which of the following numbers is divisible by both 2 and 3
Answers
Answer:
Since the number 4,608 is both divisible by 2 and 3 then it must also be divisible by 6. The answer is YES. A number is divisible by 9 if the sum of the digits is divisible by 9.
Note: the given question is incomplete as it should be as follows:
Which of the following numbers is divisible by both 3 and 2?
a) 1023 b) 4029 c) 1062 d) 2512
Solution:
Know that,
Divisibility rule for 2 states that A number is divisible by 2 if the last digit of the number is an even number.
Divisibility rule for 3 states that a number is completely divisible by 3 if the sum of its digits is divisible by 3.
Check option (a).
Given number: .
Observe that last digit of the given number is 3, which is an odd number.
Therefore, it is not divisible by 2.
Find the sum of the given number.
Observe that the sum of the given number is 6 which is divisible by 3.
Therefore, the given number is divisible by 3 but not by 2.
Hence, this is not the correct option.
Check option (b).
Given number: .
Observe that last digit of the given number is 9, which is an odd number.
Therefore, it is not divisible by 2.
Find the sum of the given number.
Observe that the sum of the given number is 15 which is divisible by 3.
Therefore, the given number is divisible by 3 but not by 2.
Hence, this is not the correct option.
Check option (c).
Given number: .
Observe that last digit of the given number is 2, which is an even number.
Therefore, it is divisible by 2.
Find the sum of the given number.
Observe that the sum of the given number is 9 which is divisible by 3.
Therefore, the given number is divisible by 3 and 2.
Hence, this is the correct option.
Check option (d).
Given number: .
Observe that last digit of the given number is 2, which is an even number.
Therefore, it is not divisible by 2.
Find the sum of the given number.
Observe that the sum of the given number is 10 which is not divisible by 3.
Therefore, the given number is divisible by 2 but not by 3.
Hence, this is not the correct option.