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The total number of four digit numbers such that each of first three digits is divisible by the last digit, is equal to ______.

Option: 1

1086


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{abcd}

1. When \mathrm{d=1}\mathrm{a} can take any value from \mathrm{1\: to\: 9} and \mathrm{b\: and \: c} can take any value from \mathrm{0\: to\: 9}

\mathrm{\therefore \quad 9 \times 10 \times 10=900} such numbers

2.  \mathrm{d=2, a \: can\: be \: 2,4,6,8, b \: and\: c \: can \: be \: 0,2,4,6,8}

\mathrm{\therefore 4 \times 5 \times 5=100}

3. \mathrm{d=3, a=3 / 6 / 9, \text { b\: and } c=0 / 3 / 6 / 9 }\\

\mathrm{\therefore 3 \times 4 \times 4=48 }

4. \mathrm{d=4, \quad a=4 / 8, b\: and \: c=0 / 4 / 8}\\

\mathrm{\therefore 2\times3\times3=18}\\

5. \mathrm{d=5, a=5, b\: and \: c=5/0}\\

\mathrm{ \therefore d=1 \times 2 \times 2=4}\\

6. \mathrm{d=6, a=6, \quad b\: and \: c=6/0}\\

\mathrm{1 \times 2 \times 2=4}

Similarly for  \mathrm{d=7,8,9 \Rightarrow \text { numbers }=4,4,4}

\mathrm{\text{Total}=900+100+48+18+4+4+4+4+4=1086}

Hence answer is \mathrm{1086}

Posted by

sudhir.kumar

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