Question

Given a set $S$ consisting of two undefined elements “pib” and “maa,” and the four postulates:

$\mathrm{P}_1:$ Every pib is a collection of maas.

$\mathrm{P}_2:$ Any two distinct pibs have one and only one maa in common.

$\mathrm{P}_3:$ Every maa belongs to two and only two pibs.

$\mathrm{P}_4:$ There are exactly four pibs.

Consider the three theorems:

$\mathrm{T}_1:$ There are exactly six maas.

$\mathrm{T}_2:$ There are exactly three maas in each pib.

$\mathrm{T}_3:$ For each maa there is exactly one other maa not in the same pib with it.

The theorems which are deducible from the postulates are:

$\mathrm{T}_3$ only

$\mathrm{T}_2$ and $\mathrm{T}_3$ only

$\mathrm{T}_1$ and $\mathrm{T}_2$ only

$\mathrm{T}_1$ and $\mathrm{T}_3$ only

all

Answer: E

Concepts:basic counting graph theory logical deduction

Difficulty rating: 2030

Small Hint:

Label the four pibs $1,$ $2,$ $3,$ and $4$

Big Hint:

Each maa corresponds to an unordered pair of pibs

Solution:

Label the four pibs $1,$ $2,$ $3,$ and $4.$ By $\mathrm P_2$ and $\mathrm P_3,$ every maa is exactly the common maa of one unordered pair of pibs. Thus there are $\binom42=6$ maas, proving $\mathrm T_1.$

A fixed pib $i$ contains the three maas $ij$ with $j\ne i,$ proving $\mathrm T_2.$ For maa $ij,$ the unique maa sharing neither pib is the one belonging to the complementary pair of pibs, proving $\mathrm T_3.$ All three follow.

Therefore, the correct answer is E.

1967 AMC 12 Problem 38

Answer: E

Difficulty rating: 2030

Small Hint:

Big Hint:

Solution:

Problem 38 in Other Years