If alpha and beta are the least positive inte | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For $n^3+\alpha n$ to be divisible by $3$ for all $n \in N$, we check $n=1$: $1^3+\alpha(1) = 1+\alpha$. For this to be divisible by $3$, the leas

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) For $n^3+\alpha n$ to be divisible by $3$ for all $n \in N$, we check $n=1$: $1^3+\alpha(1) = 1+\alpha$. For this to be divisible by $3$, the least positive integer $\alpha$ is $2$ (since $1+2=3$).For $n^3-\beta n$ to be divisible by $6$ for all $n \in N$, we check $n=2$: $2^3-\beta(2) = 8-2\beta$. For this to be divisible by $6$, $8-2\beta = 6k$. If $\beta=1$, $8-2=6$, which is divisible by $6$. Thus, the least positive integer $\beta$ is $1$.Therefore, $\alpha+\beta = 2+1 = 3$.

If $\alpha$ and $\beta$ are the least positive integers such that for all $n \in N$, $n^3+\alpha n$ is divisible by $3$ and $n^3-\beta n$ is divisible by $6$, then $\alpha+\beta=$

Similar Questions

If $a, b$ and $c$ are in arithmetic progression, then the roots of the equation $a x^{2}-2 b x+c=0$ are

Find the number of real solutions for the equation $\left( \frac{5}{7} \right)^x = -x^2 + 2x - 3$.

Let $f(x) = x^3 + a x^2 + b x + c$ be a polynomial with integer coefficients. If the roots of $f(x)$ are integers and are in Arithmetic Progression,then '$a$' cannot take the value:

If the sum of the two roots of the equation $4x^3 + 16x^2 - 9x - 36 = 0$ is zero,then the roots are

$A$ number is the reciprocal of the other. If the arithmetic mean of the two numbers is $\frac{13}{12}$,then the numbers are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module