Question

Five digit number divisible by 3 are formed using 0, 1, 2, 3, 4, 5, 6, 7 without repetition. Total of such numbers are

• A. $312$
• B. $3125$
• C. $120$
• D. $1680$

Answer 1

The correct option is D $1680$

The maximum possible sum can be $25.$ $($when the number is formed from $7,6,5,4,3)$ and the minimum can be $10$ (when the number is formed from $0,1,2,3,4)$

Now the 5 digit number with digits' sum $12$ is formed from combination of $\{5,4,2,1,0\}$ or $\{6,3,2,1,0\}$

Hence the no. of such numbers =$(5!-4!)+(5!-4!)=192$
5 digit number with digits' sum 15 is formed from combination of $\{5,4,3,2,1\}$ or $\{6,4,3,2,0\}$ or $\{7,4,3,1,0\}$ or $\{7,5,2,1,0\}$

hence no. of such numbers=$(5!)+(5!-4!)+(5!-4!)+(5!-4!)=408$

5 digit number with digits' sum 18 is formed from combination of {6,5,4,3,0} or {7,5,4,2,0} or {7,5,3,2,1} or {7,6,3,2,0} or {7,6,4,1,0} or {6,5,4,2,1}.

hence no. of such numbers=$4\times (5!-4!)+2\times (5!)=624$

5 digit number with digits' sum 21 is formed from combination of {7,6,5,2,1} or {7,6,5,3,0} or {7,6,4,3,1}.

hence no. of such numbers =$2\times (5!)+(5!-4!)=336$

5 digit number with digits' sum 24 is formed from combination of {7,6,5,4,2}
hence the no. of such numbers =$(5!)=120$

Hence the total such numbers (according to question)$=(192+408+624+336+120)=1680$