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, 2008

Number theory Difficulty 3.0 AMC 10/12 Find the answer Canada

The sum of all the digits of the integers from $98$ to $101$ is
$9 + 8 + 9 + 9 + 1 + 0 + 0 + 1 + 0 + 1 = 38$
The sum of all of the digits of the integers from $1$ to $2008$ is

(A) $30\,054$ (B) $27\,018$ (C) $28\,036$ (D) $30\,036$ (E) $28\,054$

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project. Solutions are the publisher's, linked not copied.