2015 AMC 10B Problem 11

P11-15~3 minPrint

Primes Basic Probability Organized listing / direct enumeration

Among the positive integers less than $100$ , each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime?

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Key insight

Only the digits 2, 3, 5, 7 are allowed, so there are just 4 + 16 = 20 numbers to check. Eight of them are prime.

Solution

Which digits are allowed? The prime digits are $2, 3, 5, 7$ . (The digit $1$ is not prime.)
Count all the numbers. One-digit numbers: $2, 3, 5, 7$ , which is $4$ numbers. Two-digit numbers: $4$ choices for the first digit and $4$ for the second, which is $4 \cdot 4 = 16$ numbers. In total there are $4 + 16 = 20$ numbers.
One-digit primes. All four of $2, 3, 5, 7$ are prime.
Two-digit primes. A two-digit number ending in $2$ is even, and one ending in $5$ is a multiple of $5$ , so neither can be prime. That leaves the eight numbers ending in $3$ or $7$ :

Number	$23$	$33$	$53$	$73$	$27$	$37$	$57$	$77$
Prime?	yes	no, $3 \cdot 11$	yes	yes	no, $3 \cdot 9$	yes	no, $3 \cdot 19$	no, $7 \cdot 11$

Four of them are prime: $23, 37, 53, 73$ .
5. Probability. There are $4 + 4 = 8$ primes out of $20$ numbers, so the probability is $\dfrac{8}{20} = \dfrac25$ .

The answer is $\boxed{\textbf{(B)}\ \frac{2}{5}}$ .

Why this works

Once you see that only four digits are allowed, the list of candidates is small enough to check by hand. The last digit throws out half the list at once. For the rest, the digit-sum test for $3$ catches $33$ , $27$ and $57$ , and $77$ is a multiple of $7$ .

Concise solution

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

The prime digits are $2, 3, 5, 7$ . The one-digit candidates are these four numbers, and the two-digit candidates have $4$ choices for each digit, giving $4 \cdot 4 = 16$ numbers. The pool has $4 + 16 = 20$ numbers.

All four one-digit candidates are prime.

Among the two-digit candidates, any number ending in $2$ or $5$ is composite, so only those ending in $3$ or $7$ can be prime: $23, 33, 53, 73, 27, 37, 57, 77$ . Checking each: $23$ prime, $33 = 3 \cdot 11$ , $53$ prime, $73$ prime, $27 = 3^3$ , $37$ prime, $57 = 3 \cdot 19$ , $77 = 7 \cdot 11$ . Exactly four are prime.

So $4 + 4 = 8$ of the $20$ numbers are prime, and the probability is $\frac{8}{20} = \frac{2}{5}$ .

The answer is $\boxed{\textbf{(B)}\ \frac{2}{5}}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 2: Primes and Composites
Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Number Theory, Chapter 4: Prime Factorization (§4.2, §4.4)