325 is divisible by 5 leaving no remainder
Yes - any number ending with a 0 or 5, including 325, is divisible by 5. In this instance, 325/5 = 65
It is divisible by 2 and 3. It isn't divisible by 5 and 10.
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.
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.
Yes - any number ending with a 0 or 5, including 325, is divisible by 5. In this instance, 325/5 = 65
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
650 is divisible by: 650 and 1 65 and 10 130 and 5 325 and 2
yep it is because 2 is divisible by 2 and 3 is divisible by 3 and 0 is divisible by 5 have fun suckers!
1, 5, 13, 25, 65, 325
It is divisible by 2 and 3. It isn't divisible by 5 and 10.
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
Yes, it is divisible by 2, 3 and 9, but is not exactly divisible by 5 or 10.
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.
20 is divisible by 2, 5 and 10. It is not exactly divisible by 3 or 9.
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.