The square of any odd integer, when divided by 8, leaves remainder
A3
B5
C0
D1
Answer & Solution
Correct answer: D. 1
An odd integer is $2k+1$; its square is $4k(k+1)+1$.
Since $k(k+1)$ is even, $4k(k+1)$ is a multiple of 8, so the square is $\equiv 1 \pmod 8$.
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