The number 56890065 is completely divisible by only:
3
The problem requires determining which number completely divides 56890065 from the given options. To solve this, we need to apply divisibility rules.
Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. For 56890065, the sum of digits is 5+6+8+9+0+0+6+5 = 39. Since 39 is not divisible by 9, 56890065 is not divisible by 9.
Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. The sum of the digits, 39, is divisible by 3 (39 ÷ 3 = 13). Therefore, 56890065 is divisible by 3.
Divisibility by 11 and 4: For 11, the rule is alternating sum of digits should be divisible by 11. Alternating sum: (5-6+8-9+0-0+6-5) = -1, which is not divisible by 11. For 4, the last two digits, 65, are not divisible by 4. Therefore, 56890065 is not divisible by both 11 and 4.
Divisibility by 6 and 7: A number is divisible by 6 if divisible by both 2 and 3. 56890065 is not divisible by 2 (it's odd). Also, no individual test needed for 7 as 56890065 is not divisible by 6. Combined divisible by 6 and 7 is not possible here.
Thus, 56890065 is completely divisible by 3.
Important Questions from Divisibility and Remainder
If all positive divisors of 132 are arranged in descending order, then what digit will be at unit place of first divisor ?
If 3 2019 is divided by 10, then what is the remainder?
The number 3798125P369 is divisible by 7. What is the value of the digit P?
Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum.
Which of the following is/are correct?
1. S is always divisible by 74.
2. S is always divisible by 9.
select the correct answer using the code given below:
How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?
