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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile!

Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section on your LinkedIn profile.

Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

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0 replies
jlacosta
Nov 1, 2024
0 replies
Binomial theorem but modified (mod 31)
IHaveNoIdea010   1
N 8 minutes ago by lbh_qys
Determine the remainder of the following expression divided by $31$: $$\sum_{j=0}^{2024} \binom{4048}{2024 + j}30^{j-1}j$$
1 reply
IHaveNoIdea010
Yesterday at 12:02 PM
lbh_qys
8 minutes ago
Absolute Value
chamons   1
N 32 minutes ago by rchokler
What is the smallest possible value of the sum $\left|x-2\right| + \left|x-4\right| + \left|x-5\right| $
1 reply
chamons
an hour ago
rchokler
32 minutes ago
x - 1/y = y - 1/z = z - 1/x (German MO)
aether_1729   2
N 43 minutes ago by rchokler
Find all real $x, y, z$ such that
\[x - \frac{1}{y} = y - \frac{1}{z} = z - \frac{1}{x}.\]
Source: Problem 520943 from 52nd German MO, final round / day 1, grade 9.
2 replies
aether_1729
Yesterday at 12:45 PM
rchokler
43 minutes ago
Trigonometry
axsolers_24   2
N 2 hours ago by axsolers_24
If $\tan x=n\cdot \tan y$ and $\sin x=m\cdot\sin y$.
Find the value of $\cos x$ in terms of $m$ and $n$.
[Show your work]
2 replies
axsolers_24
Yesterday at 3:46 PM
axsolers_24
2 hours ago
No more topics!
Summation
Curves50   12
N Dec 13, 2021 by fungarwai
Find the units digit of $\sum_{k=1}^{2021} k^k$
12 replies
Curves50
Jun 30, 2021
fungarwai
Dec 13, 2021
Summation
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Curves50
125 posts
#1 • 2 Y
Y by Vietvalkyries, fungarwai
Find the units digit of $\sum_{k=1}^{2021} k^k$
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Sprites
478 posts
#2
Y by
Curves50 wrote:
Find the units digit of $\sum_{k=1}^{2021} k^k$
Taking $\sum_{k=1}^{2021} k^k$ (mod 10) we have $202+9*202+1 \equiv \boxed{1} (mod 10)$
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Facejo
2807 posts
#3
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Unrelated, but there is no closed form for $\sum_{k=1}^{n} k^k$, right?
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Sprites
478 posts
#4
Y by
Facejo wrote:
Unrelated, but there is no closed form for $\sum_{k=1}^{n} k^k$, right?

I don't think so.
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Facejo
2807 posts
#5
Y by
Sprites wrote:
Facejo wrote:
Unrelated, but there is no closed form for $\sum_{k=1}^{n} k^k$, right?

I don't think so.

Is there a form that only uses $+,-,\times,\div,\sqrt[n]{},$ and $!$? (basically like closed form except also factorials)
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jhu08
1949 posts
#6
Y by
Facejo wrote:
Unrelated, but there is no closed form for $\sum_{k=1}^{n} k^k$, right?

Nope!
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MathNoob1x
55 posts
#7
Y by
/bump
How to do this question? Any hints :wacko:
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Michael1129
285 posts
#8
Y by
Here's a sketch to the solution:
You can try to calculate the units digit for each term of the expansion and then look for a pattern. For example, notice that $2^2$ and $22^{22}$ has the same units digit. This units digit will continue to go on until $2002^{2002}$, with $k$ increasing by $20$ each time. This gives you a total of $101$ terms and each term gives you $4$ for the units digit, so this is $404$. But the thing is is that you might need to do this to quite a few different patterns which will turn out into a pretty long bash, so...
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fungarwai
859 posts
#9
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I researched this sum before on A spin off an old USAMTS

general results

Click to reveal hidden text
This post has been edited 1 time. Last edited by fungarwai, Dec 12, 2021, 1:40 PM
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BackToSchool
1629 posts
#10 • 1 Y
Y by fungarwai
Note that for $\forall k=10a+b, k^k \equiv b^k \pmod {10}$.
Meanwhile, for $\forall k=4m+n, k^k \equiv k^n \pmod {10}$.
Thus, the units digit of $k^k$ has a repeated pattern with period of $20=lcm(10, 4)$.
\begin{align*}
\sum_{k=1}^{2021} k^k & \equiv 101\cdot \sum_{k=1}^{20}k^k + 2021^{2021} \\
& \equiv 101 \cdot (1^1+2^2+3^3+4^4+5^5+6^6+7^7+8^8+9^9+10^{10}+11^{11}+12^{12}+13^{13}+14^{14}+15^{15}+16^{16}+17^{17}+18^{18}+19^{19}+20^{20})+1 \\
& \equiv 101 \cdot (1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 + 9 + 0 + 1 + 6 + 3 + 6 + 5 + 6 + 7 + 4 + 9 + 0) + 1 \\
& \equiv 101 \cdot 4 +1 \\
& \equiv \boxed{5} \pmod {10}
\end{align*}
This post has been edited 1 time. Last edited by BackToSchool, Dec 13, 2021, 2:45 AM
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fungarwai
859 posts
#11
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BackToSchool wrote:
Note that for $\forall k=10a+b, k^k \equiv b^k \pmod {10}$.
Meanwhile, for $\forall k=4m+n, k^k \equiv k^n \pmod {10}$.
Thus, the units digit of $k^k$ has a repeated pattern with period of $20$.
\begin{align*}
\sum_{k=1}^{2021} k^k & \equiv 101\cdot \sum_{k=1}^{20}k^k + 2021^{2021} \\
& \equiv 101 \cdot (1^1+2^2+3^3+4^4+5^5+6^6+7^7+8^8+9^9+10^{10}+11^{11}+12^{12}+13^{13}+14^{14}+15^{15}+16^{16}+17^{17}+18^{18}+19^{19}+20^{20})+1 \\
& \equiv 101 \cdot (1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 + 9 + 0 + 1 + 6 + 3 + 6 + 5 + 6 + 7 + 4 + 9 + 0) + 1 \\
& \equiv 101 \cdot 4 +1 \\
& \equiv \boxed{5} \pmod {10}
\end{align*}

This is right. It has a period of 20 in this case.
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fungarwai
859 posts
#12
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  1. sum=0;n=2021;m=10;
  2. for k in range(1,n+1):
  3. p=1
  4. for t in range(1,k+1):
  5. p=(p*k) % m
  6. if k<=40:
  7. print("No.",k,":",p)
  8. sum=(sum+p) % m
  9. print(sum)
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fungarwai
859 posts
#13
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I post a new entry for this problem on my blog now.

modular arithmetic on 1^1+2^2+3^3+...+n^n
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