Anong pinakamahirap na problem sa math ang na-solve ninyo?

[eveready battery]

Durog sa application ng D.E.

musta na pala engr. maximus,pasado ba sa board exam ngayon?

[xperryment]

ahmmmm so far for me mathematical model of a 2 wheel balancing robotA(inverted pendulum)

[treacher88]

di pa kasi tumatanda si math kaya sa atin pinapasolve problems nia 

[a.out (drako65)]

nag try ako sa christmas yr2ooo challenge ni queena lee, "eureka" columnist ng phil daily inq. dun lang na reproduce ng 100,000X(?) ang pangalan ko nationwide(?). hint: ang ginamit/ pinadala kong answer sheet lumang dyaryo with barbed/jagged edge. pinunit ko lang just to make an impression. hindi laway pinandikit ko sa sobre- kanin, pinalamanan ko pa ng kanin yung contact edge ng sobre. before pa to ng internet na 24hrs. kasi ng mga sumunod na christmas challenge niya w/c appears on the sat issue of the PDI, may internet na dyan sa metro na 24hrs, yung mga ng submit- 15-30 mins after publication. wala ng laban mga tiga probinsiya.

astig ka! nabigyan nya din ako ng book nya may dedication at signature nya, collector's item Cute pa sya noon, he he kaya inspired akong magsolve

[pabzjose]

pan po ito e solve?

(2^345)/29. find the remainder?

[motion55]

pan po ito e solve?

(2^345)/29. find the remainder?

2^345 sa binary is just 1 followed by 345 zeroes. Brute force long division kaya naman.

Wala lang but the answer is 21.

(2^345)/29 = 17186261501689712417186261501689712417186261501689712417186261501689712417186261501689712417186261501689712417 remainder 21.

Actually a short C language program did it in a flash.

Code:

#include <stdio.h>

#define DIVISOR 29

typedef unsigned char INT8;

void main ()
{
INT8 INT345[44];
INT8 OUT345[44];

int i, input;

input = 0;

// initialize 2^345
for (i=0; i<44; i++)
{
INT345[i] = 0;
OUT345[i] = 0;
}

INT345[43] = 2; // 1 followed by 345 zeroes

for (i=43; i>=0; i--)
{
input = (input<<8) + INT345[i];
OUT345[i] = input/DIVISOR;
input = input%DIVISOR;  //Save the remainder for carry.
}

printf("\n");
for (i=43; i>=0; i--)
{
printf("%i", OUT345[i]);
}
printf("\n Remainder is %i", input);
}

[motion55]

Oops I rechecked my code and the correct answer is 19.

(2^345)/29 =

11A7B9611A7B9611A7B9611A7B9611A7B9611A7B9611A7B9611A7B9611A7B9611A7B9611A7B9611A7B9611 (hex)

remainder is 19.

Code:

#include <stdio.h>

#define DIVISOR 29

typedef unsigned char INT8;

void main ()
{
INT8 INT345[44];
INT8 OUT345[44];

int i, input;

input = 0;

// initialize
for (i=0; i<44; i++)
{
INT345[i] = 0;
OUT345[i] = 0;
}

INT345[43] = 2; // 2 followed by 86 zeroes in hexadecimal

for (i=43; i>=0; i--)
{
input = (input<<8) + INT345[i];
OUT345[i] = input/DIVISOR;
input = input%DIVISOR;
}

printf("\n");
for (i=43; i>=0; i--)
{
printf("%X", OUT345[i]);
}
printf("\n Remainder is %i", input);
}

[a.out (drako65)]

> echo '2^345%29' | bc

[zrehtiek/keitherz]

^ walang thrill.
galing talaga ni si motion55. gagamit pa sana ako ng BigInteger library ng java e. +1!
di ko naisip mag mano mano ng long division.

[a.out (drako65)]

sige mano manohin mo, marami pa akong gagawin, he he

[zrehtiek/keitherz]

^ ginawa na nga ni sir motion55.

[a.out (drako65)]

Try this in C (if you can)
1. Calculate pi to 5000 decimal places.
2. Calculate 6^6^6. (fyi: it is the number of parallel universes in Heinlein's "The Number of the Beast")