Question

There are K people in a room, each person picks a day of the year to get a free dinner at a fancy restaurant.K is such that there must be at least one group of six people who select the same day. What is the smallest such K if the year is a leap year (366days)?
(a) 1829 (b) 1831 (c) 1830 (d) 1832 (e) 1833. Answer is c.please explain how

Answer 1

According to me it is 366*5=1830 but how 1831

Answer 2

With 1830, you do not necessarily have a single group of SIX people that choose the same day of the year.
With 1831, you will, by the pigeon-hole theorem.

Answer 3

Ask in math, not computers.

Answer 4

Ok can u please explain it by pigeonhole principle....

Answer 5

The pigeon hole principle says that if there are n+1 one objects to be placed in n boxes, at least one of the boxes must contain two objects.

Similarly, if you have 366 days, and 1830 people, the "best" you can do to spread it out is to have 366 groups of 5 people, but still will not make a single group of six people. However, if you have 1831 people and 366 days, by the pigeon hole principle, to place the 1831st person, you need to make a group of 6 people, which is the minimum number of people to create at least one group of 6 people.