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Number theory Difficulty 3.8 AMC 10/12 Find the answer Canada

The digits from 1 to 9 are written in order so that the digit $n$ is written $n$ times. This forms the block of digits $1223334444\cdots 999999999$ . The block is written 100 times. What is the 1953 $^{rd}$ digit written?

Pick one

Solution

In each block $1223334444\cdots999999999$ , there is 1 digit 1, 2 digits 2, 3 digits 3, and so on.

The total number of digits written in each block is $1+2+3+4+5+6+7+8+9=45$ .

We note that $1953\div45$ gives a quotient of 43 and a remainder of 18 (that is, $1953=45\times43+18$ ).

Since each block contains 45 digits, then 43 blocks contain $43\times 45=1935$ digits.

Since $1953-1935=18$ , then the 18 $^{th}$ digit written in the next block (the 44 $^{th}$ block) will be the 1953 $^{rd}$ digit written.

Writing out the first 18 digits in a block, we get 122333444455555666, and so the 1953 $^{rd}$ digit written is a 6.

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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.