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Is 45 divisible by 6?
No,because 45 is not divisible by both 2 and 3. In order for a number to be divisible by 6 it has to be divisible by both 2 and 3.
Is 90 divisible by 2 or 3?
The number 90 is divisible by both 2 and 3.
Is 78 divisible by 2 3 or both?
Both.
What determines whether a number is divisible by both 2 and 6?
Something divisible by 6 has to be divisible by both 2 and 3. and that just answers that anything disible by 6 is divisible by 2
Are 132 and 198 relatively prime?
No, both are divisible by 2.No, both are divisible by 2.No, both are divisible by 2.No, both are divisible by 2.
Is 45 divisible by 6?
No,because 45 is not divisible by both 2 and 3. In order for a number to be divisible by 6 it has to be divisible by both 2 and 3.
If a number is divisble by both 2 and 3 is it divisible by 6?
Any number that is divisible by both 2 and 3 is divisible by 6.
Is 1142 divisible by 2 and 3?
2 yes, 3 no.
Is 432 divisible by 2 or 3?
432 is divisible evenly by both 2 and 3
Is 90 divisible by 2 or 3?
The number 90 is divisible by both 2 and 3.
Is 5232 divisible by 6?
Since 5232 is divisible by both 2 and 3, it is divisible by 6.A number must be divisible by both 2 and 3 to be divisible by 6.The number 5232 is even, so it is divisible by 2.If you add the individual digits in the number (5+2+3+2=12) you get a number that is divisible by 3, meaning the original number (5232) is also divisible by 3.
Is 78 divisible by 2 3 or both?
Both.
How do you know if 54132 is divisible by 6?
To determine if a number is divisible by 6, it must be divisible by both 2 and 3. To determine if a number is divisible by 2, it should be even - in other words, it should end with 0, 2, 4, 6, or 8. To determine if a number is divisible by 3, the sum of its digits should be divisible by 3. 54,132 is an even number, so it is divisible by 2. 5 + 4 + 1 + 3 + 2 = 15, which is divisible by 3, so 54,132 is divisible by 3. Since 54,132 is divisible by both 2 and 3, it is divisible by 6.
What determines whether a number is divisible by both 2 and 6?
Something divisible by 6 has to be divisible by both 2 and 3. and that just answers that anything disible by 6 is divisible by 2
Are 132 and 198 relatively prime?
No, both are divisible by 2.No, both are divisible by 2.No, both are divisible by 2.No, both are divisible by 2.
Is 293 divisible by 6?
It is not divisible by 6. Note:If a number id divisible by 6 then it must be divisible by both 2 and 3.The above number is not divisible by 2 and 3 either.
What are the divisibility rule for 6?
To be divisible by 6, the number must be divisible by both 2 and 3:To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8};To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3.As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3.If the number is not divisible by 2 or 3 (or both) then the number is not divisible by 6.examples:126Last digit is even so it is divisible by 2 1 + 2 + 6 = 9 which is divisible by 3, so it is divisible by 3→ 126 is divisible by both 2 and 3, so it is divisible by 6124Last digit is even so it is divisible by 2 1 + 2 + 4 = 7 which is not divisible by 3, so it is not divisible by 3→ 126 is divisible by 2 but not divisible by 3, so it is not divisible by 6123Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 3 = 6 which is divisible by 3, so it is divisible by 3→ 123 is divisible by 3 but not divisible by 2, so it is not divisible by 6121Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 1 = 4 which is not divisible by 3, so it is not divisible by 3→ 121 is not divisible by either 2 or 3, so it is not divisible by 6
