No.
35 is not divisible by 3.
753 is not divisible by 35. The factors of 753 are 1, 3, 251, 753.
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.
Yes it is, and the answer to this is 35.
35 remainder 3
No. 35 is divisible by: 1, 5, 7, 35.
this # is divisible by 3, 5, and 9 :-) ;-P
It is: 315
21 and 35 are both divisible by 7 21/35 = 3/5
No. 5, 20, 25, 35, etc
315 is divisible by these numbers: 1 3 5 7 9 15 21 35 45 63 105 and 315.
It is 315.