Including the section prefix characters method

The job number used in this example is HIF13297I?.
	The high order digit in the Rotation No. is multiplied by 6
	In this example 1 * 6 = 6
	The second high order digit in the Rotation No. is multiplied by 5
	In this example 3 * 5 = 15
	The third high order digit in the Rotation No. is multiplied by 4
	In this example 2 * 4 = 8
	The fourth high order digit in the Rotation No. is multiplied by 3
	In this example 9 * 3 = 27
	The fifth high order digit in the Rotation No. is multiplied by 2
	In this example 7 * 2 = 14
The aggregate total of the result of each multiplication is then calculated: (6+15+8+27+14) = 70
	The third letter of the section prefix (in this case F) is searched for in the alphabet. The position at which the letter appears in the alphabet is (ie F is position 6), is multiplied by 7: F (position 6) * 7 = 42
	This is added to the total above: 70 + 42 = 112
	The new total is 112
	The second letter of the section prefix (in this case I) is searched for in the alphabet. This letter's position is multiplied by 8: I (position 9) * 8 = 72
	This is added to the total above: 112 + 72 = 184
	The new total is 184.
	The first letter of the section prefix (in this case H) is searched for in the alphabet. This letter's position is multiplied by 9:
		H (position 8) * 9 = 72
	This is added to the total above:
		184 + 72 = 256
	The new total is 256
	If the total is below 11 it is moved straight into the check digit, otherwise, 11 is divided into this total giving an integer result and an integer remainder (256 / 11) = 23 with a remainder of 3.
	The total is 23 but this is not required, only the remainder is used (3), which is subtracted from 11 giving a check-digit of 8. If the result is 10, 3 is moved to the check digit If the result is 11, 0 is moved to the check digit. If the result is not 10 or 11 then the remainder is moved straight into the check digit.