Answer aati hogi to krna.
Answers
Answer:
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1.
We have
a and b are two odd positive integers such that a & b
but we know that odd numbers are in the form of 2n+1 and 2n+3 where n is integer.
so, a = 2n + 3, b = 2n + 1, n ∈1
Given ⇒ a > b
now, According to given question
Case I:
a + b 2n+3+2n+1
-------- = ------------------
2 2
= 4n + 4
-------------
2
=2n + 2 = 2 ( n + 1 )
put let m= 2n + 1 then,
a + b
------- = 2m ⇒ even number.
2
Case II:
a − b 2n+3−2n−1
------- = ------------------
2 2
2
-- = 1 ⇒ odd number.
2
Hence we can see that, one is odd and other is even.
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2.
n(n + 1) = n² + 1
= ((n+ 1) – 1)(n + 1)
= (n+ 1)² - (n+ 1)
So if (n + 1) is an even no (n + 1)² is even and diff of even is always even is (n + 1).
If odd then (n + 1)² is odd and diff of odd is always even
So n(n + 1) is always even and divisible by 2
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3.
Let the three consecutive positive integers be n, n+1 and n+2.
Whenever a number is divided by 3, the remainder obtained is either 0,1 or 2.
Therefore, 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+2=3p+1+2=3p+3=3(p+1) 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+2 is always divisible by 3 that is:
n(n+1)(n+2) is divisible by 3.
Similarly, whenever a number is divided by 2, the remainder obtained is either 0 or 1.
Therefore, n=2q or 2q+1, where q is some integer.
If n=2q, then n and n+2=2q+2=2(q+1) is divisible by 2.
If n=2q+1, then 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+2 is always divisible by 2.
Since, n(n+1)(n+2) is divisible by 2 and 3.
Hence, n(n+1)(n+2) is divisible by 6.
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4.
n³-n = n(n²-1) = n(n -1)(n + 1) is divided by 3 then possible reminder is 0, 1 and 2
[if P = ab +r, then 0 < r <a by Euclid lemma ]
Let n = 3r, 3r +1,3r + 2, where r is an integer
Case 1:- when n = 3r
Then, n³-n is divisible by 3 [n³ -n = n(n-1)(n+1) = 3r(3r-1)(3r+1), clearly shown it is divisible by 3]
Case2:- when n = 3r +1
e.g., n - 1= 3r +1-1= 3r
Then, n-n = (3r + 1)(3r)(3r + 2), it is divisible by 3
Case 3:- when= 3r - 1
e.g., n +1= 3r - 1+1= 3r
Then, n³ -n = (3r -1)(3r -2)(3r) ,
it is divisible by 3
From above explanation we observed n³ -n is divisible by 3, where n is any positive integers
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5.
Let n= 6q+5 be a positive integer for some integer q.
We know that any positive integer can be of the form 3k, or 3k+1, or 3k+2. (From Euclid’s division lemma for b= 3)
∴ q can be 3k or, 3k+1 or, 3k+2. (From Euclid’s division lemma for b= 3)
If q= 3k, then
⇒ n= 6q+5
⇒ n= 6(3k)+5
⇒ n= 18q+5 = (18q+3)+ 2
⇒ n= 3(6q+1)+2
⇒ n= 3m+2, where m is some integer
If q= 3k+1, then
⇒ n= 6q+5
⇒ n= 6(3k+1)+5
⇒ n= 18q+6+5 = (18q+9)+ 2
⇒ n= 3(6q+3)+2
⇒ n= 3m+2, where m is some integer
If q= 3k+2, then
⇒ n= 6q+5
⇒ n= 6(3k+2)+5
⇒ n= 18q+12+5 = (18q+15)+ 2
⇒ n= 3(6q+5)+2
⇒ n= 3m+2, where m is some integer
Hence, if a positive integer is of form 6q + 5, then it is of the form 3q + 2 for some integer q.
Conversely,
Let n= 3q+2
And we know that a positive integer can be of the form 6k, or 6k+1, or 6k+2, or 6k+3, or 6k+4, or 6k+5. (From Euclid’s division lemma for b= 6)
So, now if q=6k then
⇒ n= 3q+2
⇒ n= 3(6k)+2
⇒ n= 6(3k)+2
⇒ n= 6m+2, where m is some integer
Now, this is not of the form 6q + 5.
Therefore, if n is of the form 3q + 2, then is necessary won’t be of the form 6q + 5.
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Here, the integer ‘n’ is of the form 5q+1.
⇒ n= 5q+1
On squaring it,
⇒ n2= (5q+1)2
⇒ n2= (25q2+10q+1)
⇒ n2= 5(5q2+2q)+1
⇒ n2= 5m+1, where m is some integer. [For m = 5q2+2q]
Therefore, the square of any positive integer of the form 5q + 1 is of the same form.
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hey mate ✔️brainly✔️ is more nice then ❌doubt nut❌
Let the 2 consecutive numbers be, x,x+1
product of these consecutive numbers, =x(x+1)
(1) even
let, x=2k
product =2k[2k+1]
from above equation it is clear that the product is divisible by 2
(2) odd
let, x=2k+1
product =(2k+1)[(2k+1)+1]
=2(2k
2
+3k+1)
from above equation it is clear that the product is divisible by 2