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Q.

The digit at the unit place in the numbers192005 + 112005 _  92005 is

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a

2

b

1

c

0

d

8

answer is B.

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Detailed Solution

(19)2005+(11)2005−(9)2000=(10+9)2005+(10+1)2005−(9)2005=92005+2005C1(9)2004×10+…+ 2005C0+2005C110+…−(9)2005 = 2005C192004×10+ multiple of 10+ (1+multipleof10) ∴ Unit digit =1
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