alleryFor any integer $n$, the expression $n^2 - 1$ is always divisible by 8 when $n$ is
Aany integer
Ban even integer
Ca prime number
Dan odd integer
Answer & Solution
Correct answer: D. an odd integer
If $n$ is odd, write $n = 2k+1$.
Then $n^2 - 1 = 4k(k+1)$.
Since $k(k+1)$ is a product of two consecutive integers it is even
so $4k(k+1)$ is divisible by 8. For even $n$, $n^2-1$ is odd, so not divisible by 8.
Keep practising Unit 1B2: Algebraic Limits and Continuity
Browse all Unit 1B2: Algebraic Limits and Continuity questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
The decimal expansion of $\dfrac{257}{2^3 \cdot 5^4}$ terminates after how many decimal plThe product of the HCF and LCM of 6 and 20 is:Which of the following is an irrational number?The remainder when $2^{100}$ is divided by 7 isHow many positive integers less than 1000 are divisible by neither 2 nor 5?The HCF of two consecutive positive integers isIf $a = 2^3 \cdot 3^4 \cdot 5^2$ and $b = 2^4 \cdot 3^2 \cdot 7$, then $\dfrac{\text{LCM}(The smallest positive integer that is divisible by every integer from 1 to 10 is