F x f f x f x 2 how many such functions with degree 2 exist

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Previous Question: how many digit of xeros at the end of product of the number from 1 to 100? Next Question: That's an interesting question! In a normal year - that is, a year with no leap day - there are 365 days, which comes to 52 weeks of 7 days, plus one more day. So if a normal year starts on a Monday, then the following year has to start on a Tuesday. Does that make sense? However, if a leap year starts on a Monday, then the following year has to start on a Wednesday, because of the extra day in February. And if two years start on the same day, and if they're both normal years or both leap years, then they have to have the same calendar. Do you see why this is true? 1990 is a normal year, and in fact, it _does_ start on a Monday. So 1991 must start on a Tuesday, and 1992 must start on a Wednesday. But 1992 is a leap year, so 1993 must start on a Friday (instead of a Thursday). I've started a table below: Year Starts on Leap? Add ---- --------- ----- --- 1990 Monday N 1 1991 Tuesday N 1 1992 Wednesday Y 2 1993 Friday N 1 1994 Saturday N 1 If you continue the table, you'll eventually find a year that starts on Monday and isn't a leap year. That's the year you're looking for.

(#M40020151) TCS question Keep an EYE

F(x)=f(f(x))=f(x^2).....how many such functions with degree>2 exist?: Asked In TCS suparna guha (11 years ago); Unsolved Read Solution (3); Is this Puzzle helpful? (2) (7) Submit Your Solution

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Latest TCS Aptitude Question SOLUTION: F(x)=f(f(x))=f(x^2).....how many such functions with degree>2 exist?

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(#M40020151) TCS question Keep an EYE

F(x)=f(f(x))=f(x^2).....how many such functions with degree>2 exist?

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