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Babbage problem: Difference between revisions
Added a C# solution
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<pre>The square of 25264 is 638269696 !
</pre>
=={{header|C sharp}}==
<lang csharp>namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0; //the base number multiplied by itself
long baseNumber = 0; //the number to be squared, this one will be iterated
do //this sets up the loop
{
baseNumber += 1; //add one to the base number
baseNumberSquared = baseNumber * baseNumber; //multiply the base number by itself and store the value as baseNumberSquared
}
while (Right6Digits(baseNumberSquared) != 269696); //this will continue the loop until the right 6 digits of the base number squared are 269,696
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString(); //this is converts the number to a different type so it can be cut up
if (numberAsString.Length < 6) { return baseNumberSquared; }; //if the number doesn't have 6 digits in it, just return it to try again.
numberAsString = numberAsString.Substring(numberAsString.Length - 6); //this extracts the last 6 digits from the number
return long.Parse(numberAsString); //return the last 6 digits of the number
}
}
}}</lang>
{{out}}
<pre>The smallest integer whose square ends in 269,696 is 25264
The square is 638269696</pre>
=={{header|Clojure}}==
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{{out}}
<pre>(25264)</pre>
=={{header|COBOL}}==
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