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Q: What number has two digits have only six factors and is divisible by fifteen?

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A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.

To know if a number is composite without listing its factors, you can use these rules:All numbers that end with 2, 4, 6, 8, and 0 (even numbers) are divisible by 2.If the sum of the digits in a number is divisible by 3, the number is divisible by 3.If the last two digits are divisible by 4, the number is divisible by 4.All numbers that have 5 in the one's place are divisible by 5.If a number is divisible by 2 AND is divisible by 3, it is divisible by 6.If the last 3 digits of a number are divisible by 8, the number is divisible by 8.If the sum of the digits of number is divisible by 9, the number is divisible by 9.If a number ends wuth 0, the number is divisible by 10.

because it has more than two factors.

A number is divisible by 3 if the sum of its digits is divisible by 3.

If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.

A number is divisible by 3 only if the sum of its digits is. 1+1+2=4, which is not divisible by 3, therefore 112 is not.

Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.

81 has factors other than 1 and itself, so it is not a prime number. For example, it is divisible by 3. (You can check this by adding the digits of the number: 8 + 1 = 9. If the sum of the digits is divisible by 3, the number is also divisible by 3.) Therefore, 81 is a composite number.

To be divisible by 15, it must also be divisible by 3 and by 5. To be divisible by 3, the sum of the digits must be divisible by 3; to be divisible by 5, the number must end with a zero or a five. Considering all these criteria, I guess that number would be 1110.

21 has factors other than 1 and itself, so it is not a prime number. For example, it is divisible by 3. (You can check this by adding the digits of the number: 2 + 1 = 3. If the sum of the digits is divisible by 3, the number is also divisible by 3.) Therefore, 21 is a composite number.

Get the last two digits of the number. If that number is divisible by 4 then that number is divisible by 4. Example : 2393844 The last two digits are 44 and since 44/4 = 11 or 44 is divisible by 4 then the number 2,393,844 is divisible by 4.

9101997 has factors other than 1 and itself, so it is not a prime number. For example, it is divisible by 3. (You can check this by adding the digits of the number: 9 + 1 + 0 + 1 + 9 + 9 + 7 = 36, which is divisible by 3. If the sum of the digits is divisible by 3, the number is also divisible by 3.) Therefore, 9101997 is a composite number.