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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$24$

Vedclass · Answer

(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:

Similar Questions

If ${}^n P_r = 30240$ and ${}^n C_r = 252$,then the ordered pair $(n, r)$ is equal to

The number of all $3$-digit numbers $abc$ (in base $10$) for which $(a \times b \times c) + (a \times b) + (b \times c) + (c \times a) + a + b + c = 29$ is:

The total number of seven-digit numbers whose sum of digits is even is

The number of four-letter words that can be formed using the letters of the word $BARRACK$ is

The number of three-digit numbers in which $9$ appears in exactly one place is

Vedclass Test Series

Exam Paper Generator

Online Exam Module