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(B) The total number of $3$-digit numbers is $999 - 100 + 1 = 900$.Let $A$ be the set of $3$-digit numbers divisible by $3$. The smallest $3$-digit nu

Vedclass · Accepted Answer

$432$

Vedclass · Answer

(B) The total number of $3$-digit numbers is $999 - 100 + 1 = 900$.Let $A$ be the set of $3$-digit numbers divisible by $3$. The smallest $3$-digit number divisible by $3$ is $102$ and the largest is $999$. Number of terms $= \frac{999 - 102}{3} + 1 = 300$.Let $B$ be the set of $3$-digit numbers divisible by $4$. The smallest $3$-digit number divisible by $4$ is $100$ and the largest is $996$. Number of terms $= \frac{996 - 100}{4} + 1 = 225$.Let $A \cap B$ be the set of $3$-digit numbers divisible by both $3$ and $4$,i.e.,divisible by $	ext{lcm}(3, 4) = 12$. The smallest $3$-digit number divisible by $12$ is $108$ and the largest is $996$. Number of terms $= \frac{996 - 108}{12} + 1 = 75$.The number of $3$-digit numbers divisible by $3$ or $4$ is $|A \cup B| = |A| + |B| - |A \cap B| = 300 + 225 - 75 = 450$.Now,we must exclude numbers divisible by $48$. The smallest $3$-digit number divisible by $48$ is $144$ $(48 	imes 3)$ and the largest is $960$ $(48 	imes 20)$. The number of such terms is $20 - 3 + 1 = 18$.Therefore,the required number of values is $450 - 18 = 432$.

The number of $3$-digit numbers that are divisible by either $3$ or $4$ but not divisible by $48$ is:

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