jakedobkin

#!/usr/local/bin/node

/* the question is a little unclear- b/c a naive reading might assume
you only need to check the five digit numbers, since the example given is 4 digits
this would be a special class of numbers called "narcissistic"- but here we're looking
for a larger case- ANY number of digits that, when ^5 and added, give the number, starting
from two (since the problem says don't count 1).  harder is to figure out upper limit.
at first, i just let it run to 1MM, hoping that would be enough.  it was.  then i read
up and found someone had figured out the exact limit with this logic:
assume each digit is a 9: 9^5~~ 60K-- so every digit we add gives us another 60K or so. 

jakedobkin

#!/usr/local/bin/node

// our basic method here is to start with the largest coin and count down to the smallest
// each time you use up a coin (ie. that loop gets to zero) you're still checking the next
// smallest coin, down to the smallest one.  there is probably a way to write this recursively
// instead of copying out the loop 8 times, but i don't know recursive functions yet.  
// remove the comments if you want an actual enumeration of each set.  

goal = 200;
count = 0; 

jakedobkin

# first we'll need a set to store results in- that means we won't need to de-dupe
h=set()
total=0;

# now, we'lll make the strings- 
# 1 digits x 4 digits will be 4 or 5 digits (9-10 total) and 2 digits x 3 digits will by 4 or 5 digits (9-10 total) so try up to 9999x99
for a in range(1,9999): 
	for b in range(1,99):
		product = str(a)+"x"+str(b)+"x"+str(a*b)

jakedobkin

# first we'll need a set to store results in- that means we won't need to de-dupe
h=set()
total=0;

# now, we'lll make the fractions 10 to 99 in numerator and denominator
for a in range(10,100): 
	for b in range(10,100):
		if (a!=b):
			numerator = str(a)
			denominator = str(b)

jakedobkin

#!/usr/local/bin/node

// these are the values of the factorials of each digit from 0-9- from earlier exercise

factorial = [1,1,2,6,24,120,720,5040,40320,362880]

sum = 0;

// you could use advanced math to figure out the upper bound- i just tried up to 10^7
for (a=3;a<=100000;a++)

jakedobkin

#!/usr/local/bin/node

// first, prep prime seive up to 1MM

n=1000000;
myPrimes = new Array();
for (i=2;i<=n;i++)
	{
	myPrimes[i]=true;
	}

jakedobkin

#!/usr/local/bin/node

function reverse(s){
    return s.split("").reverse().join("");
}

function ispali(a)
	{
	a = a.toString();
	for (i=0;i<=a.length-1;i++)

jakedobkin

#!/usr/local/bin/node

// first, prep prime seive up to 1MM

n=1000000;
myPrimes = new Array();
for (i=2;i<=n;i++)
	{
	myPrimes[i]=true;
	}

jakedobkin

# first we'll need a set to store results in- that means we won't need to de-dupe
h=set()
total=0;

# obviously i would be less than < 5 digits long, since it's at least n = 2 (number cat number*2)
# and b is less than 7- because with a > 2, that would be 10 digits
for a in range(1,9999):
	product = ""
	for b in range(1,7): 
		product += str(a*b)

jakedobkin

# initialize an array to hold results
l=[0]*1001

# brute force to calculate a,b,c
for a in range(1,1000):
	for b in range(1,1000): 
		c = (a**2 + b**2)**0.5

# verify c is a perfect square and p is < 1000, if it is, increment count
		if (a+b+c<=1000 and c-round(c,0)==0):