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Q: Is 2561 divisible by 5 and 3?

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No, look at 5 it is odd and not divisible by 3. false,because if you look at the 5 it is not divisible by 3.

2390 is not divisible by 3 but is divisible by 5. 2 + 3 + 9 + 0 = 14; 1 + 4 = 5: 5 is not divisible by 3, so 14 is not divisible by 3, so 2390 is not divisible by 3. 2390 ends with a 0, so 2390 is divisible by 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

15 is divisible by 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.

Yes 5 + 5 + 5 = 15 which is divisible by 3, so 555 is divisible by 3. (555 ÷ 3 = 185.)

yep it is because 2 is divisible by 2 and 3 is divisible by 3 and 0 is divisible by 5 have fun suckers!

It is divisible by 2 and 3. It isn't divisible by 5 and 10.

3*8*5=120, therefore 120 is divisible by 3, 8 and 5.

It is divisible by 2 & 3. It is not divisible by 5, 9 & 10. 534 is even → divisible by 2 5 + 3 + 4 = 12 → divisible by 3 534 does not end in 0 or 5 → not divisible by 5 5 + 3 + 4 = 12 → not divisible by 9 534 does not end in 0 → not divisible by 10

15 is divisible by 3 and 5, because 15/3=5, and 15/5=3.

780 equally divisible by both 3 and 5. 320 by 5 not by 3

It is divisible by 3.

Only 315. To be divisible by 3 numbers have to add to a number divisible by 3 (3+1+5 =9). To be disable by 5, number has to end in a 5 or 0

A number is divisible by 15 if it is divisible by 5, and by 3. For example, 165 is divisible by 15, since it ends in 5, and the sum of its digits is divisible by 3.

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.

It is divisible by 3 and 5 leaving no remainders

The sum of its digits is 1+2+0 = 3 which is divisible by 3. So 120 is divisible by 3. The number 120 ends in a 0 and so it is divisible by 5.

It is divisible by 3 because 381/3 = 127

It not divisible by 5 because it does not end 5 or 0 . . . . . . . . . . It is not divisable by 6 because it is not even . . . . . . . . . It is divisible by 3 because all the digits add up to a multiple of 3 (1+5+9= 15) . . . . . . . . So really 159 is only divisible by 3

True. Since 615 ends in 5, it is divisible by 5. Since the sum 12, of the digits of 615, is divisible evenly by 3, 615 is divisible by 3.

No. divisible by 3 4 5 is 60

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

One of the infinitely many integers divisible by 2, 3 and 5 is 3000.

It is divisible by 2, 5 and 10, but not 3 and 9.