Proving problem?

1.
Since 3a is divisible by n, let 3a = k₁n …… [1]
where k₁ is an integer.

Since (12a + 5b) is divisible by n, let (12a + 5b) = k₂n …… [2]
where k₂ is an integer.

[2] - [1] × 4:
(12a + 5b) - (3a) × 4 = k₂n - (k₁n) × 4
12a + 5b - 12a = k₂n - 4k₁n
5b = (k₂ - 4k₁)n
10b = 2(k₂ - 4k₁)n
Since 2(k₂ - 4k₁) is an integer, 10b is divisible by n.

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2.
Since (3a + 7b) is divisible by n, let (3a + 7b) = k₁n …… [1]
where k₁ is an integer.

Since (2a + 5b) is divisible by n, let (2a + 5b) = k₂n …… [2]
where k₂ is an integer.

[1] × 5 - [2] × 7:
(3a + 7b) × 5 - (2a + 5b) × 7 = (k₁n) × 7 - (k₂n) × 5
15a + 35b - 14a - 35b = 7k₁n - 5k₂n
a = (7k₁n - 5k₂)n
Since (7k₁ - 5k₂) is an integer, a is divisible by n.

[2] × 3 - [1] × 2
(2a + 5b) × 3 - (3a + 7b) × 2 = (k₂n) × 3 - (k₁n) × 2
6a + 15b - 6a - 14b = 3k₂n - 2k₁n
b = (3k₂ - 2k₁)n
Since (3k₂ - 2k₁) is an integer, b is divisible by n.

Hence, a is divisible; b is divisible by n.

1.
It is given that 3a is divisible by n
Then that means 12a is divisible by n.

(12a+5b) divisible by n
(12a+5b) - 12a divisible by n
5b divisible by n
10b divisible by n.



2.
3(3a+7b) - 4(2a+5b) = a+b is divisible by n

(3a+7b) - 3(a+b) = 4b is divisible by n
(2a+5b) - 2(a+b) = 3b is divisible by n

Since 4 and 3 are coprime that means b is divisible by n.

There is probably a more straightforward way of getting there...

1. 3a is d'ble by n...(1)., (12a+b) is d'ble by n...(2).;
By (1), 12a is d'ble by n.;
Then, if (12a+b) is d'ble by n, then b must also be d'ble by n. Clearly any integer multiple of b
must be d'ble by n. So 10b is d'ble by n.

2. (3a+7b) is d'ble by n...(1)., (2a+5b) is d'ble by n...(2).;
By (1), there exists some integer k1 such that 3a+7b = k1*n...(3);                                                  By (2), there exists some integer k2 such that 2a+5b = k2*n...(4);
3(4)-2(3) implies (15-14)b = 3*k2*n -2*k1*n, ie., b = n(3k2-2k1)...(5). Clearly, b is d'ble by n.;
Inserting n(3k2-2k1) for b in (4) gives 2a + 5n(3k2-2k1) = k2*n. Then 2a  = k2*n -5n(3k2-2k1),*
ie., 2a = n(k2-15k2+10k1), ie., a = n(5k1-7k2). Clearly, a is d'ble by n.