Nine hundred distinct n digit numbers are to be formed using only the three digits 2,5 and 7. the smallest value of n for which this is possible is
A
4
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B
8
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C
7
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D
9
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Solution
The correct option is B7 Using 2,5 and 7 with repetition, each place of n digits number can be chosen in 3 ways. So total of n digit number is equal to =3×3×3×....×3(ntimes)=3n
According to the question, 3n≥900⇒3n−2≥100⇒n−2≥5⇒n≥7