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[SC2B Key Contest]Math challenge

Blogs > yh8c4
Post a Reply
Normal
yh8c4
Profile Blog Joined July 2009
108 Posts
Last Edited: 2010-05-02 01:01:04
May 01 2010 22:18 GMT
#1
contest is over, winner is hamster1800

Original problem was taken from here: http://projecteuler.net/index.php?section=problems&id=290

Solution:

+ Show Spoiler +


ok, the key is yours:

xxxxxx-xxxx-xxxxxx-xxxx-xxxxxx

have fun

-----------------------------------------
Original Message:
Sure thing
9090754242258600
1710153888867000

-----------------------------------------
Original Message:
could you give me 2 examples for x for which ds(x) = ds(x * 137)?

-----------------------------------------
Original Message:
Answer =
20444710234716473

Solution is a dynamic programming algorithm. State is based on the last k digits of the number, lazily computing the digit sum of 137*x digit by digit, storing the next 4 digits to the left of the last k digits (using the fact that 9*137 + 99 < 10000). I iterate through the values of k from 0 to 17, increasing it by one each time (so that at the end we have the values for 18), computing the number of numbers x for which the next 4 digits to the left are "a" and the difference between ds(x) and ds(137*x) (up to the last k digits) is "b".

Code below:

#include<cstdio>
#include<cstdlib>

using namespace std;

int main() {
long long **now, **next;
now = new long long*[10000];
next = new long long*[10000];
for(int i = 0; i<10000; i++) {
now[i] = new long long[401];
next[i] = new long long[401];
now[i] += 200;
next[i] += 200;
for(int j = -200; j<=200; j++) {
now[i][j] = next[i][j] = 0;
}
}
now[0][0] = 1;
for(int i = 0; i<18; i++) {
for(int j = 0; j<10000; j++) {
for(int k = -200; k<=200; k++) {
next[j][k] = 0;
}
}
for(int a = 0; a<10000; a++) {
for(int b = -162; b<=162; b++) {
for(int d = 0; d<=9; d++) {
int na = a/10 + 137*d, nb = b - a%10 + d;
next[na][nb] += now[a][b];
}
}
}
long long **t = now;
now = next;
next = t;
}
long long ans = 0;
for(int i = 0; i<10000; i++) {
for(int j = -200; j<=200; j++) {
int a = i, b = j;
while(a != 0) {
b -= a%10;
a/=10;
}
if(b == 0) {
ans += now[i][j];
}
}
}
printf("%lld\n",ans);
}




----------------------


I dont know whether there is still a high demand for beta keys, but I received another friend invite key, which you can win by solving the following challenge:

Let x be an integer equal to or greater than 0 and smaller than 1000000000000000000 (10^18).

Let ds(x) be the digit sum of x, i.e. ds(123) = 1 + 2 + 3 = 6

For how many distinct x is ds(x) = ds(137 * x)?

The first one to pm me a) the correct answer and b) a little explanation (e.g. source code) how to find the solution will receive a beta key.

Once I receive a solution, I'll reply whether it's correct or not. If it's not correct you can try again.

Good luck!

edit 1: The solution is NOT 1!
edit 2: ds(solution) = 59


zealing
Profile Blog Joined January 2009
Canada806 Posts
May 01 2010 22:21 GMT
#2
can't you just give me the key? i skiped school to play BW so i don't know math, could i just beat you 1v1 Destination BW for the key? its all i got
Think you got lag? It took Jesus 3 days to respawn.
yh8c4
Profile Blog Joined July 2009
108 Posts
May 01 2010 22:25 GMT
#3
On May 02 2010 07:21 zealing wrote:
can't you just give me the key? i skiped school to play BW so i don't know math, could i just beat you 1v1 Destination BW for the key? its all i got


unfortunately, beating me in bw is anything but a challenge and thus I can't reward you with a beta key for that
Sadistx
Profile Blog Joined February 2009
Zimbabwe5568 Posts
May 01 2010 22:27 GMT
#4
Bah, I wish a java SDE lying around, i could tell you in less than a minute :/

