39 is a whole number as in not a fraction or decimal.
sqrt(39) = ±6, to the nearest whole number.
yes if you include decimal places.... any number is divisible except zero. The answer is 7.8
39
No, it is divisible by 1, 3, 13, 39.
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.
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
39 is a whole number as in not a fraction or decimal.
sqrt(39) = ±6, to the nearest whole number.
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.
yes it is divisible by 1,2,10,5 and 39
39 rounds to 40.
The next composite number after 38 is 39. It is divisible to 3 and 13 in addition to 1 and 39.
yes if you include decimal places.... any number is divisible except zero. The answer is 7.8
39
No, it is divisible by 1, 3, 13, 39.