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

a.31

b. 63

c. 75

d. 91

Since in all three cases the last digit is 1,

the number should give remainder 1 when divided individually by 2,3,5 .

So the no. may be 31 or 91 .

Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 gives exactly two out of the three cases the leading digit as 1

Hence option d

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