The number of 3-digit odd numbers divisible b | vedclass

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(C) number is divisible by $3$ if the sum of its digits is divisible by $3$. We need to form a $3$-digit odd number using digits from ${1, 2, 3, 4, 5,

Vedclass · Accepted Answer

$24$

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(C) number is divisible by $3$ if the sum of its digits is divisible by $3$. We need to form a $3$-digit odd number using digits from ${1, 2, 3, 4, 5, 6}$ without repetition. The last digit must be $1, 3,$ or $5$.Case $1$: Last digit is $1$. The sum of the other two digits $x+y$ must be $3k-1$. Possible pairs ${x, y}$ from ${2, 3, 4, 5, 6}$ are ${2, 3}, {2, 6}, {3, 5}, {4, 5}$. Each pair gives $2$ permutations. Total $= 4 	imes 2 = 8$.Case $2$: Last digit is $3$. The sum of the other two digits $x+y$ must be $3k-3$. Possible pairs ${x, y}$ from ${1, 2, 4, 5, 6}$ are ${1, 2}, {1, 5}, {2, 4}, {4, 5}$. Each pair gives $2$ permutations. Total $= 4 	imes 2 = 8$.Case $3$: Last digit is $5$. The sum of the other two digits $x+y$ must be $3k-5$. Possible pairs ${x, y}$ from ${1, 2, 3, 4, 6}$ are ${1, 3}, {1, 6}, {2, 4}, {3, 6}$. Each pair gives $2$ permutations. Total $= 4 	imes 2 = 8$.Total numbers $= 8 + 8 + 8 = 24$.

The number of $3$-digit odd numbers divisible by $3$ that can be formed using the digits $1, 2, 3, 4, 5, 6$ when repetition is not allowed is:

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