Yes, because evenly divisible by 5 w/no remainder
840=2*2*2*3*5*7
32 is not a factor of 840.
actually 7 is a factor of 840, because when you divide 7 into 840 you get 120 with no remainders therefore making it a factor of 840.🤡
7*5*4*3*2 Is the longest factor string of 840
Stings with four factor that make 840
840 2, 420 2, 2, 210 2, 2, 2, 105 2, 2, 2, 5, 25 2, 2, 2, 5, 5, 5
2 x 420 3 x 280 4 x 210 5 x 168 6 x 140
The three factor strings of 840 refer to sets of three factors that multiply together to equal 840. For example, one possible factor string is (2, 3, 140), since (2 \times 3 \times 140 = 840). Other combinations include (7, 12, 10) and (5, 12, 14). The total number of distinct factor strings can vary based on the factors chosen.
The three-factor strings of 840 refer to the different ways to express 840 as a product of three integers. The factorization of 840 is (2^3 \times 3^1 \times 5^1 \times 7^1). Some examples of three-factor strings include (1 \times 1 \times 840), (2 \times 2 \times 210), and (4 \times 5 \times 42). There are multiple combinations, and to find all possible strings, you would generate all sets of three factors that multiply to 840.
840 = 2 x 2 x 2 x 3 x 5 x 7 OR 23 x 3 x 5 x 7
The GCF of 320 and 840 is 40.
Multiples of 840