1, 2, 3, 6, 107, 214, 321, 642
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60, 107, 214, 321, 428, 535, 642, 1070, 1284, 1605, 2140, 3210, 6420
Yes.
This statement is not always true. While it is true that if a number is divisible by 4, it is also divisible by 2, the reverse is not necessarily true. For example, the number 6 is divisible by 2 but not by 4. In general, being divisible by 2 is a necessary but not a sufficient condition for being divisible by 4.
7 + 4 + 1 = 12 which is divisible by 3, so 741 is divisible by 3 741 is odd, all multiples of 4 are even, 741 is not divisible by 4 (Alternatively 4 x 2 + 1 x 1 = 9 which is not divisible by 4, so 741 not divisible by 4) All multiples of 5 end with 5 or 0, 741 ends with 1, so 741 not divisible by 5 To be divisible by 6, the number must be divisible by 2 (even) and divisible by 3.741 is odd, so not divisible by 2, so 741 is not divisible by 6 7 + 4 + 1 = 12 which is not divisible by 9, so 741 is not divisible by 9 All multiples of 10 end with 0, 741 ends with 1 so 741 is not divisible by 10 Summary: 741 is divisible by 3, but not by 4, 5, 6, 9 nor 10.
Yes and it is 107
You can add up the digits, 6+4+2=12 and if that number is divisible by three, so it the original number. So 12 is certainly divisible by 3 so that means 642 is also.
No. 642 is not evenly divisible by nine.
Just by looking at it, without trying any factors to see if they work, you can see-- 642 is even, i.e. divisible by 2, since it ends in '2'-- 642 is divisible by 3, since its digits add up to (6+4+2) = 12 which is divisible by 3.Just looking at 642, you already know of at least 2 factors that it has besides '1' and itself,so it can't be a prime number.
Yes because 6+4+2 is divisible by three.The simple way to decide if any number is entirely divisible by three is to first discard any digit in the number that is divisible by three (3, 6, 9) then add any of the digits remaining together and if at any point you get something that is divisible by three you can disregard those digits. If at the end you have a remainder, the number is not evenly divisible by three. If there is no remainder the entire number is divisible by three.In your example 642, I threw away the 6 leaving 4 and 2 , added those = 6 which is divisible by three (actually I remembered that 42 was divisible by 6 so I knew right after I had taken away the first 6 that your number was divisible by 3.
1, 2, 3, 6, 107, 214, 321, 642
Yes, evenly
5
642 is composite. It is divisible by 2, 3, 107, 214, and 321.
here are some 2,3,4,6,8,12,13,16,24,26,32,48,52,321&642
4- If the last two digits are divisible by 4, the whole number is divisible by 4. 6- If the number is even and also divisible by 3, it is divisible by 6.
No, 642 divided by 4 is 160.5