2024 AMC 12A Problem 11

Attempt Problem 11 of the 2024 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

There are exactly

K

positive integers

b

with

5\le b\le2024

such that the base-

b

integer

2024_b

is divisible by

16

(where

16

is in base ten). What is the sum of the digits of

K?

16

17

18

20

21

2024_b=2b^3+2b+4,

so you need

2(b^3+b+2)

to be divisible by

16

Reduce

b^3+b+2\pmod 8

and count

b

in each valid residue class of

[5,2024]

Here

2024_b=2b^3+2b+4

=2(b^3+b+2),

2024_b

is divisible by

16

exactly when

b^3+b+2

is divisible by

8.

Checking residues

\bmod 8,

b^3+b+2\equiv0

precisely for

b\equiv3,6,7\pmod8.

Counting

b

[5,2024]:

residue

3

gives

11,\ldots,2019

(

252

values), residue

6

gives

6,\ldots,2022

(

253

values), and residue

7

gives

7,\ldots,2023

(

253

values). So

K=252+253+253=758,

and its digit sum is

7+5+8=20.

Thus, the correct answer is D.

Problem 11 in other years