An n digit positive number with exactly n digits. 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) 6
(b) 7
(c) 8
(d) 9
Answers
Answered by
2
Answer:
Using 2,5 and 7 with repetition each place of n digit number can be chosen in 3 ways.
Hence, total number of n−digit numbers =3×3×3...n times =3
n
.
According to given condition 3
n
≥900⇒3
n−2
≥100
∴n−2≥5⇒n≥7
Answer is 7.
Answered by
8
Given that,
An n digit positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7.
Since, we have to formed n digit number, so it means there are n places to fill the given three digits 2, 5 and 7.
So, it means,
.
.
.
.
.
So, total number of ways in which n digit number can be formed with the help of the digits 2, 5 and 7 is
Now, According to statement,
So, using hit and trial method,
So, it implies
Hence,
- The smallest value of n is 7
So,
- Option (b) is correct.
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