Question

Please Solve these Questions with proper explabation.
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Answer 1

Answer:

1.6q + r is a positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5

Then, the positive integers are of the form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.

Taking cube of each term, we have, (6q)3 = 216 q3 = 6(36q)3 + 0 = 6m + 0, where m is an integer (6q+1)3 = 216q3 + 108q2 + 18q + 1 = 6(36q3 + 18q2 + 3q) + 1 = 6m + 1, where m is an integer

(6q+2)3 = 216q3 + 216q2 + 72q + 8 = 6(36q3 + 36q2 + 12q + 1) +2 = 6m + 2, where m is an integer

(6q+3)3 = 216q3 + 324q2 + 162q + 27 = 6(36q3 + 54q2 + 27q + 4) + 3 = 6m + 3, where m is an integer

(6q+4)3 = 216q3 + 432q2 + 288q + 64 = 6(36q3 + 72q2 + 48q + 10) + 4 = 6m + 4, where m is an integer

(6q+5)3 = 216q3 + 540q2 + 450q + 125 = 6(36q3 + 90q2 + 75q + 20) + 5 = 6m + 5, where m is an integer

Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

2.let n be any positive integer and b=3

n =3q+r

where q is the quotient and r is the remainder

0_ <r<3

so the remainders may be 0,1 and 2

so n may be in the form of 3q, 3q=1,3q+2

CASE-1

IF N=3q

n+4=3q+4

n+2=3q+2

here n is only divisible by 3

CASE 2

if n = 3q+1

n+4=3q+5

n+2=3q=3

here only n+2 is divisible by 3

CASE 3

IF n=3q+2

n+2=3q+4

n+4=3q+2+4

=3q+6

here only n+4 is divisible by 3

HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE

3. Let n,n+1,n+2 be three consecutive positive integers.

We know that n is of the form 3q,3q+1 or, 3q+2

So, we have the following

Case I When n=3q

In this case, n is divisible by 3 but n+1 and n+2 are not divisible by 3.

Case II When n=3q+1

In this case, n+2=3q+1+2=3 is divisible by 3 but n and n+1 are not divisible by 3.

Case III When n=3q+2

In this case, n+1=3q+1+2=3(q+1) is divisible by 3 but n and n+2 are not divisible by 3.

Hence one of n,n+1 and n+2 is divisible by 3.

4.(n³-n)

Put n = 1,

(1³-1) = 1-1 = 0 is divisible by 6

Put n = 2,

(2³-2) = 8-2 = 6 is divisible by 6

Put n = 3,

(3³-3) = 27-3 = 24 is divisible by 6

Put n = 4,

(4³-4) = 64-4 = 60 is divisible by 6

Therefore,For any positive integer 'n', (n³ -n) is divisible by 6.

5. Any positive integer is of the form 5q , 5q + 1 , 5q + 2

here ,

b = 5

r = 0 , 1 , 2 , 3 , 4

when r = 0 , n = 5q

n  = 5q              ----> divisible by 5               ===> [1]

n + 4 = 5q + 4    [ not divisible by 5 ]

n + 8 = 5q + 8 [ not divisible by 5 ]

n + 6 = 5q + 6 [ not divisible by 5 ]

n + 12 = 5q + 12 [ not divisible by 5 ]

Answer 2

Answer:

Answer is here:-

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