If its divisible by 5 AND 2 it must be divisible by 10
So you just have to pick the only number between 21 and 39 that's divisible by 10
All real numbers are divisible by four. However, no numbers with decimal extensions are evenly divisible by four. The only numbers evenly divisible by four between 30 and 39 are 32 and 36. Otherwise, including decimals, there would be an infinite number of rational numbers between those limits divisible by four, and infinitely more than that if the real numbers are included (numbers that cannot be represented by terminating or repeating continued decimal expansions).
There are 39 natural numbers between 1,000 and 1,500, inclusive, that are divisible by 13.
39 and 40
39
NB 3,9,21,39,63. We note that #1;The difference between 3 & 9 = 6 #2 ;The difference between 9 & 21 = 12 ( 2 x 6) #3 ; The difference between 21 & 39 = 18 ( 3 x 6) #4 ; The difference between 39 & 63 = 24 ( 4 x 6) So it follows #5 ; The difference between 63 & X = ( 5 x 6 = 30 ) So 63 + 30 = 93 The answer !!!!!
There are eight numbers between 0 and 50 that are divisible by three but not by two: 3, 9, 15, 21, 27, 33, 39, 45.
3,9,15,21,27,33, 39 and 45
40 is nearest to 39 and is exactly by divisible 4
Because it is divisible by numbers other than 1 and itself. 39 is divisible by 1, 3, 13, 39.
All multiples of 39, which is an infinite number.
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48.
The number 9 is the smallest, but there are an infinite number of answers to the question. 15, 21, 27, 33, 39, and 45 are some more.
9, 21, and 39 are divisible by three, so they are composite. 3, 17, and 41 have only two factors, 1 and itself, so they are prime.
yes it is divisible by 1,2,10,5 and 39
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.
12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, and 48. As you may have noticed, all you have to do is count every third number.
numbers divisible by 3; 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 ect...