- ✓$9$
- B$18$
- C$27$
- D$36$
As $N$ be the least positive integer and when a non-zero $\operatorname{digit} C$ is written af ter the last digit of $N$, the resulting number is divisible by $C$.
So, $10 N+C$ is divisible by $C$
$\therefore 10 N$ must be divisible by $C$.
Now, the least integer $(N)$ which is divisible by digit ' $C$ ' i.e. ( 1 to 9$)$ must be L.C.M of $\{1,3,4,6,7,9\}$.
$= L.C.M$ $of$ $\{4,7,9\}$
$=252=N$
and sum of digits of number ' $N$ ' is
$2+5+2=9$
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$L _1: x \sqrt{2}+ y -1=0$ and $L _2: x \sqrt{2}- y +1=0$
For a fixed constant $\lambda$, let $C$ be the locus of a point $P$ such that the product of the distance of $P$ from $L_1$ and the distance of $P$ from $L_2$ is $\lambda^2$. The line $y=2 x+1$ meets $C$ at two points $R$ and $S$, where the distance between $R$ and $S$ is $\sqrt{270}$.
Let the perpendicular bisector of $RS$ meet $C$ at two distinct points $R ^{\prime}$ and $S ^{\prime}$. Let $D$ be the square of the distance between $R ^{\prime}$ and $S ^{\prime}$.
($1$) The value of $\lambda^2$ is
($2$) The value of $D$ is