Question
A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals:
Answer: Option D
:
D
Since the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of numbers are 61, 91.
Among these only 31 and 91 are a part of the answer choices.
Among these, (31)10=(11111)2=(1011)3=(111)5. Thus, all three forms have leading digit 1.
Hence the answer is 91.
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:
D
Since the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of numbers are 61, 91.
Among these only 31 and 91 are a part of the answer choices.
Among these, (31)10=(11111)2=(1011)3=(111)5. Thus, all three forms have leading digit 1.
Hence the answer is 91.
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