all perfect cubes (any number powered by 3) by adding or subtracting 1 become factor of 7 except base number as 7 and its factors.
7 is a factor of which number?
7
A factor of a number is any number which the original number can be divided by evenly. In this case, 29 divides by 7 to give 4 remainder 1. This remainder means that 29 does not divide evenly by 7. Thus 7 is not a factor of 29.
1 is the least common factor of the number 2, number 4 and number 7. 28 is the least common multiple of the number 2, number 4 and 7.
7
7 and/or 13
For 2 to be a factor, the number has to be even. If you multiply two odd numbers together, you get an odd number. 7 is odd, so if I multiply it by any odd number, I will get a number that doesn't have 2 as a factor (e.g. 7x5=35 which doesn't have 2 as a factor). Thus, the answer to the question is "no," not all multiples of 7 have 2 as a factor.
A factor = a number which can be divided into the given number with no remainder ie 1, 7, 49 7 * 7 = 49
The only number that could be both a factor and a multiple is 7 itself.
The prime factors of 147 are 3, 7 and 7. From those we can create all the other factors ... (3 x 7) and (7 x 7) which are 21 and 49. 1 is a factor of every number, and every number is a factor of itself. So all the factors of 147 are 1, 3, 7, 7, 21, 49 and 147
Since the number 7 is a prime number and not a factor of 26, the greatest common positive integer factor of 7 and 26 is 1.
7 because 7and 1 are the only factor since 7 is a prime number. 1 isn't a prime or composite number so it is 7.
2 is a factor of 14 but not a multiple of 7.
Factor each number into its prime factors. 14 = 2 * 7 49 = 7 * 7 We take all the prime factors they have in common, and multiply. In this case the only prime factor they have in common is 7, so that is the answer.
Yes. Since 7 is a prime number, you have found all the prime numbers, with 7 being the final prime factor.
1 and 7
7-c
7