Given :  n is  a positive integer with all digits equal to 5  

n is divisible by 2003

To Find :   last SIX digit of n/2003

Solution:

n/2003  =     ________ABCDEF

ABCDEF being last 6 digit

n = 5555555______55555...

n = 2003 * (______ABCDEF)

=>  2003 * ABCDEF  last digits  = 555555555

3 * 0 = 0

3* 1 = 3

3 *2 = 6

3 * 3 = 9

3 * 4 = 12

3 * 5 = 15

2 * 6 = 18

3 * 7 = 21

3 * 8 = 24

3 * 9 = 27

F must be  5    ∵ 3 * 5 = 15

2003 * 5  = 10015    

5 is last Digit   and 1001 is carried forward

Hence E must be  8       ∵ 3 * 8 + 1 = 25

so that 2003 * 8 + 1001  ends with 5

2003 * 8 + 1001  = 17025

5 is last Digit   and  1702 is carried forward

D must be 1      ∵ 3 * 1 + 2 =  5

2003 * 1 + 1702  = 3705  

5 is last Digit   and  370 is carried forward  

C must be 5     ∵ 3 * 5 + 0 =  15

2003 * 5 + 370 = 10385

5 is last Digit   and  1038 is carried forward    

B must be 9       ∵ 3 * 9 + 8 =  15

2003 * 9 + 1038  = 19065

5 is last Digit   and   1906 is carried forward    

A must be 3          ∵ 3 * 3 + 6 =  15

2003 * 3  + 1906  = 7915

5 is last Digit   and    791  is carried forward    

last six digit of n/2003   = 395185

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