No it's not.
As 4 is less than 621, 4 is not divisible by 621. Nor is 621 divisible by 4: 4 is even and all multiples of 4 are even, 621 is odd and so cannot be a multiple of 4. To check divisibility by 4, if the the last 2 digits are divisible by 4, the original number is divisible by 4. For 621, the last 2 digits are 21 and 21 is not divisible by 4, thus 621 is not divisible by 4; nor is 622 since the last 2 digits 22 is not divisible by 4.
47 is a prime number 21 & 87 divisible by 3, and 4 is divisible by 2
85 is not exactly divisible by 4. Dividing 85 by 4 gives 21 with a remainder of 1.
Both 21 and 32 are divisible by 1. Additionally, 21 is divisible by 3 and 7, while 32 is divisible by 2, 4, 8, and 16. However, they have no common divisors other than 1, making them relatively prime.
To determine if 10534341 is divisible by 9, add the digits (1+0+5+3+4+3+4+1 = 21); since 21 is divisible by 9, the number is divisible by 9. For divisibility by 3, the sum of the digits (21) is also divisible by 3. However, for divisibility by 2, the number must be even, and since 10534341 ends in 1, it is not divisible by 2. Therefore, 10534341 is divisible by 9 and 3, but not by 2.
The numeral '21' is evenly divisible by 21, 7, 3, and 1.
To determine if 4908 is divisible by 3, we need to sum its digits. 4 + 9 + 0 + 8 = 21. Since 21 is divisible by 3 (21 ÷ 3 = 7), 4908 is also divisible by 3. Therefore, 4908 is divisible by 3.
To check if 10,534,341 is divisible by 3, add the digits: 1 + 0 + 5 + 3 + 4 + 3 + 4 + 1 = 21, which is divisible by 3, so 10,534,341 is divisible by 3. For divisibility by 9, since 21 is not divisible by 9, the number is not divisible by 9. Lastly, since the last digit is 1 (an odd number), it is not divisible by 2. Therefore, 10,534,341 is divisible by 3 only.
They're both divisible by 1,3,7, and 21.
4, 8, 12, 16, and 20.
There is nothing that is divisible in 37 and 21. Because 37 is a prime number. However, 21 is divisible by 3 and 7.
No, it is divisible by: 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126