This is so easy if you have any programming thing lying around.
Lightwip
Profile Blog Joined April 2010
United States5497 Posts
May 01 2010 22:28 GMT
#5
I might try to write a program to do this.
If you are not Bisu, chances are I hate you.
Shrine
Profile Blog Joined April 2010
Australia107 Posts
May 01 2010 22:37 GMT
#6
11
Hell is empty, All the devils are here.
yh8c4
Profile Blog Joined July 2009
108 Posts
May 01 2010 22:37 GMT
#7
yes, writing a program is a good idea, but be aware that 10^18 is really big
Lightwip
Profile Blog Joined April 2010
United States5497 Posts
May 01 2010 22:48 GMT
#8
Yeah, I'm not sure that would even fit in a long int.
If you are not Bisu, chances are I hate you.
Sharkified
Profile Joined January 2009
Canada254 Posts
May 01 2010 22:52 GMT
#9
On May 02 2010 07:27 Sadistx wrote:
Bah, I wish a java SDE lying around, i could tell you in less than a minute :/

This is so easy if you have any programming thing lying around.


Nah it takes forever to do it, would have to find a complicated way of doing it, then rather just learn maths.
yh8c4
Profile Blog Joined July 2009
108 Posts
May 01 2010 22:56 GMT
#10
On May 02 2010 07:37 Shrine wrote:
11


wrong
JonnyWhy
Profile Blog Joined May 2010
Canada94 Posts
May 01 2010 22:57 GMT
#11
43?
Say something juicy.
Sharkified
Profile Joined January 2009
Canada254 Posts
May 01 2010 23:00 GMT
#12
Is it a one try thing ?
JonnyWhy
Profile Blog Joined May 2010
Canada94 Posts
May 01 2010 23:03 GMT
#13
On May 02 2010 08:00 Sharkified wrote:
Is it a one try thing ?


On May 02 2010 07:18 yh8c4 wrote:
Once I receive a solution, I'll reply whether it's correct or not. If it's not correct you can try again.


Say something juicy.
yh8c4
Profile Blog Joined July 2009
108 Posts
May 01 2010 23:04 GMT
#14
no, this is not limited to 1 try, but stop just posting or pming a single number. to receive the key you must have the correct answer AND an explanation why your answer is correct.
yh8c4
Profile Blog Joined July 2009
108 Posts
Last Edited: 2010-05-01 23:12:07
May 01 2010 23:06 GMT
#15
here's a little hint:

let cs be the correct solution. ds(cs) = 59.
Synthesis
Profile Joined August 2008
Canada167 Posts
May 01 2010 23:08 GMT
#16
42.

Source
Sadistx
Profile Blog Joined February 2009
Zimbabwe5568 Posts
May 01 2010 23:09 GMT
#17
On May 02 2010 08:08 Synthesis wrote:
42.

Source


That joke got old the first 1000 times people used it.
Nehsb
Profile Joined May 2009
United States380 Posts
May 01 2010 23:09 GMT
#18
Hmmm, 137 | 10001, 99999999, and 10^12 + 1.

Might be useful?

Might be able to use that to cut down the numbers you have to search...
TwilightStar
Profile Blog Joined August 2009
United States649 Posts
May 01 2010 23:19 GMT
#19
3.76
(5)Twilight Star.scx --------- AdmiralHoth: There was one week when I didn't shave for a month.
Sadistx
Profile Blog Joined February 2009
Zimbabwe5568 Posts
May 01 2010 23:24 GMT
#20
From what I understand, it will be a pretty big number, around 29000000000000000.

I have no means of actually solving this without getting a programming language that handles huge numbers and writing code for it.

