Solved ProjectEuler/038,104
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ProjectEuler/038/desc.yml
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ProjectEuler/038/desc.yml
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title: What is the largest 1 to 9 pandigital that can be formed by multiplying a fixed number by 1, 2, 3, ... ?
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url: http://projecteuler.net/problem=38
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desc: |
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Take the number 192 and multiply it by each of 1, 2, and 3:
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192 x 1 = 192
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192 x 2 = 384
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192 x 3 = 576
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By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
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The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
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What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n 1?
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solution: |
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Bruteforce
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solutions:
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solve.php:
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desc: Basic solution
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language: php
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ProjectEuler/038/solve.php
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ProjectEuler/038/solve.php
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<?php
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function pandigital($number) {
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$array = count_chars($number,1);
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ksort($array);
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if($array == array(49=>1,50=>1,51=>1,52=>1,53=>1,54=>1,55=>1,56=>1,57=>1)) { return true;} else { return false; }
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}
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while(true) {
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$value++;
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$test = array(1,2);
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$key = 2;
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while(true) {
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$new = array();
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foreach($test as $p) {
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$new[] = $value * $p;
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}
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$var = implode('',$new);
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if(strlen($var) != 9) { break; }
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if(strlen($var) == 9 AND pandigital($var)) { $good[] = $var; break; }
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$test[] = $key++;
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}
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if($value > 9999) { break; }
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}
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rsort($good);
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echo $good[0];
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ProjectEuler/104/desc.yml
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ProjectEuler/104/desc.yml
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title: Finding Fibonacci numbers for which the first and last nine digits are pandigital.
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url: http://projecteuler.net/problem=104
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desc: |
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The Fibonacci sequence is defined by the recurrence relation:
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Fn = Fn1 + Fn2, where F1 = 1 and F2 = 1.
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It turns out that F541, which contains 113 digits, is the first Fibonacci number for which the last nine digits are 1-9 pandigital (contain all the digits 1 to 9, but not necessarily in order). And F2749, which contains 575 digits, is the first Fibonacci number for which the first nine digits are 1-9 pandigital.
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Given that Fk is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.
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solution: |
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Bruteforce
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solutions:
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solve.php:
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desc: Bruteforce - Sooooo slow
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language: php
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ProjectEuler/104/solve.php
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ProjectEuler/104/solve.php
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<?php
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function pandigital($num) {
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for ($i = 1; $i <= 9; $i++) {
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if(strpos($n,(string)$i) === false) {
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return false;
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}
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}
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return true;
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}
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$var = 1;
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$k = 1;
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while(true) {
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$k++;
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$new = bcadd($var,$prev);
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$prev = $var;
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$var = $new;
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if(pandigital(substr($var,-9)) AND pandigital(substr($var,0,9))) { echo $k; die; }
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}
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