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JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • question_answer
    An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

    A) 72 (7!)                  

    B) 18 (7!)

    C) 40 (7!)                  

    D) 36 (7!)

    Correct Answer: D

    Solution :

                    We know that any number is divisible by 9 if sum of the digits of the number is divisible by 9. Now sum of the digits from 0 to 9 = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 Hence to form 8 digits numbers which are divisible by 9, a pair of digits either 0 and 9, 1 and 8, 2 and 7, 3 and 6 or 4 and 5 are not used.
    Digits which are not used to form 8 digits number divisible by Number of 8 digits numbers which are divisible by9
    0 and 9 \[8\times 7!\]
    1 and 8 \[7\times 7!\]
    2 and 7 \[7\times 7!\]
    3 and 6 \[7\times 7!\]
    4 and 5 \[7\times 7!\]
    Hence total number of 8 digits numbers which are divisible by 9 \[=8\times (7!)+7\times (7!)+7\times (7!)+7\times (7!)+7\times \text{(}7!\text{)}\] \[=36\times (7!)\]


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