Number of 4-digit numbers (the repetition of | vedclass

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(D) For a number to be divisible by $15$,it must be divisible by both $3$ and $5$.Since the number must be divisible by $5$,the last digit must be $5$

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(D) For a number to be divisible by $15$,it must be divisible by both $3$ and $5$.Since the number must be divisible by $5$,the last digit must be $5$.Let the $4-$digit number be $d_1 d_2 d_3 5$.For the number to be divisible by $3$,the sum of its digits $(d_1 + d_2 + d_3 + 5)$ must be divisible by $3$.This implies $(d_1 + d_2 + d_3 + 5) \equiv 0 \pmod{3}$,or $(d_1 + d_2 + d_3) \equiv 1 \pmod{3}$.Possible combinations of $(d_1, d_2, d_3)$ using digits ${1, 2, 3, 5}$ such that their sum is $1 \pmod{3}$ are:$1$. $(1, 2, 1) \rightarrow 3$ permutations: $1215, 2115, 1125$$2$. $(2, 2, 3) \rightarrow 3$ permutations: $2235, 2325, 3225$$3$. $(3, 3, 1) \rightarrow 3$ permutations: $3315, 3135, 1335$$4$. $(1, 1, 5) \rightarrow 3$ permutations: $1155, 1515, 5115$$5$. $(2, 3, 5) \rightarrow 6$ permutations: $2355, 2535, 3255, 3525, 5235, 5325$$6$. $(3, 5, 5) \rightarrow 3$ permutations: $3555, 5355, 5535$Total numbers $= 3 + 3 + 3 + 3 + 6 + 3 = 21$.

Number of $4-$digit numbers (the repetition of digits is allowed) which are made using the digits $1, 2, 3$ and $5$,and are divisible by $15$,is equal to $............$.

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