Nothing much new here. All you really have to know is how indexes work. Here’s the Python code:

the_num = '''73167176531330624919225119674426574742355349194934
 96983520312774506326239578318016984801869478851843
 85861560789112949495459501737958331952853208805511
 12540698747158523863050715693290963295227443043557
 66896648950445244523161731856403098711121722383113
 62229893423380308135336276614282806444486645238749
 30358907296290491560440772390713810515859307960866
 70172427121883998797908792274921901699720888093776
 65727333001053367881220235421809751254540594752243
 52584907711670556013604839586446706324415722155397
 53697817977846174064955149290862569321978468622482
 83972241375657056057490261407972968652414535100474
 82166370484403199890008895243450658541227588666881
 16427171479924442928230863465674813919123162824586
 17866458359124566529476545682848912883142607690042
 24219022671055626321111109370544217506941658960408
 07198403850962455444362981230987879927244284909188
 84580156166097919133875499200524063689912560717606
 05886116467109405077541002256983155200055935729725
 71636269561882670428252483600823257530420752963450'''

the_num = the_num.replace('\n ','')  #To make it look like a normal number
largest_product = 0
curdigit = 0

for digit in the_num:
    product = 1

    for x in range(5):
        try:
            product *= int(the_num[curdigit + x])
        except:
            pass
    if product > largest_product: largest_product = product
    curdigit += 1

print largest_product

Answer: 40824
Elapsed time: 0.007169008255 seconds.

All of my solutions have been less than one second (Actually, I think they were all less than a hundredth of a second). How long can I keep this up?

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

One Possible Solution: Go

package main

import "fmt"
import "strconv"

const fileIn = "73167176531330624919225119674426574742355349194934" +
	"96983520312774506326239578318016984801869478851843" +
	"85861560789112949495459501737958331952853208805511" +
	"12540698747158523863050715693290963295227443043557" +
	"66896648950445244523161731856403098711121722383113" +
	"62229893423380308135336276614282806444486645238749" +
	"30358907296290491560440772390713810515859307960866" +
	"70172427121883998797908792274921901699720888093776" +
	"65727333001053367881220235421809751254540594752243" +
	"52584907711670556013604839586446706324415722155397" +
	"53697817977846174064955149290862569321978468622482" +
	"83972241375657056057490261407972968652414535100474" +
	"82166370484403199890008895243450658541227588666881" +
	"16427171479924442928230863465674813919123162824586" +
	"17866458359124566529476545682848912883142607690042" +
	"24219022671055626321111109370544217506941658960408" +
	"07198403850962455444362981230987879927244284909188" +
	"84580156166097919133875499200524063689912560717606" +
	"05886116467109405077541002256983155200055935729725" +
	"71636269561882670428252483600823257530420752963450"

func main() {
	largest := 0
	for i := 0; i < (len(fileIn) - 4); i++ {
		fiveDigits := fileIn[i:(i + 5)]

		s1 := fiveDigits[0:1]
		s2 := fiveDigits[1:2]
		s3 := fiveDigits[2:3]
		s4 := fiveDigits[3:4]
		s5 := fiveDigits[4:5]

		x1, _ := strconv.Atoi(s1)
		x2, _ := strconv.Atoi(s2)
		x3, _ := strconv.Atoi(s3)
		x4, _ := strconv.Atoi(s4)
		x5, _ := strconv.Atoi(s5)

		product := x1 * x2 * x3 * x4 * x5

		if product > largest {
			largest = product
		}

	}
	fmt.Println("Greatest product is: ", largest)
}

Motivation

This is another one of those slightly interesting problems that comes up in my day-to-day life, and I thought I’d share. Roughly the problem shows the very interesting result that the only automorphisms of the extension where is the rational function field in one variable (i.e. the quotient field of the polynomial ring in one variable) are the fractional linear transformations (i.e. that ) where . This gives us a mapping and we shall determine what it’s kernel is, giving us an alternate description of of the automorphism group of this extension.

Automorphisms of

Before we begin we prove a certain lemma concerning the degrees of certain

Lemma: Let be a field and consider the ratioanl function field . Let then with and . Then, .

