2021 AMC 10A Problem 7

P1-10~2 minPrint

Logic Puzzles Set up the equation/formula and compute; no special trick needed

Tom has a collection of $13$ snakes, $4$ of which are purple and $5$ of which are happy. He observes that

all of his happy snakes can add,

none of his purple snakes can subtract, and

all of his snakes that can't subtract also can't add.

Which of these conclusions can be drawn about Tom's snakes?

Show solution without answering

Key insight

Follow the chain: a purple snake can't subtract, so it can't add. Every happy snake can add, so no happy snake can be purple.

Solution

1. Rewrite each fact as an "if ... then ..." rule:
- If a snake is happy, then it can add.
- If a snake is purple, then it can't subtract.
- If a snake can't subtract, then it can't add.
2. The second and third rules link up. A purple snake can't subtract, and a snake that can't subtract can't add. So a purple snake can't add.
3. Now think about a happy snake. By the first rule it can add. But step 2 says purple snakes can't add. So a happy snake cannot be purple. That is statement (D).
4. Check why the others fail:
- (A) says purple snakes can add, the opposite of step 2.
- (B) cannot be true: no happy snake is purple, and Tom does have $4$ purple snakes, so those snakes are not happy.
- (C) turns the first rule around. The rule says nothing about a snake that can add but is not happy.
- (E) Nothing in the facts says a happy snake can't subtract.

The answer is $\boxed{\textbf{(D)}\ \text{Happy snakes are not purple.}}$

Why this works

A sentence like "all happy snakes can add" is a rule of the form "if happy, then can add". Rules can be chained: if A leads to B and B leads to C, then A leads to C. A rule can also be read backwards with both parts negated: "if it can't add, then it isn't happy". That backwards reading is exactly step 3. The numbers $13$ , $4$ and $5$ play no part in the reasoning.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Write the three facts as implications:
1. happy $\Rightarrow$ can add;
2. purple $\Rightarrow$ cannot subtract;
3. cannot subtract $\Rightarrow$ cannot add.

Chaining 2 and 3: purple $\Rightarrow$ cannot add. Combining with 1, a happy snake can add, so it is not one of the snakes that cannot add, so it is not purple. That is exactly statement (D).

The other choices fail: (A) contradicts the chain just derived; (B) contradicts (D) unless there are no purple snakes, but there are four; (C) reverses fact 1's direction and is not implied; (E) is not implied, since nothing says a snake that can add cannot subtract.

The answer is $\boxed{\textbf{(D)}\ \text{Happy snakes are not purple.}}$

Review in the AoPS books

Intermediate Counting & Probability, Chapter 2: Sets and Logic (§2.4)