How many positive integers less than 2*10^4 are there in which each digit is a prime number?

BTGmoderatorDC wrote: ↑

Mon May 25, 2020 6:39 pm

How many positive integers less than 2*10^4 are there in which each digit is a prime number?

(A) 256
(B) 326
(C) 340
(D) 625
(E) 775

OA C

Source: Magoosh

So, we have to find the count of numbers that are less than 2*10^2 or 20,000 and in which each digit is prime or one among 2/3/5/7 (four digits).

Numbers less than 20,000 must be 4-digit, 3-digit, 2-digit and 1-digit since a 5-digit number less than 20,000 or less than equal to 19,999 would have its ten-thousands digit = 1, which is not prime.

• # of 4-digit numbers such that each digit is prime = 4*4*4*4 = 256; each place has four choices;
• # of 3-digit numbers such that each digit is prime = 4*4*4 = 64; each place has four choices;
• # of 2-digit numbers such that each digit is prime = 4*4 = 16; each place has four choices;
• # of 1-digit numbers such that each digit is prime = 4

Total count of such numbers = 256 + 64 + 16 + 4 = 340

The correct answer: C

Hope this helps!

-Jay
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BTGmoderatorDC wrote: ↑

Mon May 25, 2020 6:39 pm

How many positive integers less than 2*10^4 are there in which each digit is a prime number?

(A) 256
(B) 326
(C) 340
(D) 625
(E) 775


OA C

Solution:

We note that 2*10^4 = 2 * 10,000 = 20,000. We need to determine the number of integers less than 20,000 in which each digit is a prime number. The prime single-digit numbers are 2, 3, 5, and 7. Since 1 is not a prime, we see that we can rule out all numbers greater than or equal to 10,000. In other words, the number must be no more than 4 digits.

If it’s a 4-digit number, then there are 4 x 4 x 4 x 4 = 4^4 = 256 such numbers.

If it’s a 3-digit number, then there are 4 x 4 x 4 = 4^3 = 64 such numbers.

If it’s a 2-digit number, then there are 4 x 4 = 4^2 = 16 such numbers.

If it’s a 1-digit number, then there are 4 such numbers.

So there are 256 + 64 + 16 + 4 = 340 such numbers.

Answer: C