The units digit of a three-digit number is prime. The hundreds digit is twice the tens digit. How many of these three digit numbers are prime?
first condition:
XX2 XX3 XX5 XX7
Second condition: hundreds is twice tens
842 632 422 212
843 ....
...
847................217
There will be 16 of those
I will leave it up to you to decide how many of those are prime numbers.
(remember, just because a number ends in an odd number, does not mean it is prime.
e.g. 633 is not prime, but 843 is prime )
Here is a list of primes:
https://primes.utm.edu/lists/small/1000.txt
To find the number of three-digit numbers satisfying the given conditions, we need to consider the possible values for the units digit, tens digit, and hundreds digit separately.
First, let's look at the units digit. A digit is prime if it is either 2, 3, 5, or 7. Therefore, there are four possibilities for the units digit.
Next, let's consider the tens digit. It is not explicitly given in the problem, but based on the fact that the hundreds digit is twice the tens digit, we can determine its possible values. If the tens digit is 1, then the hundreds digit would be 2, which contradicts the problem statement that the number is three digits. Thus, the tens digit must be 2, 3, 4, 5, 6, 7, 8, or 9, giving us eight possibilities.
Finally, we come to the hundreds digit. Since it is twice the tens digit, the hundreds digit can take on any value from 2 to 9, giving us eight possibilities.
To find the total number of three-digit numbers satisfying the given conditions, we multiply the number of possibilities for each digit together:
Number of three-digit numbers = (number of possibilities for units digit) x (number of possibilities for tens digit) x (number of possibilities for hundreds digit)
Number of three-digit numbers = 4 x 8 x 8 = 256
Therefore, there are 256 three-digit numbers satisfying the given conditions.