Question

please answer these questions...
I'll mark brainliest...NO SPAMMING , ELSE WILL BE REPORTED...

Answer 1

For 1st number:-

We apply Euclid Division algorithm on n which is   b=3.
a = bq +r  on putting a = n and b = 3
n = 3q +r  
i.e n = 3q   -------- (1),
n = 3q +1 --------- (2),
 n = 3q +2  -----------(3)
n = 3q is divisible by 3
or n +2  = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
From this we can infer that n, n+2 , n+4 are divisible by 3.

For 2nd part:-
117 = 65 × 1 + 52 ----> [ 2 ]
65 = 52 x 1 + 13 -----> [1]
52 = 13 x 4 + 0
HCF = 13
13 = 65m + 117n
From [ 1] ,13 = 65 - 52 x 1
From [2] ,52 = 117 - 65 x 1 ----> [3]

Hence ,
13 = 65 - [ 117 - 65 x 1 ] ------> from [3]
= 65 x 2 - 117
= 65 x 2 + 117 x [-1 ]
m = 2 and n = -1

Answer 2

Hope this might help u......