How many four-digit numbers abcd exist such that a is even, b is divisible by 5, c is prime, and d is odd?
a = 2, 4, 6, 8,
b = 0, 5
c = 2, 3, 5, 7
d= 1, 3, 5, 7, 9
Ms Sue
you are Simply the best with all caps
B E S T. I keept looking over this problem.
You're welcome.
What answer did you get?
432
To find the number of four-digit numbers that satisfy the given conditions, we can analyze each condition separately and then multiply the results together.
Condition 1: a is even
An even number is divisible by 2. In a four-digit number, the thousands place (a) can be any even number from 0 to 9, excluding 0 (since it cannot be the leading digit).
So, there are 4 choices for the value of a: 2, 4, 6, or 8.
Condition 2: b is divisible by 5
A number is divisible by 5 if its units digit is either 0 or 5. In a four-digit number, the hundreds place (b) can be any digit from 0 to 9.
So, there are 10 choices for the value of b: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9.
Condition 3: c is prime
Prime numbers are positive integers greater than 1 that have no positive divisors other than 1 and themselves. Prime numbers less than 10 include 2, 3, 5, and 7.
So, there are 4 choices for the value of c: 2, 3, 5, or 7.
Condition 4: d is odd
An odd number is not divisible by 2. In a four-digit number, the units place (d) can be any odd number from 1 to 9.
So, there are 5 choices for the value of d: 1, 3, 5, 7, or 9.
To find the total number of four-digit numbers satisfying all the conditions, we multiply the number of choices for each condition together:
Total number of four-digit numbers = Number of choices for a * Number of choices for b * Number of choices for c * Number of choices for d
Total number of four-digit numbers = 4 * 10 * 4 * 5 = 800
Therefore, there are 800 four-digit numbers abcd that satisfy the given conditions.