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, 2008

Number theory Difficulty 5.0 AIME, harder Prove it Slovenia

a. Show that there is no positive integer $n$ such that the sum of the digits of $10^n + 9n$ is divisible by $2007$ .

b. Find at least one positive integer $n$ such that the sum of the digits of $10^n + 9n$ equals $2008$ .

Solution

a. An integer is divisible by $9$ if and only if the sum of its digits is divisible by $9$ . Assume that the sum of the digits of $10^n + 9n$ is divisible by $2007$ . Since $2007$ is a multiple of $9$ , the sum of the digits of $10^n + 9n$ must be divisible by $9$ . This implies that $10^n + 9n$ is divisible by $9$ , which is clearly not the case.

b. Let $n = 1\ldots1$ (223 ones). Then $9n = 9\ldots9$ (223 nines) and $10^n + 9n$ is equal to $10\ldots09\ldots9$ (223 nines and $n - 223 = 1\ldots1 - 223$ zeroes). The sum of the digits of this number is $1 + 9 \cdot 223 = 2008$ .

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