# Project Euler in PL/SQL: Problem 8

Another Project Euler problem and solution that utilizes loops.

## Problem 8:

### Largest Product in a Series

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

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

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

The hardest part of this problem was setting up how we will loop through the 13-digits and multiply them, which ended up not being all that difficult. There’s not a built-in function as nice as Ruby’s `collect` or `map` for arrays, so I just kept the data as a string and coupled a SUBSTR with a LOOP to get the job done, which happens to work very efficiently as the solution is found almost instantly.

```
DECLARE
problem_num VARCHAR2(2000) := '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450';
first_digit PLS_INTEGER := 0; -- this counts our position in the string for SUBSTR
checking_digits VARCHAR2(13); --current set of 13 digits to check
digits_product NUMERIC;
BEGIN
WHILE first_digit < LENGTH(problem_num) LOOP
checking_digits := SUBSTR(problem_num, first_digit, 13);
digits_product := 1;
first_digit := first_digit + 1;

--skip to next set of digits if a 0 exists; will always multiply to 0
IF REGEXP_LIKE(checking_digits, '0')
THEN
CONTINUE;
END IF;

FOR i IN 1 .. LENGTH(checking_digits) LOOP
digits_product:= SUBSTR(checking_digits,i,1) * digits_product;
END LOOP;