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
4 choices for a
2 choices for b
4 choices for c
4 choices for d
Mulitply to find how many 4 digit numbers exist under these conditions.
there are 4 even digits
only 1 digit, 5 , is divisible by 5
(although you could make an argument for 0)
1, 3, 5 , 7 or four digits are prime
1,3,5,7,9 or 5 digits are odd
so 4*1*4*5 = 80
I agree with Damon, I was thinking about zero and in 10 , 20 , etc.
I got 80 to thanks Damon and Dr Jane
To find the number of four-digit numbers that satisfy the given conditions, we can break down the problem step by step.
Step 1: Count the number of options for each digit.
a) For the thousands digit (a):
Since a needs to be even, the possible options are 2, 4, 6, or 8. There are four options.
b) For the hundreds digit (b):
Since b needs to be divisible by 5, the possible options are 0, 5. There are two options.
c) For the tens digit (c):
Since c needs to be a prime number, the possible options are 2, 3, 5, 7. There are four options.
d) For the units digit (d):
Since d needs to be odd, the possible options are 1, 3, 5, 7, 9. There are five options.
Step 2: Multiply the number of options for each digit.
To find the total number of four-digit numbers that satisfy all the conditions, we need to multiply the number of options for each digit together.
Total number of options = Number of options for a * Number of options for b * Number of options for c * Number of options for d
= 4 * 2 * 4 * 5
= 160
Therefore, there are 160 four-digit numbers that satisfy the given conditions.