Doctor Chapatin

Number Theory Level 4

When we write Doctor Chapatin’s birthday in the format dd/mm (2 digits for the day and 2 for the month), we only need the digits 0 , 1 , 2 0, 1, 2 and 3. 3.

How many dates could there be with these conditions ?

S o u r c e : F i n a l F F J M Source: Final FFJM


The answer is 75.

Antony Diaz by Antony Diaz , 25, Canada

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2 solutions

Ranik Chakraborty
Mar 20, 2015

the possible days can be 1,2,3,10,11,12,13,20,21,22,23,30,31...and months 01,02,03,10,11,12...adding possible combinations (13+11+13+13+12+13) we get 75.

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Aran Pasupathy
May 1, 2015

We first need to determine the number of ways of permuting two out of the four digits 0, 1, 2 and 3, including 11 and 22 and excluding 32 since 11 and 22 denote two particular days in a particular month while 32 does not.

(4P2)+(2-1)= 13

We then need to determine the number of ways of permuting two out of the four digits 0, 1, 2 and 3, including 11 and excluding 13, 20, 21, 23, 30, 31 and 32 since 11 denotes the month of November while 13, 20, 21, 23, 30, 31 and 32 do not denote any particular month(s) in a year.

(4P2)+(1-7)= 6

The number of dates there could be= (13*6), excluding 30/02, 31/02 and 31/11 since no such dates exist.

Thus, the number of dates there could be= (13*6)-3= 75

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