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How many numbers between 100 and 3000 are divisible by 3?
One-third of them
What numbers are divisible by 3 and 9?
Oh, dude, numbers divisible by both 3 and 9 are multiples of the least common multiple of 3 and 9, which is 9. So, any number that is a multiple of 9 is also divisible by 3 and 9. Like, it's just basic math, nothing to lose sleep over.
Is 4908 divisible by 3?
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.
What are all the positive integers divisible by 2 or 3?
Well, darling, the positive integers divisible by 2 are all the even numbers, and the ones divisible by 3 are those multiples of 3. So, to find the numbers divisible by either 2 or 3, you just need to combine these two sets. In other words, it's the set of all even numbers and multiples of 3. Voilà!
624 divisible by 3?
yes.
Is 693 divisible by 3?
693 is evenly divisible by 3. You can tell that at a glance because every one of its digits is evenly divisible by 3.
Are all numbers divisible by -3 divisible by 3?
No, just every third one.
Which one of these are divisible by 3 94 73 123?
only 123 is divisible by 3. 123 / 3 = 41
Is 5193 a prime number?
No, it is divisible by 3.No, it is divisible by 3.No, it is divisible by 3.No, it is divisible by 3.
Explain why the product of any four consecutive integers is divisible by 24?
At least one of the four will be divisible by 4. One other will be divisible by 2 (but not by 4) One other will be divisible by 3. Hence the product will be divisible by 4 x 2 x 3 = 24.
What is 339 divisible by?
It is divisible by 3, for example.It is divisible by 3, for example.It is divisible by 3, for example.It is divisible by 3, for example.
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
What is divisible by 2 3 5?
One of the infinitely many integers divisible by 2, 3 and 5 is 3000.
Are all numbers that are divisible by 9 also divisible by 3 are all numbers that are divisible by 3 also divisible by 9 how do you know?
All numbers divisible by 9 are divisible by 3; since 9 = 3 x 3 all multiples of 9 are also multiples of 3. However, all numbers divisible by 3 are not divisible by 9, eg 6 = 2 x 3 but 6 is not divisible by 9 (since 6 is not a multiple of 9) - it only takes one counter example to disprove a theory.
What is 15345 divisible by?
3 is one factor.
Does 6 go evenly into 35?
No. To be divisible by 6 the number must be divisible by both 2 and 3. To be divisible by 2, the last digit of the number must be even (ie one of {0, 2, 4, 6, 8}). The last digit of 35 is 5 which is not even and so 35 is not divisible by 2. To be divisible by 3, sum the digits of the number and if this sum is divisible by 3 then so is the original number. As the test can be applied to the sum, repeatedly summing the digits of the sums until a single digit remains, then the original number is divisible by 3 only if this single digit is one of {3, 6, 9}. For 35 3 + 5 = 8 which is not divisible by 3 (nor is is one of {3, 6, 9}, thus 35 is not divisible by 3. As 35 is not divisible by both 2 and 3 (in fact it is divisible by neither 2 nor 3) it is not divisible by 6.
What a number between 400 and 500 that is divisible by 3 not 2?
To find out if a number is divisible by 3 add up all it's digest and check if the total is divisible by 3. to check if a number is divisible by 2 check if the last digit is divisible by 2. So one such number is 405.
