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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
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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 44 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
44 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!
A spin off an old USAMTS
Plops   26
N Dec 18, 2019 by Joker7902
Determine the rightmost three digits of the number
\[1^1+2^2+3^3+...+999^{999}+1000^{1000}.\]
26 replies
Plops
Jul 4, 2019
Joker7902
Dec 18, 2019
A spin off an old USAMTS
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G H BBookmark kLocked kLocked NReply
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Plops
946 posts
#1 • 2 Y
Y by fungarwai, Adventure10
Determine the rightmost three digits of the number
\[1^1+2^2+3^3+...+999^{999}+1000^{1000}.\]
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aops5234
530 posts
#2 • 3 Y
Y by Williamgolly, Adventure10, Mango247
Let $x=1^1+2^2+3^3+...+999^{999}+1000^{1000}.$ It is evident that
$1000^{1000} < x < 1000^{1} + 1000^{2} + 1000^{3} + … + 1000^{999} + 1000^{1000} $
or that $x$ is between $1000000... $ and $100100100...$ . Since both these numbers have exactly $3001$
digits, the first three digits of $x$ must be ${\boxed{100}}$.
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chocolatelover111
1815 posts
#3 • 1 Y
Y by Adventure10
Wait isn't right-most the last three digits?
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serichaoo
370 posts
#4 • 1 Y
Y by Adventure10
@2above, those are the leftmost 3 digits. The question is asking for the rightmost 3 digits.
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dragonrider1
461 posts
#5 • 2 Y
Y by Adventure10, Mango247
Remember to always reread the question to make sure you are answering the right thing!
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chocolatelover111
1815 posts
#7 • 2 Y
Y by Adventure10, Mango247
Rightmost is furthest from the right while I was thinking that it was at the right... :P
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fungarwai
859 posts
#9 • 3 Y
Y by Adventure10, Mango247, Mango247
consideration of module 8

I haven't complete module 125 yet. Let's check the answer from my program first.

ans=700 by program
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fungarwai
859 posts
#10 • 3 Y
Y by Mixer_V, Math5000, Adventure10
for module 125

conclusion of CRT
This post has been edited 2 times. Last edited by fungarwai, Jul 4, 2019, 10:45 PM
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gasgous
887 posts
#11 • 1 Y
Y by Adventure10
$3001$ digits.
Attachments:
k.pdf (312kb)
This post has been edited 1 time. Last edited by gasgous, Jul 4, 2019, 12:03 PM
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gasgous
887 posts
#12 • 1 Y
Y by Adventure10
fungarwai wrote:
for module 125

conclusion of CRT
Nice solution.
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JustKeepRunning
2958 posts
#14 • 1 Y
Y by Adventure10
aops5234 wrote:
Let $x=1^1+2^2+3^3+...+999^{999}+1000^{1000}.$ It is evident that
$1000^{1000} < x < 1000^{1} + 1000^{2} + 1000^{3} + … + 1000^{999} + 1000^{1000} $
or that $x$ is between $1000000... $ and $100100100...$ . Since both these numbers have exactly $3001$
digits, the first three digits of $x$ must be ${\boxed{100}}$.

How do you understand that $1000^{1000}$ has $3001$ digits?
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gasgous
887 posts
#15 • 1 Y
Y by Adventure10
JustKeepRunning wrote:
aops5234 wrote:
Let $x=1^1+2^2+3^3+...+999^{999}+1000^{1000}.$ It is evident that
$1000^{1000} < x < 1000^{1} + 1000^{2} + 1000^{3} + … + 1000^{999} + 1000^{1000} $
or that $x$ is between $1000000... $ and $100100100...$ . Since both these numbers have exactly $3001$
digits, the first three digits of $x$ must be ${\boxed{100}}$.

How do you understand that $1000^{1000}$ has $3001$ digits?
Let $x=1000^{1000}$
$\quad \Rightarrow 1+\log _{10} x=1+\log _{10} 1000^{1000}+1=1+1000 \log _{10} 1000=1+1000 \log _{10} 10^{3}=3001$
The number of digits of $x=\lfloor 3001\rfloor= 3001$
This post has been edited 1 time. Last edited by gasgous, Jul 4, 2019, 12:33 PM
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Chimphechunu
233 posts
#16 • 1 Y
Y by Adventure10
Plz explain me what is right most digits of the number?It means right most three digits of the number obtained after summing the sequence?or it mean three digits of 1000^1000??
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dragonrider1
461 posts
#17 • 2 Y
Y by Chimphechunu, Adventure10
It means the rightmost numbers after we sum it up.
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TRex_1
2715 posts
#18 • 2 Y
Y by Adventure10, Mango247
fungarwai wrote:
for module 125

conclusion of CRT

How did you do conclusion of CRT section? I'm confused how you converted it to a matrix, and the only values I recognize from the beginning are $125$ and $8$ from the mods.
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LTE
233 posts
#19 • 2 Y
Y by Adventure10, Mango247
so is this problem just intended to a massive bash?

the title says "spin off an old USAMTS," so I assuming there is a nice way to do this (my logic is that if a user bothered to post a spin off of a problem, there is some interesting thing about it).
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JustKeepRunning
2958 posts
#20 • 2 Y
Y by Adventure10, Mango247
LTE wrote:
so is this problem just intended to a massive bash?

the title says "spin off an old USAMTS," so I assuming there is a nice way to do this (my logic is that if a user bothered to post a spin off of a problem, there is some interesting thing about it).

What is a "spin off" of a problem?
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Plops
946 posts
#21 • 1 Y
Y by Adventure10
Before more people try and do this problem (and also because I now feel guilty), I posted this problem just to see if it was possible to find a nice way to do this....but now, I don't think so. It is easy to find it mod $8$, but finding it mod $125$ is more difficult. It was mostly a question which I did not know the answer to and was hoping someone else could solve.

