No, it is divisible by 1, 3, 13, 39.
390 is divisible by 3: 390/3 = 130 390 is not divisible by 9: 390/9 = 43.3 recurring (that is, 43.3333..) 390 is divisible by 10: 390/10 = 39
It is divisible by 2 and 3. It isn't divisible by 5 and 10.
2, 3, 6, and 39
9 is divisible by 3 10 is divisible by 2, 5 and 10. Everything else, no.
No, it is divisible by 1, 3, 13, 39.
390 is divisible by 3: 390/3 = 130 390 is not divisible by 9: 390/9 = 43.3 recurring (that is, 43.3333..) 390 is divisible by 10: 390/10 = 39
It is divisible by 2 and 3. It isn't divisible by 5 and 10.
2, 3, 6, and 39
It is evenly divisible by 2 and 10 but not by 3 and 9.
Yes, it is divisible by 2, 3 and 9, but is not exactly divisible by 5 or 10.
9 is divisible by 3 10 is divisible by 2, 5 and 10. Everything else, no.
20 is divisible by 2, 5 and 10. It is not exactly divisible by 3 or 9.
It is divisible by 2, 5 and 10, but not 3 and 9.
composite 3 goes into 3 or 30 3 goes into 9 so 3 goes into 39 13 + 13 + 13 = 39 = 3 x 13 ------------------------------------- Applying a divisibility test: 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. For 39 this gives: 39 → 3 + 9 = 12 12 → 1 + 2 = 3 3 is one of {3, 6,9} so 39 is divisible by 3. As 39 is divisible by 3 (that is 3 is a factor of 39 that is less than 39) and 3 is greater than 1 then 39 is a composite number.
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
Both 26 & 39 are exactly divisible by 13. 26/13=2 and 39/13=3. Therefore 26/39 in its simplest form is 2/3