Show that one and only one out of n, n + 3, n + 6, or n + 9 is divisible by 4.

One and only one of the numbers n , n + 3 , n + 6 , or n + 9 is divisible by 4 based on the remainder of n when divided by 4. Each case shows that precisely one of the four expressions results in a 0 remainder when divided by 4. Therefore, the statement holds true for any integer n .
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Answered by Anonymous | 2025-07-04

To prove that one and only one of n , n + 3 , n + 6 , or n + 9 is divisible by 4, let's consider the possible remainders when an integer n is divided by 4.
For any integer n , one of the following must be true:

n ≡ 0 ( mod 4 )

If n is divisible by 4, then the remainder is 0.
Therefore, n is divisible by 4.


n ≡ 1 ( mod 4 )

If n ≡ 1 ( mod 4 ) , then n + 3 ≡ 4 ≡ 0 ( mod 4 ) .
Therefore, n + 3 is divisible by 4.


n ≡ 2 ( mod 4 )

If n ≡ 2 ( mod 4 ) , then n + 6 ≡ 8 ≡ 0 ( mod 4 ) .
Therefore, n + 6 is divisible by 4.


n ≡ 3 ( mod 4 )

If n ≡ 3 ( mod 4 ) , then n + 9 ≡ 12 ≡ 0 ( mod 4 ) .
Therefore, n + 9 is divisible by 4.



In each case, exactly one of the numbers n , n + 3 , n + 6 , or n + 9 is divisible by 4. Since these four cases account for all possible values of n mod 4, this proves that exactly one of these numbers is divisible by 4.

Answered by OliviaLunaGracy | 2025-07-06