Proof: We first note that the polynomial is irreducible over . Indeed, by Gauss’s lemma we need merely check that it is irreducible in but and this is clearly irreducible in the latter since it’s linear. Moreover, it’s easy to see that is a root of this polynomial and thus we see that is a minimal polynomial for over and thus

and so the result follows.

Using this we can complete the problem then:

Theorem: Let be any field. Then, is equal to the set of all linear fractional transformations of the form where .

Proof: Let then we know that if (where we can clearly assume that neither nor is constant) then

so that evidently if is going to be surjective we must have that . Thus, we see automatically that our only hope for an automorphism is a linear fractional transformation.

Assume now that are such that . Since is the free commutative -algebra on one generator and is a commutative -algebra we know that we can extend the set map to a -algebra map . If we can show that this map is injective (i.e. has trivial kernel) then by the universal characterization of quotient fields we may extend this to a map . To see that this map is injective assume that

and that for some . This then tells us that is algebraic over . If this is all easy so assume not then note that and since the algebraic elements of in form a field we have that we may subtract, multiply, and divide any algebraic elements to/from . That said, constants are all algebraic and and so we may clearly perform the following operations

to conclude that is algebraic over , but this is clearly ridiculous. Thus we see that all the ‘s must be zero, and thus has trivial kernel. Thus, (by previous comment) we are granted an extension which is clearly a -algebra map and which satisfies . Since this is a ring (field) map on a field we know that must be injective, and since we have (by the lemma) that we must have that is surjective, and thus we may conclude that is bijective. Thus, we then see that .

Now, to see that such a map won’t work if we merely note that if were an element of with then necessarily but then we see that isn’t even well-defined.

Combining the above results tells us that are precisely the fractional linear transformations coming from invertible matrices as desired.

From the above we clearly are able to create a mapping given by . Moreover, from the above (an a little elbow grease) we actually know that is an epimorphism, and so the first isomorphism theorem tells us that . That said it’s fairly easy to see that the matrix determines the corresponding automorphism up to scaling, so that (where we have proven before this last equality, where the stands for the center of a ring) and thus we have that where the stands for the projective linear group.

References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

“Hei, apa yang terjadi?” tanya seseorang di antara kerumunan orang itu.

“Aku dengar ada pembunuhan di sana,” jawab orang yang berada di dekatnya.

Benar, sebuah pembunuhan terjadi di taman rekreasi alam mayang. Petugas polisi tampak sibuk memeriksa korban dan juga tempat kejadiannya. Korban bernama Hadrian, berusia 27 tahun. Penyebab kematiannya adalah pukulan yang keras di kepala bagian belakang. Pembunuh memukul korban secara diam-diam dari belakang, tanpa diketahui oleh korban. Waktu kematiannya adalah antara pukul 14.00 – 14.10. Korban ditemukan tewas oleh salah seorang teman baiknya ketika SMA, Irwan. Irwan lah orang yang menghubungi polisi untuk datang ke tempat kejadian.

Siang itu, korban dan ketiga teman baiknya berencana akan melaksanakan acara reuni. Tiga teman baiknya itu ialah Irwan, Joko, dan Kaka. Acara reuni itu akan dilaksanakan di Alam Mayang, tepatnya di tempat kejadian. Mereka bertiga berjanji akan bertemu di tempat itu pada pukul 14.30.

Korban sedang menggunakan laptop miliknya sebelum dibunuh. Sebuah laptop berada di dekat korban dan dalam keadaan menyala. Pada tombol keyboard laptop tersebut, ada bekas darah korban berbentuk jari, tepatnya pada tombol “Ctrl” dan tombol “C”. Darah ini berasal dari tangan kirinya yang berdarah. Diduga tangan kirinya berdarah karena korban memegangi kepalanya yang berdarah setelah dipukul pembunuh. Tangan kanan korban menggenggam mouse laptop miliknya. Apakah maksud dari semua ini? Benar, ini adalah pesan kematian korban dan pembunuhnya tidak menyadari hal ini.

