@Applied Course there can be multiple values for k, for example you replace k with 15, and 18 can still divide the number hence option(D) should be the correct option, right?

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$\frac{5k89k2}{18}$

$18=9 \times 2$

So $5k89k2$ should be divided by $9$ as well as $2$.

As unit digit is even number $5k89k2$ is divisible by $2$.

$\textit{Divisibility rule 9:}$ Sum of digits should be multiple of $9$

$5+k+8+9+k+2 = 24 + 2k$

Now check with options

(A) $\rightarrow$ If $k = 8$ then $24 + 2*8 = 40$ not divisible by $9$.

(B) If $k = 6$ then $24 + 2*6 = 36$ divisible by $9$. Therefore $k=6$ is answer.

$18=9 \times 2$

So $5k89k2$ should be divided by $9$ as well as $2$.

As unit digit is even number $5k89k2$ is divisible by $2$.

$\textit{Divisibility rule 9:}$ Sum of digits should be multiple of $9$

$5+k+8+9+k+2 = 24 + 2k$

Now check with options

(A) $\rightarrow$ If $k = 8$ then $24 + 2*8 = 40$ not divisible by $9$.

(B) If $k = 6$ then $24 + 2*6 = 36$ divisible by $9$. Therefore $k=6$ is answer.

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