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Question

The remainder, if 1+2+22+23+....+21999 is divided by 5 is?

A
0
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B
1
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C
2
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D
3
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Solution

The correct option is A 0
1+2+22+.....+21999 is in geometric progression
So we get 1+2+22+.....+21999=22000121=220001
220001 can be written as 165001=(15+1)5001=500C0×15500+500C1×15499+500C2×15498+.....+500C500×1501=15×constant
Therefore the given series is divisible by 5
So the remainder is zero
Therefore option A is correct

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