Question

please answer the above question

Answer 1

Consider how many digits you have to choose at each spot: for the one's spot, you have 33digits; for the tens spot, you have three digits; all the way to the nnth spot, you have 33 digits to choose from. As such, the total number of ways to make an nn-digit number with only 2,52,5and 77 is going to be 3n3n.

Thus we're looking for the smallest integer nnsuch that 3n≥9003n≥900, or in other words, since the log function is increasing, nln(3)≥ln(900)nln⁡(3)≥ln⁡(900) or, dividing by ln(3)ln⁡(3), n≥6.19n≥6.19. Thus, n=7n=7

so 2 is the answer.
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