I'm sure one of the math majors on here has a better way than brute forcing it.
Nemesis
Profile Blog Joined May 2009
Canada2568 Posts
May 01 2010 23:26 GMT
#21
the answer is 59!
Lee Young Ho fighting! KT P are just CHINTOSSTIC.
yh8c4
Profile Blog Joined July 2009
108 Posts
May 01 2010 23:27 GMT
#22
On May 02 2010 08:24 Sadistx wrote:
From what I understand, it will be a pretty big number, around 29000000000000000.

I have no means of actually solving this without getting a programming language that handles huge numbers and writing code for it.

I'm sure one of the math majors on here has a better way than brute forcing it.


that's a good start, the correct solution has the same number of digits
Sadistx
Profile Blog Joined February 2009
Zimbabwe5568 Posts
May 01 2010 23:28 GMT
#23
On May 02 2010 08:26 Nemesis wrote:
the answer is 59!


59! gives you a number that's more than 10^18 so its obviously wrong.
ghrur
Profile Blog Joined May 2009
United States3786 Posts
May 01 2010 23:30 GMT
#24
23473423423423427
:D
darkness overpowering
Nehsb
Profile Joined May 2009
United States380 Posts
Last Edited: 2010-05-01 23:50:19
May 01 2010 23:47 GMT
#25
i think you can solve it by splitting it up into two overlapping pieces, an 10^11 size piece and a 10^10 size piece, write arrays for "certain values", and then calculate the number that sum to 0....

That shouldn't take too long...

edit: I forgot my comp can't store int arr[10000000000]

Does anyone know how to store that?

edit2: nvm you only need to store long long arr[199];
jannis
Profile Joined April 2010
Netherlands38 Posts
May 02 2010 00:01 GMT
#26
just to be sure: do you mean the digital sum or the reduced digital sum (that is, applying the digital sum until the number is in the range 0-9)?
hello 8]
Thratur
Profile Blog Joined June 2008
Canada917 Posts
May 02 2010 00:02 GMT
#27
My source code is running but is taking quite some time
Nehsb
Profile Joined May 2009
United States380 Posts
May 02 2010 00:03 GMT
#28
On May 02 2010 09:01 jannis wrote:
just to be sure: do you mean the digital sum or the reduced digital sum (that is, applying the digital sum until the number is in the range 0-9)?


it says ds(solution) = 59, so I'm pretty sure it's digit sum...
yh8c4
Profile Blog Joined July 2009
108 Posts
May 02 2010 00:05 GMT
#29
On May 02 2010 09:01 jannis wrote:
just to be sure: do you mean the digital sum or the reduced digital sum (that is, applying the digital sum until the number is in the range 0-9)?


ds(999) = 27
Nehsb
Profile Joined May 2009
United States380 Posts
May 02 2010 00:11 GMT
#30
k I finished writing the program, and it should run in time, but it crashes...

do you accept it if we send you the code, without the answer, but with a description of how we tried to find the answer and why it should run in < a minute?
Hamster1800
Profile Blog Joined August 2008
United States175 Posts
May 02 2010 00:18 GMT
#31
I PMed my answer and a description. Code runs in 1.128s on my box
D is for Diamond, E is for Everything Else
Nehsb
Profile Joined May 2009
United States380 Posts
May 02 2010 00:20 GMT
#32
On May 02 2010 09:18 Hamster1800 wrote:
I PMed my answer and a description. Code runs in 1.128s on my box




What was your code btw?