Spin off: the original problem wanted us to find the leftmost digits, which is really simple. I misread and when doing it, tried to find the rightmost digits. Not succeeding, I posted this to see if there was a nice way of doing this.
This post has been edited 1 time. Last edited by Plops, Jul 4, 2019, 8:09 PM
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TRex_1
2715 posts
#22 • 1 Y
Y by Adventure10
Sorry for asking again, even though it is a massive bash, can someone please explain the conclusion of CRT part of fungarwai's solution?
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LTE
233 posts
#23 • 1 Y
Y by Adventure10
TRex_1 wrote:
Sorry for asking again, even though it is a massive bash, can someone please explain the conclusion of CRT part of fungarwai's solution?

I couldn't explain his solution (I know 0 about non-trivial matrices), but I just want to point out that you don't need to use matrices to finish the CRT. You just need to find a number between 0 and 1000 that is 75(mod 125) and 4 (mod 8), which is pretty easy to find by going by 125s.
This post has been edited 1 time. Last edited by LTE, Jul 4, 2019, 8:33 PM
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Plops
946 posts
#24 • 2 Y
Y by Adventure10, Mango247
In fungarwai's solution, $p(k)=r^2+5(k-1)(5k-5+r)r+25(k-1)^2\binom{r}{2}=r^2+5r(k-1)+25(k-1)^2\left(1+\binom{r}{2}\right)$.

Doesn't $p(k)=r^2+5(k-1)(5k-5+r)r+25(k-1)^2=r^2+25(k-1)^2(r)+5r(k-1)r+25(k-1)^2 \binom{r}{2}$ though, which does not equal $r^2+5(k-1)(5k-5+r)r+25(k-1)^2\binom{r}{2}=r^2+5r(k-1)+25(k-1)^2\left(1+\binom{r}{2}\right)$.

Can someone point out anything I am missing.
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fungarwai
859 posts
#25 • 2 Y
Y by Adventure10, Mango247
TRex_1 wrote:
fungarwai wrote:
for module 125

conclusion of CRT

How did you do conclusion of CRT section? I'm confused how you converted it to a matrix, and the only values I recognize from the beginning are $125$ and $8$ from the mods.

Application of elementary operations in Euclidean algorithm
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fungarwai
859 posts
#26 • 1 Y
Y by Adventure10
Plops wrote:
In fungarwai's solution, $p(k)=r^2+5(k-1)(5k-5+r)r+25(k-1)^2\binom{r}{2}=r^2+5r(k-1)+25(k-1)^2\left(1+\binom{r}{2}\right)$.

Doesn't $p(k)=r^2+5(k-1)(5k-5+r)r+25(k-1)^2=r^2+25(k-1)^2(r)+5r(k-1)r+25(k-1)^2 \binom{r}{2}$ though, which does not equal $r^2+5(k-1)(5k-5+r)r+25(k-1)^2\binom{r}{2}=r^2+5r(k-1)+25(k-1)^2\left(1+\binom{r}{2}\right)$.

Can someone point out anything I am missing.

Thanks for pointing out.
I didn't notice that as the mistakes offset finally. I've corrected #10.
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fungarwai
859 posts
#27 • 1 Y
Y by Adventure10
There is a coincidence $\sum_{k=1}^{200} (5k-5+r)^{5k-5+r}\equiv 0 \pmod{5^3}$ when $r\neq 1$

I think the general case should be like that
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fungarwai
859 posts
#28 • 1 Y
Y by Adventure10
$p^{s-1}|N\Rightarrow \sum_{k=1}^N (pk-p+1)^{pk-p+1}
\equiv (1+\frac{p}{2})N\equiv 
\begin{cases} 0 \pmod{p^s} & p=2\\
N\pmod{p^s} & p\neq 2\end{cases}$

Proof

With $\varphi(p^s)=p^{s-1}(p-1)|N \Rightarrow \sum_{k=1}^N (pk-p+r)^{pk-p+r}
\equiv 0 \pmod{p^s} (2\le r\le p-1)$ from above

$4|500, \sum_{k=1}^{1000} k^k\equiv 
\sum_{r=1}^2 \sum_{k=1}^{500} (2k-2+r)^{2k-2+r}
\equiv 0+\sum_{k=1}^{500} (2k)^{2k}\equiv 4\pmod{8}$

$25|200, \sum_{k=1}^{1000} k^k\equiv 
\sum_{r=1}^5 \sum_{k=1}^{200} (5k-5+r)^{5k-5+r} \equiv 200+0+0\equiv 75\pmod{125}$

program check
This post has been edited 3 times. Last edited by fungarwai, Nov 25, 2020, 12:46 PM
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Mixer_V
451 posts
#29 • 1 Y
Y by Adventure10
fungarwai wrote:
for module 125

conclusion of CRT

Nice solution.
Is there an easier one tho?
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Joker7902
6 posts
#30 • 1 Y
Y by Adventure10
JustKeepRunning wrote:
aops5234 wrote:
Let $x=1^1+2^2+3^3+...+999^{999}+1000^{1000}.$ It is evident that
$1000^{1000} < x < 1000^{1} + 1000^{2} + 1000^{3} + … + 1000^{999} + 1000^{1000} $
or that $x$ is between $1000000... $ and $100100100...$ . Since both these numbers have exactly $3001$
digits, the first three digits of $x$ must be ${\boxed{100}}$.

How do you understand that $1000^{1000}$ has $3001$ digits?

take log base 10. It's trivial
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