Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Digits and Integers.

Digits and Integers – AIME I, 1990

Let T={\(9^{k}\): k is an integer, \(0 \leq k \leq 4000\)} given that \(9^{4000}\) has 3817 digits and that its first (leftmost) digit is 9, how many elements of T have 9 as their leftmost digit?

  • is 107
  • is 184
  • is 840
  • cannot be determined from the given information

Key Concepts




Check the Answer

But try the problem first…

Answer: is 184.

Suggested Reading

AIME I, 1990, Question 13

Elementary Number Theory by David Burton

Try with Hints

First hint

here \(9^{4000}\) has 3816 digits more than 9,

Second Hint

or, 4000-3816=184

Final Step

or, 184 numbers have 9 as their leftmost digits.

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