CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find the highest power of 2 in [10301010+30]; where [x] is the greatest integer function.


A

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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 a1r)
a=1020 and r=(3109)
Given series is 1020+1020×(3109)+1020×(3109)2+1020×(3109)3+.........
1020+1011×(3)+102×(3)2+107×(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)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Highest Power and Summation
QUANTITATIVE APTITUDE
Watch in App
Join BYJU'S Learning Program
CrossIcon