GO Classes CS/DA 2025 | Weekly Quiz 1 | Fundamental Course | Question: 8

Answer 1

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

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

Answer 2

Correct Options: A, C, D.

 

Comments

in the optn: D it will be 8n ie. (5n-(-3n))

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Yes you are correct it is a calculation mistake, thank you for pointing it out!

Answer 3

Ans. A,C,D

Answer 4

use the property that n|(a-b) whenever a≅b(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