Find the highest power of 2 in [10301010+30]; where [x] is the greatest integer function.
A
1
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B
2
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C
3
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D
4
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Solution
The correct option is B
2
[10301010+30]=[1020×10101010+30]=⎡⎢
⎢
⎢
⎢⎣1020×10101010(1+301010)⎤⎥
⎥
⎥
⎥⎦=⎡⎢
⎢
⎢
⎢⎣10201+301010⎤⎥
⎥
⎥
⎥⎦ Given expression is sum of infinite G.P. (Compare it with a1−r) a=1020 and r=(−3109) Given series is 1020+1020×(−3109)+1020×(−3109)2+1020×(−3109)3+......... 1020+1011×(−3)+102×(−3)2+10−7×(−3)3+....... Other terms are very small, we can neglect them. Only the first three terms need to be considered. In them, the highest power of 2 is 2, as the numbers are a multiple of 4. (Sum of first three terms of given series = 99999999700000000900)