Mine:

#include<iostream>
#include<stdlib.h>

using namespace std;

int digitSum(long long num);

int main(){
long long arr1[301][137];
long long arr2[301][137];
for(int i = 0; i < 301; i++){
for(int j = 0; j < 137; j++){
arr1[i][j] = arr2[i][j] = 0;
}
}
for(long long i = 0; i < 100000000; i++){
long long num = digitSum((137 * i) % 1000000000) - digitSum(i);
long long num2 = (137 * i) / 1000000000;
arr1[150+num][num2]++;
}
for(long long i = 0; i < 1000000000; i++){
for(int j = 0; j < 137; j++){
int num = digitSum(137 * i + j) - digitSum(i);
arr2[150+num][j]++;
}
}
long long answer = 0;
for(int i = 0; i <= 1000; i++){
for(int j = 0; j < 137; j++){
answer += arr1[i][j]*arr2[300 - i][j];
}
}
cout << answer << endl;
return 0;
}

int digitSum(long long num){
if(num == 0)return 0;
return (num%10) + digitSum(num/10);
}
yh8c4
Profile Blog Joined July 2009
108 Posts
May 02 2010 00:22 GMT
#33
On May 02 2010 09:11 Nehsb wrote:
k I finished writing the program, and it should run in time, but it crashes...

do you accept it if we send you the code, without the answer, but with a description of how we tried to find the answer and why it should run in < a minute?


why do you think it should run in < 1 minute?
Sharkified
Profile Joined January 2009
Canada254 Posts
May 02 2010 00:23 GMT
#34
On May 02 2010 09:18 Hamster1800 wrote:
I PMed my answer and a description. Code runs in 1.128s on my box


I doubt it's your box, you either have an amazing or amazingly bad code.
I have an i7-860 and it takes forever to run my code.
Nehsb
Profile Joined May 2009
United States380 Posts
Last Edited: 2010-05-02 00:26:50
May 02 2010 00:25 GMT
#35
On May 02 2010 09:22 yh8c4 wrote:
Show nested quote +
On May 02 2010 09:11 Nehsb wrote:
k I finished writing the program, and it should run in time, but it crashes...

do you accept it if we send you the code, without the answer, but with a description of how we tried to find the answer and why it should run in < a minute?


why do you think it should run in < 1 minute?


NVM, sorry for some reason I thought my comp ran 200 billion operations per second.

my code probably is more efficient if you replace the arr1 thing with 10^10 and the arr2 thing witih 10^8

edit: that should be like.... (10^10 + 10^8 * 137) * a small number about efficiency?,
Thratur
Profile Blog Joined June 2008
Canada917 Posts
May 02 2010 00:25 GMT
#36
I calculated mine would take appromatively 1000 hours lol

Need to find a way to optimize this...
Hamster1800
Profile Blog Joined August 2008
United States175 Posts
May 02 2010 00:36 GMT
#37
Okay, I got the key, so I'll just put the solution up for people who want to see
+ Show Spoiler +

Answer =
20444710234716473

Solution is a dynamic programming algorithm. State is based on the last k digits of the number, lazily computing the digit sum of 137*x digit by digit, storing the next 4 digits to the left of the last k digits (using the fact that 9*137 + 99 < 10000). I iterate through the values of k from 0 to 17, increasing it by one each time (so that at the end we have the values for 18), computing the number of numbers x for which the next 4 digits to the left are "a" and the difference between ds(x) and ds(137*x) (up to the last k digits) is "b".

Code below:
#include<cstdio>
#include<cstdlib>

using namespace std;

int main() {
long long **now, **next;
now = new long long*[10000];
next = new long long*[10000];
for(int i = 0; i<10000; i++) {
now[i] = new long long[401];
next[i] = new long long[401];
now[i] += 200;
next[i] += 200;
for(int j = -200; j<=200; j++) {
now[i][j] = next[i][j] = 0;
}
}
now[0][0] = 1;
for(int i = 0; i<18; i++) {
for(int j = 0; j<10000; j++) {
for(int k = -200; k<=200; k++) {
next[j][k] = 0;
}
}
for(int a = 0; a<10000; a++) {
for(int b = -162; b<=162; b++) {
for(int d = 0; d<=9; d++) {
int na = a/10 + 137*d, nb = b - a%10 + d;
next[na][nb] += now[a][b];
}
}
}
long long **t = now;
now = next;
next = t;
}
long long ans = 0;
for(int i = 0; i<10000; i++) {
for(int j = -200; j<=200; j++) {
int a = i, b = j;
while(a != 0) {
b -= a%10;
a/=10;
}
if(b == 0) {
ans += now[i][j];
}
}
}
printf("%lld\n",ans);
}
D is for Diamond, E is for Everything Else
yh8c4
Profile Blog Joined July 2009
108 Posts
May 02 2010 00:37 GMT
#38
Hamster1800 is the winner, his solution:


ok, the key is yours:

xxxxxx-xxxx-xxxxxx-xxxx-xxxxxx

have fun

-----------------------------------------
Original Message:
Sure thing
9090754242258600
1710153888867000

-----------------------------------------
Original Message:
could you give me 2 examples for x for which ds(x) = ds(x * 137)?

-----------------------------------------
Original Message:
Answer =
20444710234716473

Solution is a dynamic programming algorithm. State is based on the last k digits of the number, lazily computing the digit sum of 137*x digit by digit, storing the next 4 digits to the left of the last k digits (using the fact that 9*137 + 99 < 10000). I iterate through the values of k from 0 to 17, increasing it by one each time (so that at the end we have the values for 18), computing the number of numbers x for which the next 4 digits to the left are "a" and the difference between ds(x) and ds(137*x) (up to the last k digits) is "b".

Code below:

#include<cstdio>
#include<cstdlib>

using namespace std;

int main() {
long long **now, **next;
now = new long long*[10000];
next = new long long*[10000];
for(int i = 0; i<10000; i++) {
now[i] = new long long[401];
next[i] = new long long[401];
now[i] += 200;
next[i] += 200;
for(int j = -200; j<=200; j++) {
now[i][j] = next[i][j] = 0;
}
}
now[0][0] = 1;
for(int i = 0; i<18; i++) {
for(int j = 0; j<10000; j++) {
for(int k = -200; k<=200; k++) {
next[j][k] = 0;
}
}
for(int a = 0; a<10000; a++) {
for(int b = -162; b<=162; b++) {
for(int d = 0; d<=9; d++) {
int na = a/10 + 137*d, nb = b - a%10 + d;
next[na][nb] += now[a][b];
}
}
}
long long **t = now;
now = next;
next = t;
}
long long ans = 0;
for(int i = 0; i<10000; i++) {
for(int j = -200; j<=200; j++) {
int a = i, b = j;
while(a != 0) {
b -= a%10;
a/=10;
}
if(b == 0) {
ans += now[i][j];
}
}
}
printf("%lld\n",ans);
}

Sharkified
Profile Joined January 2009
Canada254 Posts
May 02 2010 00:38 GMT
#39
Congratz, you sir are a better programmer than I am.
eddoo
Profile Joined March 2010
30 Posts
Last Edited: 2010-05-02 00:59:43
May 02 2010 00:56 GMT
#40
Argh, I managed to hand count that number using combinatorics and on the verge of victory refresh to see a dynamic algorithm win.

edit: I used the ds(x) = 59 hint though. Good job hamster
yh8c4
Profile Blog Joined July 2009
108 Posts
May 02 2010 01:00 GMT
#41
by the way, i took this problem from here: http://projecteuler.net/index.php?section=problems&id=290
jannis
Profile Joined April 2010
Netherlands38 Posts
May 02 2010 01:06 GMT
#42
congratulations, although the method is still not so clear to me
hello 8]
flamewheel
Profile Blog Joined December 2009
FREEAGLELAND26781 Posts
May 02 2010 01:07 GMT
#43
LOL HAMSTER.
(I know that dude)
Grats~
Writerdamn, i was two days from retirement
topspinserve
Profile Blog Joined October 2008
United States147 Posts
May 02 2010 02:09 GMT
#44
Hamster Math/CS imba ><
Normal
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