)}}} \right) + y\left( {\frac{{2ac}}{{(b - a)(b + a)}}} \right){/tex} {tex}- \frac{{2a}}{{a + b}} = 0{/tex}
{tex} ⇒ \frac{1}{{a + b}}\left( {\frac{{2abx}}{{a - b}} - \frac{{2acy}}{{a - b}} - 2a} \right) = 0{/tex}
{tex}⇒ \frac{{2abx}}{{a - b}} - \frac{{2acy}}{{a - b}} - 2a = 0{/tex}
2abx - 2acy - 2a(a - b) = 0 ....(iv)
From equation (iii) and (iv), we get
a1 = b, b1 = c and c1 = - (a + b)
and a2 = 2ab , b2 = -2ac and c3 = -2a(a - b)
by cross-multiplication, we get
{tex}\frac{x}{{ - 4{a^2}c}} = \frac{{ - y}}{{4a{b^2}}} = \frac{{ - 1}}{{4abc}}{/tex}
Now, {tex}\frac{x}{{ - 4{a^2}c}} = \frac{{ - 1}}{{4abc}} {/tex}
{tex}⇒ x = \frac{a}{b}{/tex}
And, {tex}\frac{{ - y}}{{4a{b^2}}} = \frac{{ - 1}}{{4abc}} {/tex}
{tex}⇒ y = \frac{b}{c}{/tex}
The solution of the system of equation are {tex}\frac{a}{b}{/tex} and {tex}\frac{b}{c}{/tex}.
Posted by Kanchan Yadav 7 years, 7 months ago
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Anshika Mittal 7 years, 7 months ago
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Kunal Rajour 7 years, 7 months ago
Posted by Ashutosh Ranjan 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Any of the prime numbers that can be multiplied to give the original number.
Example: The prime factors of 15 are 3 and 5 (because 3×5=15, and 3 and 5 are prime numbers).
Posted by Kanika Bhardwaj 7 years, 7 months ago
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Posted by Vishakha Thakur 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Here we have to find out HCF of 2160 and 847 by Using Euclid’s division Lemma, we get
2160 = 847{tex}\times{/tex}2 + 466
Also 847 = 466{tex}\times{/tex}1 + 381
466 = 381{tex}\times{/tex}1 + 85
381 = 85{tex}\times{/tex}4 + 41
85 = 41{tex}\times{/tex}2 + 3
41=3{tex}\times{/tex}13 + 2
3 = 2{tex}\times{/tex}1 + 1
2 = 1{tex}\times{/tex}2 + 0
{tex}\therefore{/tex}HCF = 1.
Hence the numbers are co-prime.
Posted by Aryan Verma 7 years, 7 months ago
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Saloni Saini 7 years, 7 months ago
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Posted by Chiku Singh 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
{tex}\alpha , \beta \text { and } \gamma{/tex} are zeroes of the polynomial 6x3 + 3x2 - 5x + 1
in the given polynomial, 6x3 + 3x2 - 5x + 1
a=6, b=3, c=-5, d=1
Sum of the roots = {tex}- \frac {b}{a}{/tex}
{tex}\alpha + \beta + \gamma = - \frac { 3 } { 6 }{/tex}
{tex}\alpha + \beta + \gamma = - \frac { 1 } { 2 }{/tex}
sum of the Product of the roots = {tex}\frac {c}{a}{/tex}
{tex}\alpha \beta + \beta \gamma + \gamma \alpha = - \frac { 5 } { 6 }{/tex}
Product of the roots = {tex}- \frac{d}{a}{/tex}
{tex}\alpha \beta \gamma = - \frac { 1 } { 6 }{/tex}
{tex}\therefore \quad \frac { 1 } { \alpha } + \frac { 1 } { \beta } + \frac { 1 } { \gamma } = \frac { \alpha \beta + \beta \gamma + \gamma \alpha } { \alpha \beta \gamma }{/tex}
{tex}= \frac { - 5 / 6 } { - 1 / 6 } = \frac { - 5 } { 6 } \times \frac { 6 } { - 1 }{/tex}
Hence, {tex}\alpha ^ { - 1 } + \beta ^ { - 1 } + \gamma ^ { -1 } = 5{/tex}
Posted by Tejal C 7 years, 7 months ago
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Vaishnavi Singh 7 years, 7 months ago
2Thank You