let n be a positive integer with all digits equal to 5 such that n is divisible by 2003. find the last six digit of n/2003
Answers
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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