Difference between revisions of "2024 AMC 10A Problems/Problem 18"
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+ | {{duplicate|[[2024 AMC 10A Problems/Problem 18|2024 AMC 10A #18]] and [[2024 AMC 12A Problems/Problem 11|2024 AMC 12A #11]]}} | ||
+ | |||
==Problem== | ==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>? | 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>? | ||
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<math>\textbf{(A) }16\qquad\textbf{(B) }17\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad\textbf{(E) }21</math> | <math>\textbf{(A) }16\qquad\textbf{(B) }17\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad\textbf{(E) }21</math> | ||
− | ==Solution== | + | ==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>\ | + | <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 <math>3</math> is too small so <math>759 - 1 = 758\implies\boxed{\textbf{(D) }20}</math>. |
− | ~OronSH ~mathkiddus | + | |
+ | ~OronSH ~[[User:Mathkiddus|mathkiddus]] ~andliu766 ~megaboy6679 | ||
+ | |||
+ | ==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== | ==See also== | ||
− | {{AMC10 box|year=2024|ab=A| | + | {{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}} | {{MAA Notice}} |
Latest revision as of 14:37, 17 November 2024
- The following problem is from both the 2024 AMC 10A #18 and 2024 AMC 12A #11, so both problems redirect to this page.
Contents
[hide]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 1
, if even then . If odd then so . Now so but is too small so .
~OronSH ~mathkiddus ~andliu766 ~megaboy6679
Solution 2
Clearly, is either even or odd. If is even, let .
Thus, one solution is for some integer , or .
What if is odd? Then let :
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
Solution 3
Note that is to be divisible by , which means that is divisible by .
If , then is not divisible by .
If , then is not divisible by .
If , then is not divisible by .
If , then is divisible by .
If , then is not divisible by .
If , then is not divisible by .
If , then is divisible by .
If , then is divisible by .
Therefore, for every values of , of them will make divisible by . Therefore, since is divisible by , values of , but this includes , which does not satisfy the given inequality. Therefore, the answer is ~Tacos_are_yummy_1
More detail by ~luckuso
Solution 4
We need take away one because is out of range, so
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
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.