Tidak ditemukan saksi yang melihat langsung pembunuhan karena lokasi reuni mereka adalah tempat yang sepi dan tidak banyak orang di sekelilingnya. Tersangka adalah ketiga teman baik korban dan ketiganya memiliki motif yang kuat, setidaknya cukup kuat untuk membuatnya membunuh korban. Berikut keterangan dari masing-masing tersangka.

Irwan, 27 tahun

Aku sangat terkejut ketika aku datang ke sini. Aku tiba di sini pada pukul 14.30 sesuai janji kami dan ketika itu pula acara reuni kami hancur. Ketika aku tiba, aku sudah menemukan Hadrian tewas di tempat reuni kami. Aku segera menelepon polisi ketika itu.

Antara pukul 14.00 dan 14.10, aku sedang pergi jalan-jalan ke mall dan aku pergi sendirian.

Aku kesal karena Hadrian pernah menggagalkan pernikahanku. Dia menjelek-jelekkan aku di depan tunanganku dan saat itu pula aku putus hubungan dengan tunanganku. Tetapi, aku sudah memaafkannya dan tidak ingin lagi membahas hal tersebut.

Masing-masing dari kami memiliki julukan Antoni dijuluki dengan Bad Boys karena perilakunya yang kurang baik. Aku sendiri dijuluki Mouse Copy karena wajahku mirip dengan tikus.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

One Possible Solution: Java

import java.io.*;

public class Problem_08 
	{
		  public static void main( String [] args )
		  {
		    try
		    {
		       FileReader fro = new FileReader( "1000.txt" );
		       BufferedReader bro = new BufferedReader( fro );
		       StringBuilder sb = new StringBuilder( );
		  
		       // declare String variable and prime the read
		       String stringfromFile = bro.readLine( );
		 
		       while( stringfromFile != null ) // end of the file
		       {
		          sb.append(stringfromFile); // append into one long string
		          stringfromFile = bro.readLine( );  // read next line
		       }
		       String str = sb.toString( ); // convert from StringBuilder to String
		       String fiveDigits = ""; // initialize fiveDigits
		       int largest = 0; // initialize largest
		       
		       for(int i = 0; i < str.length() - 4; i++){
		    	    fiveDigits = str.substring( i, ( i + 5 )); // substring of str for fiveDigits
		    	    
		    	    // convert each of the five digits into integers
		    	    int v = Integer.parseInt(fiveDigits.substring(0, 1));
		    	    int w = Integer.parseInt(fiveDigits.substring(1, 2));
		    	    int x = Integer.parseInt(fiveDigits.substring(2, 3));
		    	    int y = Integer.parseInt(fiveDigits.substring(3, 4));
		    	    int z = Integer.parseInt(fiveDigits.substring(4, 5));
		    	    
		    	    // find the product of each of the five digits
		    	    int product = v * w * x * y * z;
		    	    
		    	    // keep track of the largest product
		    	    if(product > largest)
		    	    {
		    	    	largest = product;
		    	    }
		       }
		       System.out.println("Largest = " + largest);
		       bro.close( );
		    }
		 
		    catch( FileNotFoundException filenotfoundexxption )
		    {
		      System.out.println( "1000.txt, does not exist" );
		    }
		 
		    catch( IOException ioexception )
		    {
		      ioexception.printStackTrace( );
		    }
		  }
	}

One Possible Solution: Python

# Python version = 2.7.2
# Platform = win32

fjf = open('1000.txt', 'rU')
bcm = fjf.readlines()
hws = ''.join(bcm)
bvc = hws.replace('\n', '')
fjf.close()

largest = 0

for i in range(0, (len(bvc) - 4)):
    fiveDigits = bvc[i:(i + 5)]

    x1 = int(fiveDigits[0:1])
    x2 = int(fiveDigits[1:2])
    x3 = int(fiveDigits[2:3])
    x4 = int(fiveDigits[3:4])
    x5 = int(fiveDigits[4:5])
    
    product = x1 * x2 * x3 * x4 * x5
    
    if product > largest:
        largest = product
        
print "Greatest product is: %s" % largest

Euler Project