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Number Theory Level 3

True or False?

There exists a 45-digit positive integer such that it is a power of 2, and its digits consist of:

one 1, two 2's, three 3's, ... , nine 9's.

False True

1 solution

Chew-Seong Cheong
Nov 27, 2016

Let the number be $N$ , then the sum of digits of $N$ is $\displaystyle \sum_{n=1}^9 n^2 = \frac {9(10)(19)}{6} = 3\cdot 5 \cdot 19$ . Since the sum of digits of $N$ is divisible by 3, it means that $N$ is divisible by 3 and it cannot be a power of 2. Therefore, the answer is $\boxed{\text{False}}$ .

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