The number of 5 digit numbers which are divisible by 3 that can be formed by using the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9, when repetition of digits is allowed, is

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$3^{9}$

${4.3}^{9}$

${5.3}^{8}$

${7.3}^{8}$

answer is A.

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Detailed Solution

_______(5 blanks)
1st blank can be filled in 9 ways
2nd blank can be filled in 9 ways.....(since repetition is allowed)
3rd blank can be filled in 9 ways
4th blank can be filled in 9 ways
Now, we have to fill the 5th blank carefully such that the number of divisible by 3. Add the 4 numbers in the first 4 blanks.
If their sum is in the form 3n, then fill the last blank by 3, 6 or 9 so that the sum of all digits is divisible by 3.
If their sum is in the form 3n+1, then fill the last blank by 2, 5 or 8.
If their sum is in the form 3n+2, then fill the last blank by 1, 4 or 7.
Therefore, in any case, the last blank can be filled in 3 ways only.
$∴$ Ans = $9 \times 9 \times 9 \times 9 \times 3 = 3^{9}$