2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

This is an interesting problem!

First thing's first, we can establish that the largest positive number that meets the condition is $1×2×3..×20$ or simply $20!$ We can work our way down by repeatedly dividing this upper boundary number by any number in the range [1,20] and seeing if it's an even division.

This approach results in a runtime complexity of O(log(n!)), better known as O(n log n)

In [16]:

```
factors = 20
upper = math.factorial(factors)
divisors = range(2, factors+1)
current = upper
#repeatedly attempt to divide current number by prime factors ordered
#from largest to smallest as long as the result has a remainder of 0
while True:
found = False
for p in reversed(divisors):
c = current / p
if c % p == 0:
found = True
current = c
break
if not found:
break
print 'divided by', p, 'got', current
```

The sum of the squares of the first ten natural numbers is, 1^{2} + 2^{2} + ... + 10^{2} = 385

The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)^{2} = 552 = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

**Method 1: brute force**

Complexity: O(N)

In [10]:

```
def squareDiff(x):
s = range(1, x+1)
sumSquares = sum([x*x for x in s])
squareSum = math.pow(sum(s),2)
diff = squareSum - sumSquares
return diff
squareDiff(100)
```

Out[10]:

Easy enough, however it's well known that the sum of a series of natural numbers up to n can be calculated as $\frac{n(n+1)}{2}$

Is it possible that the sum of a series of natural numbers squared up to n can be calculated in constant time as well? I didn't know the answer and cheated by using a genetic algorithm to attempt to fit an equation to match *sumSquares* for the first 140 inputs.

Amazingly, it came back with a polynomial that had 0 residual error: $\frac{1}{6}n + \frac{1}{2}n^2 + \frac{1}{3}n^3$

Let's plot this polynomial to double check

In [11]:

```
brute = lambda n: sum([x*x for x in xrange(1,n+1)])
poly = lambda n: round(1./6 * n + 1./2 * pow(n, 2) + 1./3 * pow(n, 3))
x = np.array(range(1,2200))
brute_y = np.array([brute(t) for t in x])
poly_y = np.array([poly(t) for t in x])
plt.plot(x, brute_y, label='Brute Force', color='blue')
plt.plot(x, poly_y, label='Polynomial', color='red')
plt.legend()
print 'max error:', max(brute_y - poly_y)
```

^{3} while brute force only deals with n^{2}. We'll switch to floats to overcome overflow issues in both cases.

In [12]:

```
brute = lambda n: sum([1.*x*x for x in xrange(1,n+1)])
poly = lambda n: round(1./6 * n + 1./2 * pow(n, 2.) + 1./3 * pow(n, 3.))
x = np.array(range(1,2200))
brute_y = np.array([brute(t) for t in x])
poly_y = np.array([poly(t) for t in x])
plt.plot(x, brute_y, label='Brute Force', color='blue')
plt.plot(x, poly_y, label='Polynomial', color='red')
plt.legend()
print 'max error:', max(brute_y - poly_y)
```

A maximum error of 0 across a small input set is promising, however I got in touch with a friend to check, and he promtly came back with a proof!

All credit to **Jonah Schreiber** for the below proof:

**Base**

Here, we show that the formula is correct for $n=1$. On the left-hand side, we have $1$, and on the right-hand side, we have $\frac{1^3}{3}+\frac{1^2}{2}+\frac{1}{6}=1$, so the base case is true.

**Assumption**

Assume that

$$\sum_{x=1}^{n}{x^2}=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$$is true.

**Induction**

Show then that it is true for $n+1$, that is,

$$\sum_{x=1}^{n+1}{x^2}=\frac{(n+1)^3}{3}+\frac{(n+1)^2}{2}+\frac{n+1}{6}$$Let us break out the last term on the left-hand side, and expand the right-hand side:

$$\sum_{x=1}^{n}{x^2}+(n+1)^2=\frac{n^3+3n^2+3n+1}{3}+\frac{n^2+2n+1}{2}+\frac{n+1}{6}$$We already know the sum on the left-hand side, which we insert.

$$\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}+n^2+2n+1=\frac{n^3+3n^2+3n+1}{3}+\frac{n^2+2n+1}{2}+\frac{n+1}{6}$$Collecting terms, we get

$$\frac{n^3}{3}+\frac{3n^2}{2}+\frac{13n}{6}+1=\frac{n^3}{3}+\frac{3n^2}{2}+\frac{13n}{6}+1$$The two sides are equal, so the formula is correct.

Thanks again to Jonah Schreiber for the above proof.

See http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm for several derivations of this formula.

And so, finally,

**Method 2: Arithmetic **

Complexity: O(1)

In [13]:

```
def squareDiff2(x):
sumSquares = round(1./6 * x + 1./2 * pow(x, 2.) + 1./3 * pow(x, 3.))
squareSum = pow(x*(x+1)/2., 2)
return squareSum - sumSquares
squareDiff2(100)
```

Out[13]: