Factors are divisors. If you know the divisibility rules, you know that 80 is divisible by 1, 2, 4, 5 and 8. If you divide 80 by those numbers, you find the other half of the factor pairs.
(80,1)(40,2)(20,4)(16,5)(10,8)
Those for 1, 2, 4, 5 and 8.
By using the divisibility rules, I can tell that 864 is divisible by 2, 3, 4, 6, 8 and 9. By dividing those numbers into 864 I can create factor pairs, any of which I can use to start the tree. 864 432,2 216,2,2 108,2,2,2 54,2,2,2,2 27,2,2,2,2,2 9,3,2,2,2,2,2 3,3,3,2,2,2,2,2
You only have to test the numbers 1 through 5. If you know the rules of divisibility, you know that 3, 4 and 5 aren't factors.
Yes. Factor pairs are always repeated across pairs since factor pairs are certain kinds of pairs.
The prime factors of 75 are 3,5,5 as 3*5*5=75. This can be determined by testing 75 for divisibility by prime numbers starting from 2. Using the prime factorization you can determine that all the factor pairs are (3*5)*5=15*5 and 3*(5*5)=3*25
the factor pairs are 1x35 & 5x7
The factor pairs of 28 are :28,114,27,4
Sixteen factor pairs.
From the rules of divisibility, we know that 40 is divisible by 1, 2, 4 and 5. Divide those numbers into 40 to get the rest. (40,1)(20,2)(10,4)(8,5)
The factor pairs of 27 are 1x27 and 3x9.
The factor pairs of 57 are 3x19 and 1x57