15
A/5 + 3 = 2Add Subtract 3 from both sides:A/5 = -1Multiply both sides by 5:A = -5
Yes. Because 135 is divisible by both 3 an 5, that is 135 can be divided by both 3 and 5 with no remainder.
3 x 5 x 41
5 x 123 = 615
True. Since 615 ends in 5, it is divisible by 5. Since the sum 12, of the digits of 615, is divisible evenly by 3, 615 is divisible by 3.
0.0081
The positive integer factors of 615 are: 1, 3, 5, 15, 41, 123, 205, 615
615 = 3 x 5 x 41
15
1, 3, 5, 15, 41, 123, 205, 615.
1, 3, 5, 15, 41, 123, 205, 615
A/5 + 3 = 2Add Subtract 3 from both sides:A/5 = -1Multiply both sides by 5:A = -5
615/5 = 123
Yes. Because 135 is divisible by both 3 an 5, that is 135 can be divided by both 3 and 5 with no remainder.
615/5 = 123 5 goes into 615 123 times.
3 x 5 x 41