2024 AMC 10A Problems/Problem 18

The following problem is from both the 2024 AMC 10A #18 and 2024 AMC 12A #11, so both problems redirect to this page.

Problem

There are exactly positive integers with such that the base- integer is divisible by (where is in base ten). What is the sum of the digits of ?

Solution

, if even then . If odd then so . Now so but is too small so . ~OronSH ~mathkiddus

Solution 2

\begin{align*} 2024_b\equiv0\pmod{16} \\ 2b^3+2b+4\equiv0\pmod{16} \\ b^3+b+2\equiv0\pmod8 \\ \end{align*}

Clearly, is either even or odd. If is even, let .

\begin{align*} (2a)^3+2a+2\equiv0\pmod8 \\ 8a^3+2a+2\equiv0\pmod8 \\ 0+2a+2\equiv0\pmod8 \\ a+1\equiv0\pmod4 \\ a\equiv3\pmod4 \\ \end{align*}

Thus, one solution is for some integer , or .

What if is odd? Then let :

\begin{align*} (2a+1)^3+2a+1+2\equiv0\pmod8 \\ 8a^3+12a^2+6a+1+2a+1+2\equiv0\pmod8 \\ 8a^3+12a^2+8a+4\equiv0\pmod8 \\ 4a^2+4\equiv0\pmod8 \\ a^2\equiv1\pmod2 \\ \end{align*}

This simply states that is odd. Thus, the other solution is for some integer , or .

We now simply must count the number of integers between and , inclusive, that are mod or mod . Note that the former case comprises even numbers only while the latter is only odd; thus, there is no overlap and we can safely count the number of each and add them.

In the former case, we have the numbers ; this list is equivalent to , which comprises numbers. In the latter case, we have the numbers , which comprises numbers. There are numbers in total, so our answer is .

~Technodoggo

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.