|
|
| Line 1: |
Line 1: |
| − | {{duplicate|[[2024 AMC 10A Problems/Problem 18|2024 AMC 10A #18]] and [[2024 AMC 12A Problems/Problem 11|2024 AMC 12A #11]]}}
| + | #redirect [[2024 AMC 12A Problems/Problem 14]] |
| − | | |
| − | ==Problem==
| |
| − | There are exactly <math>K</math> positive integers <math>b</math> with <math>5 \leq b \leq 2024</math> such that the base-<math>b</math> integer <math>2024_b</math> is divisible by <math>16</math> (where <math>16</math> is in base ten). What is the sum of the digits of <math>K</math>?
| |
| − | | |
| − | <math>\textbf{(A) }16\qquad\textbf{(B) }17\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad\textbf{(E) }21</math>
| |
| − | | |
| − | ==Solution 1==
| |
| − | <math>2b^3+2b+4\equiv 0\pmod{16}\implies b^3+b+2\equiv 0\pmod 8</math>, if <math>b</math> even then <math>b+2\equiv 0\pmod 8\implies b\equiv 6\pmod 8</math>. If <math>b</math> odd then <math>b^2\equiv 1\pmod 8\implies b^3+b+2\equiv 2b+2\pmod 8</math> so <math>2b+2\equiv 0\pmod 8\implies b+1\equiv 0\pmod 4\implies b\equiv 3,7\pmod 8</math>. Now <math>8\mid 2024</math> so <math>\frac38\cdot 2024=759</math> but one of the answers we got from that, <math>3</math>, is too small, so <math>759 - 1 = 758\implies\boxed{\textbf{(D) }20}</math>.
| |
| − | | |
| − | ~OronSH ~[[User:Mathkiddus|mathkiddus]] ~andliu766 ~megaboy6679 ~trevian1(minor edit for clarity)
| |
| − | | |
| − | ==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, <math>b</math> is either even or odd. If <math>b</math> is even, let <math>b=2a</math>.
| |
| − | | |
| − | \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 <math>b=2(4x+3)=8x+6</math> for some integer <math>x</math>, or <math>b\equiv6\pmod8</math>.
| |
| − | | |
| − | What if <math>b</math> is odd? Then let <math>b=2a+1</math>:
| |
| − | | |
| − | \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 <math>a</math> is odd. Thus, the other solution is <math>b=2(2x+1)+1=4x+3</math> for some integer <math>x</math>, or <math>b\equiv3\pmod4</math>.
| |
| − | | |
| − | We now simply must count the number of integers between <math>5</math> and <math>2024</math>, inclusive, that are <math>6</math> mod <math>8</math> or <math>3</math> mod <math>4</math>. 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 <math>6,14,22,30,\dots,2022</math>; this list is equivalent to <math>8,16,24,32,\dots,2024\cong1,2,3,4,\dots,253</math>, which comprises <math>253</math> numbers. In the latter case, we have the numbers <math>7,11,15,19,\dots,2023\cong4,8,12,16,\dots,2020\cong1,2,3,4,\dots,505</math>, which comprises <math>505</math> numbers. There are <math>758</math> numbers in total, so our answer is <math>7+5+8=\boxed{\textbf{(D) 20}}</math>.
| |
| − | | |
| − | ~Technodoggo
| |
| − | | |
| − | ==Solution 3==
| |
| − | Note that <math>2024_b=2b^3+2b+4</math> is to be divisible by <math>16</math>, which means that <math>b^3+b+2</math> is divisible by <math>8</math>.
| |
| − | | |
| − | If <math>b=0</math>, then <math>b^3+b+2 \equiv (0)^3 + (0) + 2 \equiv 2</math> is not divisible by <math>8</math>.
| |
| − | | |
| − | If <math>b=1</math>, then <math>b^3+b+2 \equiv (1)^3 + (1) + 2 \equiv 4</math> is not divisible by <math>8</math>.
| |
| − | | |
| − | If <math>b=2</math>, then <math>b^3+b+2 \equiv (2)^3 + (2) + 2 \equiv 4</math> is not divisible by <math>8</math>.
| |
| − | | |
| − | If <math>b=3</math>, then <math>b^3+b+2 \equiv (3)^3 + (3) + 2 \equiv (8+1)\cdot3 + (3) + 2 \equiv 8 </math> is divisible by <math>8</math>.
