"Find the greatest product of five consecutive digits in the 1000-digit number.

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

The first step here is to find a representation for that fairly humungous number. Obviously it's not going to fit into a paltry 32-bit int...but then we don't need it to. The problem description requires us to think in terms of smaller (5-digit) numbers, not one giant 1000-digit number.

So, it is sufficient for us to consider the number as an enumerable stream of
single digits, which we can conveniently represent as `IEnumerable<int>`

. I
could use a macro to convert the number into a collection initialiser, but
it's much easier to treat the string as an `IEnumerable<char>`

and let LINQ do
the heavy lifting.

```
var nums = Enumerable.AsEnumerable(
"73167176531330624919225119674426574742355349194934" +
"96983520312774506326239578318016984801869478851843" +
// ..... etc etc ......
"05886116467109405077541002256983155200055935729725" +
"71636269561882670428252483600823257530420752963450"
).Select(x => Convert.ToInt32(x.ToString()));
```

This gives us an `IEnumerable<int>`

containing every digit in the 1000-digit
number. Now, the 'obvious' way to solve the problem is to iterate through the
collection, and at each index multiply the value against the next four
indexes. A simple loop should deal with it:

```
private static int SimpleSolver(int[] ints)
{
int max = 0;
for (int i = 0; i < ints.Length - 4; i++)
{
int tmp = ints[i] * ints[i + 1] * ints[i + 2]
* ints[i + 3] * ints[i + 4];
max = Math.Max(max, tmp);
}
return max;
}
```

As ever, though, that's pretty ugly - the loop condition and product calculation is tied to the sequence size of 5, and messing with an index variable is tedious.

An alternative approach is to take advantage of LINQ's Skip and Take methods to split the problem domain into overlapping 'slices'. Similar to the for loop above, the core of the approach is to iterate through the digits, and at each digit grab a number of subsequent digits and calculate the product.

Lets look at the 5-digit slices available from the first 10 digits:

```
7 3 1 6 7 1 7 6 5 3
| 73167 |
| 31671 |
| 16717 |
| 67176 |
| 71765 |
| 17653 |
```

We can use Skip to progressively move the starting index forward, and Take to grab the 5 digits we need. So, starting with i=0, each successive slice can be sliced from the whole with:

```
var slice = ints.Skip(i++).Take(5);
```

To calculate the product of the digits in the slice, we can use the Aggregate operation:

```
slice.Aggregate(1, (curr, next) => curr*next);
```

We've met Aggregate before - it's basically a fold, which collapses a sequence to a single item by repeatedly applying an operation to an accumulating result.

This can all be wrapped up as an iterator block, like so:

```
private static IEnumerable<int> EnumerateSlices(
IEnumerable<int> ints, int sliceSize)
{
int i = 0;
while (true)
{
var slice = ints.Skip(i++).Take(sliceSize);
if (slice.Count() < sliceSize)
yield break; // end
yield return slice.Aggregate(1,
(curr, next) => curr*next);
}
}
```

Note the termination condition - when we have enumerated every slice, our next slice will contain only 4 elements (3, 4, 5, and 0 from the end of the sequence) - that's our cue to exit the loop.

Also note that this approach makes the algorithm trivial to parameterize - it will work just as well with slice sizes other than 5.

This iterator will produce an `IEnumerable<int>`

containing the products of all
slices, so the final step is to select the largest:

```
var result = EnumerateSlices(nums, 5).Max();
```

Blast, that didn't go well, posting-wise....

[...] at Basildon Coder he has chosen to represent the input as an list of integers, that means he will not have to convert [...]

[...] I have chose to represent the input number as a string, and then converting them on the fly.Over at Basildon Coder he has chosen to represent the input as an list of integers, that means he will not have to convert [...]