def main(limit):
count = 2
fibs = [0, 1]
while count < limit:
val = sum(fibs) % (10**9)
fibs[0] = fibs[1]
fibs[1] = val
val_str = str(val)
if is_pandigital(val_str):
if is_pandigital(str(top9_digits(count))):
print count
count += 1
#!/usr/bin/env python
import time
from Euler import is_pandigital
t_0 = time.clock()
for i in xrange(9876,1,-1):
n = i*100000 + i*2
if is_pandigital(n):
print n
break
print time.clock() - t_0, 'seconds to complete'
#!/usr/bin/python -Wall
# -*- coding: utf-8 -*-
"""
<div id="content">
<div style="text-align:center;" class="print"><img src="images/print_page_logo.png" alt="projecteuler.net" style="border:none;" /></div>
<h2>Pandigital multiples</h2><div id="problem_info" class="info"><h3>Problem 38</h3><span>Published on Friday, 28th February 2003, 06:00 pm; Solved by 37585; Difficulty rating: 5%</span></div>
<div class="problem_content" role="problem">
<p>Take the number 192 and multiply it by each of 1, 2, and 3:</p>
<blockquote>192 × 1 = 192<br />
192 × 2 = 384<br />
192 × 3 = 576</blockquote>
<p>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)</p>
<p>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).</p>
<p>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, ... , <var>n</var>) where <var>n</var> > 1?</p>
</div><br />
<br /></div>
"""
from Euler import is_pandigital
for i in range(9999, 9000, -1):
a1 = i * 1
a2 = i * 2
buf ="%d%d" % (a1, a2)
if (is_pandigital(int(buf))):
print buf
from Euler import is_pandigital
def top_digits(n): # Gets last 9 digits.
t = n * 0.20898764024997873 + (-0.3494850021680094)
t = int((pow(10, t - int(t) + 8)))
return t
fk, f0, f1 = 2, 1, 1
while not is_pandigital(f1) or not is_pandigital(top_digits(fk)):
f0, f1 = f1, (f1 + f0) % 10**9
fk += 1
print fk
#!/usr/bin/python -Wall
# -*- coding: utf-8 -*-
"""
<div id="content">
<div style="text-align:center;" class="print"><img src="images/print_page_logo.png" alt="projecteuler.net" style="border:none;" /></div>
<h2>Pandigital multiples</h2><div id="problem_info" class="info"><h3>Problem 38</h3><span>Published on Friday, 28th February 2003, 06:00 pm; Solved by 37585; Difficulty rating: 5%</span></div>
<div class="problem_content" role="problem">
<p>Take the number 192 and multiply it by each of 1, 2, and 3:</p>
<blockquote>192 × 1 = 192<br />
192 × 2 = 384<br />
192 × 3 = 576</blockquote>
<p>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)</p>
<p>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).</p>
<p>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, ... , <var>n</var>) where <var>n</var> > 1?</p>
</div><br />
<br /></div>
"""
from Euler import is_pandigital
for i in range(9999, 9000, -1):
a1 = i * 1
a2 = i * 2
buf = "%d%d" % (a1, a2)
if (is_pandigital(int(buf))):
print buf
from Euler import is_pandigital
p = set()
for i in range(2,100):
start = 1234
if i>9: start = 123
for j in range(start, 10000/i+1):
if is_pandigital(str(i)+str(j)+str(i*j)): p.add(i*j)
print "Answer to PE32 = ",sum(p)
#!/usr/bin/env python
import time
from Euler import is_pandigital
t_0 = time.clock()
for i in xrange(9876, 1, -1):
n = i * 100000 + i * 2
if is_pandigital(n):
print n
break
print time.clock() - t_0, 'seconds to complete'