| |
| − | | |
| − | If <math>b=4</math>, then <math>b^3+b+2 \equiv (4)^3 + (4) + 2 \equiv 0 + 4 + 2 \equiv 6 </math> is not divisible by <math>8</math>.
| |
| − | | |
| − | If <math>b=5</math>, then <math>b^3+b+2 \equiv (-3)^3 + (-3) + 2 \equiv (8+1)\cdot 3 + (-3) + 2 \equiv 2 </math> is not divisible by <math>8</math>.
| |
| − | | |
| − | If <math>b=6</math>, then <math>b^3+b+2 \equiv (-2)^3 + (-2) + 2 \equiv -8 + (-2) + 2 \equiv 0 </math> is divisible by <math>8</math>.
| |
| − | | |
| − | If <math>b=7</math>, then <math>b^3+b+2 \equiv (-1)^3 + (-1) + 2 \equiv -1 + (-1) + 2 \equiv 0 </math> is divisible by <math>8</math>.
| |
| − | | |
| − | Therefore, for every <math>8</math> values of <math>b</math>, <math>3</math> of them will make <math>b^3+b+2</math> divisible by <math>8</math>. Therefore, since <math>2024</math> is divisible by <math>8</math>, <math>\dfrac{3}{8}\cdot2024=759</math> values of <math>b</math>, but this includes <math>b=3</math>, which does not satisfy the given inequality. Therefore, the answer is <cmath>759-1=758\rightarrow7+5+8=\boxed{\text{(D) }20}</cmath> ~Tacos_are_yummy_1
| |
| − | | |
| − | More detail by ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
| |
| − | | |
| − | ==Solution 4==
| |
| − | | |
| − | <math>2024_b=2\ast\ b^3+2\ast\ b+4\ \\
| |
| − | {2024}_{\left(b+8\right)}=2\ast\left(b+8\right)^3+2\ast\left(b+8\right)+4</math>
| |
| − | <math>{2024}_{\left(b+8\right)}-{2024}_b=2*\left(8\right)*\left(b^2+8b+64\right)+2*8\ =16*\left(b^2+8b+64\right)+16</math>
| |
| − | | |
| − | \begin{align*}
| |
| − | 2024_{(b+8)}-2024_b\equiv0\ (mod\ 16)\\
| |
| − | 2024_{(b+8)}\ \ \equiv2024_b\ \ (mod\ 16)\\
| |
| − | 2024_0\equiv4\ (mod\ 16)\\
| |
| − | 2024_1\equiv8\ (mod\ 16)\\
| |
| − | 2024_2\equiv6\ (mod\ 16)\\
| |
| − | 2024_3\equiv0(mod\ 16)\\
| |
| − | 2024_4\equiv12(mod\ 16)\\
| |
| − | 2024_5\equiv8(mod\ 16)\\
| |
| − | 2024_6\equiv0(mod\ 16)\\
| |
| − | 2024_7\equiv0(mod\ 16)\\
| |
| − | \end{align*}
| |
| − | | |
| − | We need
| |
| − | <math>b\ \equiv3\ (mod\ 8)\ \ or\ b\ \equiv6\ (mod\ 8)\ \ or\ b\ \equiv7\ (mod\ 8) \\
| |
| − | \lfloor(2024-3)/8\rfloor+\lfloor(2024-6)/8\rfloor+\lfloor(2024-7)/8\rfloor+3=759</math>
| |
| − | take away one because <math>b=3</math> is out of range, so <math>758\Rightarrow7+8+5=\boxed{\text{(D) }20}</math>
| |
| − | | |
| − | | |
| − | == Video Solution by Power Solve ==
| |
| − | https://www.youtube.com/watch?v=qtFvaD9TEaA
| |
| − | | |
| − | == Video Solution by Pi Academy ==
| |
| − | | |
| − | https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
| |
| − | | |
| − | ==Video Solution by SpreadTheMathLove==
| |
| − | https://www.youtube.com/watch?v=6SQ74nt3ynw
| |
| − | | |
| − | ==See also==
| |
| − | {{AMC10 box|year=2024|ab=A|num-b=17|num-a=19}}
| |
| − | {{AMC12 box|year=2024|ab=A|num-b=10|num-a=12}}
| |
| − | {{MAA Notice}}
| |
