None of them. 15 is divisible by 3 and 5
3, 9, 15, 21, 27, 33
Yes Example 1+5=6 and 6 is a multiple of 3
Yes, it is. The digital root of 1430061, that is, the sum of all its digits is 1+4+3+0+0+6+1 = 15 The digital root of 15 is 1+5 = 6 which is divisible by 3. So 1430061 is divisible by 3.
To determine which number is divisible by 3, 6, and 9, we need to check if the sum of the digits of each number is divisible by 3. For 369: 3+6+9 = 18, which is divisible by 3, 6, and 9. Therefore, 369 is divisible by 3, 6, and 9. For 246: 2+4+6 = 12, which is divisible by 3 but not by 6 or 9. Therefore, 246 is divisible by 3 but not by 6 or 9. For 468: 4+6+8 = 18, which is divisible by 3, 6, and 9. Therefore, 468 is divisible by 3, 6, and 9. For 429: 4+2+9 = 15, which is divisible by 3 but not by 6 or 9. Therefore, 429 is divisible by 3 but not by 6 or 9. Therefore, the numbers 369 and 468 are divisible by 3, 6, and 9.
If this is a T-F question, the answer is false. It is true that if a number is divisible by 6, it also divisible by 3. This is true because 6 is divisible by 3. However, the converse -- If a number is divisible by 3, it is divisible by 6, is false. A counterexample is 15. 15 is divisible by 3, but not by 6. It becomes clearer if you split the question into its two parts. A number is divisible by 6 if it is divisible by 3? False. It must also be divisible by 2. A number is divisible by 6 only if it is divisible by 3? True.
Multiples of 9 and 6 are also divisible by three, the reverse is not true. 15 is divisible by 3, but not 6 or 9. 27 is divisible by 3 and 9, but not 6. 12 is divisible by 3 and 6, but not 9. 54 is divisible by 3, 6 and 9.
Yes. 90 / 6 = 15
15
They are divisible by 2, 3, 4, 5, 6, 10, 12, 15, 30 and 60 at least.
How can the following definition be written correctly as a biconditional statement? An odd integer is an integer that is not divisible by two. (A+ answer) An integer is odd if and only if it is not divisible by two
6 of them.
None of them. 15 is divisible by 3 and 5
No. But 870 is divisible by 3 (because 8+7+0 = 15 -> 1 + 5 = 6 and 6 is divisible by 3). 870/3 = 290.
105
a number is 6. All numbers divisible by both 4 and 15 are the multiples of 4 and 15. Since 4 and 15 have no prime factor in common (4 = 22, 15 = 3 · 5), then the least common multiple of 4 and 15 is equal to their product, namely 4 · 15 = 60. Every other multiple of 4 and 15 is divisible by 60. Thus, if 60 is divisible by a number then all the multiples of 4 and 15 are divisible by that number. Therefore, it is enough to check by which number in the given options is 60 divisible. Only 6 divides 60 and none of the other do. The answer must be (A).
3, 9, 15, 21, 27, 33