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(B) number is divisible by $6$ if it is divisible by both $2$ and $3$. Since the number must be divisible by $2$,the unit digit must be $2$ or $4$. Si

Vedclass · Accepted Answer

$16$

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(B) number is divisible by $6$ if it is divisible by both $2$ and $3$. Since the number must be divisible by $2$,the unit digit must be $2$ or $4$. Since the number must be divisible by $3$,the sum of its digits must be a multiple of $3$. Case $1$: Unit digit is $2$. The sum of the first two digits must be $3k - 2$. Possible pairs $(d_1, d_2)$ such that $d_1 + d_2 + 2$ is a multiple of $3$: $(1, 0)$ is not possible,$(1, 3) \rightarrow 132, 312$; $(2, 2) \rightarrow 222$; $(2, 5) \rightarrow 252, 522$; $(3, 1) \rightarrow 312, 132$; $(3, 4) \rightarrow 342, 432$; $(4, 3) \rightarrow 432, 342$; $(5, 2) \rightarrow 522, 252$; $(5, 5) \rightarrow 552$. Unique numbers ending in $2$: $132, 312, 222, 252, 522, 342, 432, 552$. (Total $8$ numbers). Case $2$: Unit digit is $4$. The sum of the first two digits must be $3k - 4$. Possible pairs $(d_1, d_2)$ such that $d_1 + d_2 + 4$ is a multiple of $3$: $(1, 1) \rightarrow 114$; $(1, 4) \rightarrow 144, 414$; $(2, 3) \rightarrow 234, 324$; $(3, 2) \rightarrow 324, 234$; $(4, 1) \rightarrow 414, 144$; $(4, 4) \rightarrow 444$; $(5, 3) \rightarrow 534, 354$. Unique numbers ending in $4$: $114, 144, 414, 234, 324, 444, 534, 354$. (Total $8$ numbers). Total numbers = $8 + 8 = 16$.

Total number of $3$-digit numbers that are divisible by $6$ and can be formed by using the digits $1, 2, 3, 4, 5$ with repetition is $.......$.

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