The total number of 5-digit numbers, formed b | vedclass

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(D) number is divisible by $6$ if it is divisible by both $2$ and $3$. For a number to be divisible by $2$, it must be even. The available even digits

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(D) number is divisible by $6$ if it is divisible by both $2$ and $3$. For a number to be divisible by $2$, it must be even. The available even digits are ${2, 6}$. For a number to be divisible by $3$, the sum of its digits must be divisible by $3$. The sum of all given digits ${1, 2, 3, 5, 6, 7}$ is $24$. Since we need a $5$-digit number, we must exclude one digit such that the sum of the remaining $5$ digits is divisible by $3$. If we exclude $x$, the sum of the remaining $5$ digits is $24 - x$. For this to be divisible by $3$, $x$ must be a multiple of $3$. The digits in the set ${1, 2, 3, 5, 6, 7}$ that are multiples of $3$ are ${3, 6}$. Case $1$: Exclude $3$. The set is ${1, 2, 5, 6, 7}$. Even digits are ${2, 6}$. If the last digit is $2$, we have $4! = 24$ ways. If the last digit is $6$, we have $4! = 24$ ways. Total for Case $1 = 24 + 24 = 48$. Case $2$: Exclude $6$. The set is ${1, 2, 3, 5, 7}$. Even digit is ${2}$. If the last digit is $2$, we have $4! = 24$ ways. Total for Case $2 = 24$. Total number of $5$-digit numbers $= 48 + 24 = 72$.

The total number of $5$-digit numbers, formed by using the digits $1, 2, 3, 5, 6, 7$ without repetition, which are multiples of $6$, is

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