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Sia ? 6 years, 4 months ago
Let a be the positive integer and b = 6.
Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 5.
So, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.
(6q)2 = 36q2 = 6(6q2)
= 6m, where m is any integer.
(6q + 1)2 = 36q2 + 12q + 1
= 6(6q2 + 2q) + 1
= 6m + 1, where m is any integer.
(6q + 2)2 = 36q2 + 24q + 4
= 6(6q2 + 4q) + 4
= 6m + 4, where m is any integer.
(6q + 3)2 = 36q2 + 36q + 9
= 6(6q2 + 6q + 1) + 3
= 6m + 3, where m is any integer.
(6q + 4)2 = 36q2 + 48q + 16
= 6(6q2 + 7q + 2) + 4
= 6m + 4, where m is any integer.
(6q + 5)2 = 36q2 + 60q + 25
= 6(6q2 + 10q + 4) + 1
= 6m + 1, where m is any integer.
Hence, The square of any positive integer is of the form 6m, 6m + 1, 6m + 3, 6m + 4 and cannot be of the form 6m + 2 or 6m + 5 for any integer m.
Posted by Akanksha Devgan 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
f(x)=ax2 - 5x + c.
as m and n are zeros of polynomial
therefore m + n = {tex}- \frac { ( - 5 ) } { a } = \frac { 5 } { a }{/tex} and mn = {tex}\frac { c } { a }{/tex}
Now, m + n = 10
{tex}\frac 5a{/tex} =10
so a={tex}\frac12{/tex}
also mn =10 ={tex}\frac { c } { a }{/tex}
10a=c
c=10{tex}\times{/tex} {tex}\frac 12=5{/tex}
hence c = 5 and a ={tex}\frac 12{/tex}
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Simran Tejasvi 7 years, 7 months ago
1Thank You