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The sum of all two digit numbers which when divided  by 4, yield unity as remainder, is

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a
1100
b
1200
c
1210
d
None of these

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detailed solution

Correct option is C

The first two digit number which when divided by 4  leaves remainder 1 is 4 .3 + 1 = 13 and last is 4 . 24 + 1 = 97.Thus, we have to find the sum13+17+21+…+97which is an A.P.∴ 97=13+(n−1)⋅4⇒n=22 and  Sn=n2[a+l]=11×[13+97]=11×110=1210.


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