The product of three consecutive natural numbers is divisible by:… - Vidyadip

MCQ

The product of three consecutive natural numbers is divisible by:

$3$
$8$
$6$
$11$

✓
$a$ and $c$
B
$b$ and $c$
C
$c$ and $d$
D
$a$ and $d$

✓

Answer

Correct option: A.

$a$ and $c$

Let $n, n + 1, n + 2$ be three consecutive natural numbers and $P(n)$ be their product. Then,
$P(n) = n(n + 1)(n + 2)$
We have,
$P(1) = 1 \times 2 \times 3 = 6,$ which is divisible by $3$ and $6.$
$P(2) = 2 \times 3 \times 4 = 24,$ which is divisible by $3, 8$ and $6.$
$P(3) = 3 \times 4 \times 5 = 60,$ which is divisible by $3$ and $6.$
$P(4)= 4 \times 5 \times 6 = 120,$ which is divisible by $3, 8$ and $6.$
Hence, $P(n)$ is divisible by $3$ and $6$ for all $n \in N.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from PRINCIPLE OF MATHEMATICAL INDUCTION→PRINCIPLE OF MATHEMATICAL INDUCTION — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Let S be the sum, P be the product and R be the sum of the reciprocals of 3 terms of a G.P. then $P^2R^3 : S^3$ is equal to:

Which of the following is a statement?

If $f(x) = \left\{ \begin{array}{l}\sin x,x \ne n\pi ,n \in Z\\\,\,\,\,\,\,0,\,\,{\rm{otherwise}}\end{array} \right.$ and $g(x) = \left\{ \begin{array}{l}{x^2} + 1,x \ne 0,\,2\\\,\,\,\,\,\,\,\,4,x = 0\\\,\,\,\,\,\,\,\,\,5,x = 2\end{array} \right.$ then $\mathop {\lim }\limits_{x \to 0} g\{ f(x)\} = $

The coefficient of $x^4$ in ${\left[ {\frac{x}{2}\,\, - \,\,\frac{3}{{{x^2}}}} \right]^{10}}$ is :

Variance of $^{10}C_0$ , $^{10}C_1$ , $^{10}C_2$ ,.... $^{10}C_{10}$ is

The set of all possible outcomes of any experiment is called:

If $\cos \theta + \cos 2\theta + \cos 3\theta = 0$, then the general value of $\theta $ is

If two mutually perpendicular lines through the point $A(1, 1)$ intersect $x$ & $y$ axis at points $B$ & $C$ respectively, then locus of centroid of $\Delta ABC$ is -

Let $\mathrm{p}$ and $\mathrm{q}$ be real numbers such that $\mathrm{p} \neq 0, \mathrm{p}^3 \neq \mathrm{q}$ and $\mathrm{p}^3 \neq-\mathrm{q}$. If $\alpha$ and $\beta$ are nonzero complex numbers satisfying $\alpha+\beta=-\mathrm{p}$ and $\alpha^3+\beta^3=\mathrm{q}$, then a quadratic equation having $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ as its roots is

Let $a, b \in \mathbb{R}$ and $a^2+b^2 \neq 0$. Suppose $S=\left\{z \in \mathbb{C}: z=\frac{1}{a+i b t}, t \in \mathbb{R}, t \neq 0\right\}$, where $i=\sqrt{-1}$, If $z=x+i y$ and $z \in S$, then $(x, y)$ lies on

($A$) the circle with radius $\frac{1}{2 a}$ and centre $\left(\frac{1}{2 a}, 0\right)$ for $a>0, b \neq 0$

($B$) the circle with radius $-\frac{1}{2 a}$ and centre $\left(-\frac{1}{2 a}, 0\right)$ for $a<0, b ; 0$

($C$) the $x$-axis for $a \neq 0, b=0$

($D$) the $y$-axis for $a=0, b \neq 0$