The intName method in the "Problem Solving: Stepwise Refinement' section accepted arguments

Using a recursive call, extend its range to $999,999.$ For example an input of $12,345$ should return "twelve thousand three hundred forty five".

Note that some number names may have an extra space at the beginning and$/$or end.

import iava.util.Scannes:

public class lategerName

$1$

Turns a number into its English name.

@param number a positive integer $<mn\; id="EditorElement\_5068294">9</mn><mn\; id="EditorElement\_5068295">9</mn><mn\; id="EditorElement\_5068296">9</mn><mn\; id="EditorElement\_5068297">.</mn><mn\; id="EditorElement\_5068298">9</mn><mn\; id="EditorElement\_5068299">9</mn><mn\; id="EditorElement\_5068300">9</mn>$

$($ireturn the name of the number $($e$.$g$.$ "two hundred seventy four"$)$

$\%$

public