How many 4 digit numbers abcd exist such that a is odd, b is divisible by 3, c is even, and d is prime?
The answer is 5×4×5×4=400
Since a={1,3,5,7,9}
B={0,3,6,9}
C={0,2,4,6,8}
D={2,3,5,7}
The answer is 400
To find the number of 4-digit numbers that satisfy the given conditions, we can break down the conditions one by one and count the possible options for each digit.
Condition 1: a is odd.
Since a is the thousands digit, it can be any odd number from 1 to 9. So there are 5 choices for a (1, 3, 5, 7, or 9).
Condition 2: b is divisible by 3.
Since b is the hundreds digit, it needs to be a multiple of 3. The possible choices for b are 0, 3, 6, or 9. So there are 4 choices for b.
Condition 3: c is even.
Since c is the tens digit, it needs to be an even number. The possible choices for c are 0, 2, 4, 6, or 8. So there are 5 choices for c.
Condition 4: d is prime.
Since d is the units digit, it needs to be a prime number. The prime numbers from 0 to 9 are 2, 3, 5, and 7. So there are 4 choices for d.
To find the total number of 4-digit numbers satisfying all the conditions, we need to multiply the number of choices for each digit together: 5 * 4 * 5 * 4 = 400.
Therefore, there are 400 possible 4-digit numbers that satisfy the given conditions.
Poli
'a' is odd, 5 choices
'b' is divisible by 3, 3 choices
'c' is even , 4 choices
'd' is prime, 4 choices.
So what is 5x3x4x4 ?