424 views

Let $k$ be some integer and it is given that $a \equiv b(\bmod n)$ then which of the following(s) is/are ALWAYS true?
More than one option can be true.

1. $a \equiv b-3 n(\bmod n)$
2. $a \equiv b+k(\bmod n)$
3. $a+k \equiv b+k(\bmod n)$
4. $a+5 n \equiv b-3 n(\bmod n)$

Answer: A, C, D

Given $a \equiv b \text{ (mod n)}$ i.e. $(a-b)\% n = 0$

Option A: True

$a \equiv b-3n \text{ (mod n)} = (a-b+3n)\%n = ((a-b)\%n +(3n)\%n)\%n = (0+0)\%n = 0$

Option B: False

$a \equiv b+k \text{ (mod n)} = ((a-b)\%n - (k\%n))\%n = (-k)\%n$

Now since $k$ can be any integer, $k$ might not be a multiple of $n$.

Option C: True

$a + k \equiv b+k \text{ (mod n)} = (a+k-b-k)\%n = (a-b)\%n = 0$

Option D: True

$a+5n \equiv b-3n \text{ (mod n)} = (a+5n-b+3n)\%n = ((a-b)\%n + (8n)\%n)\%n = (0+0)\%n = 0$

by

Correct Options: A, C, D.

by

Ans. A,C,D

by

use the property that n|(a-b) whenever ab(modn),

1. n|(a-b) and n|(3n) so A. is always true.
2. n|(a-b) but n|(-k) may or may not be always true
3. n|(a-b) and n|(k-k) therefore C is always true
4. n|(a-b) and n|(5n+3n) therefore D is always true