What an easy one!

What is the smallest number of people Mohhawk would need to have in a room if he wanted to be certain that at least 25 of them have the same birth month?

from Mathcounts Trainer


The answer is 289.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Matheus Rezende
Aug 3, 2015

Eu pensei inicialmente em 301, pela fórmula do 25 × 12 25 \times 12 +1. Mas após raciocinar um pouco, lembrei do fato de que Mohhawk sabia seu aniversário, não precisando de pessoas para preencher 12 meses. Ou seja, a fórmula final é 24 × 12 24 \times 12 +1.

OBS: O nome dessa matéria é "Princípio das casas dos pombos", caso queira estudá-la.

Initially, I thought in 301, by using 25 × 12 25 \times 12 +1. But this wasn't right, then I remembered that Mohhawk already knew his birthday. So he didn't need 12 people, only himself. So it's not 25 × 12 25 \times 12 +1, but 25 × 12 25 \times 12 +1-12, which is equal to 24 × 12 24 \times 12 +1.

1 year has 12 months. If 12 people have different birth months, the 13rd person WILL CERTAINLY fit one of the twelve months [of a year]. So you need 12x+1 people to be certain that at least x of them have the same birth month.

*PS: The image don't have Mohhawk represented. He is the (+1). Each box represents one month and each month has 24 dots. To have 25, you you need one more person (Mohhawk).

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...