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Q: What number is divisible by 2 3 5 9?

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The smallest number divisible by 2 3 5 is 30.

30 is divisible by, and is the least common multiple of, 2, 3, and 5.

Is 5225 divisible by 3? A number is divisible by 3, if the sum of its digits is evenly divisible by 3. 5 + 2 + 2 + 5 = 14. Since 14 is not divisible by 3, neither is 5225.

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.

30

30

no. in order for a number to be divisible by 3 its numbers have to have a sum of a number divisible by 3. EX. (302) 3+0+2 = 5. 5 is not divisible by 3. Therefore 302 is not divisible by 3.

If you want your answer to be a whole number, it is divisible by 2 and 5.

A number that is divisible by 15 is divisible both by 5 and 3 A number is divisible by 5 if it ends with 0 or 5 A number is divisible by 3 if the sum of its digits is divisible by 3 e.g. 4035 is divisible by 15 as it ends with a 5 and 4+0+3+5=12 which is divisible by 3

No, just 2, 3 and 5.

120 is one such number.

There are no numbers that satisfy this. If a number is divisible by both 2 and 5, then it must also be divisible by 10.

There is none. Any number that is divisible by 2 and 5, for example, must also be divisible by 10.

Of 2, 3, 4, 5, and 8, the smallest number divisible by 2 is 2.

60 is divisible by 3, 4, 5, and 6.

no.. heres a trick: if you add together the numbers (5+2+4) and that answer equals a number that is divisible by 3 then the number is divisible by three 5+2+4=11 11 is not divisible by 3 therefore, 524 is not divisible by 3

7 is not divisible by any of them.

How about 27

15 is divisible by 3 and 5.

60

60

30!

The number 44 is divisible by: 2 - 44/2 = 22 4 - 44/4 = 11 The number is not divisible by 3, 5, 6, or 10.

No. To be divisible by 6, the number must be divisible by both 2 and 3. To be divisible by 2 the number must be even, ie its last digit must be one of {0, 2, 4, 6, 8}; 105's last digit is 5 which is odd so 105 is not divisible by 2. To be divisible by 3, sum the digits of the number and if the result is divisible by 3, then so is the original number. For 1-5: 1 + 0 + 5 = 6 which is divisible by 3 therefore 105 is divisible by 3. Although 105 is divisible by 3 it is not divisible by 2, thus it is not divisible by 6.

180 is divisible by 2, 3, 4, 5, 9, and 10.