53000 in 2 part .both part divide by 6 like 9 two part 6 3 both dive by 3 like this
+33615 Assuming you mean that the two parts must be integers, then there is no way of doing this.
Assume the two integers are a and b.
a + b = 53000
Both a and b are exactly divisible by 6, so let a = 6m and b = 6n where m and n are also integers. Then
6m + 6n = 53000
m + n = 53000/6
Because m and n are integers the LHS must be an integer. However, the RHS, 53000/6, is not an integer (53000/6 ≈ 8833.333).
Hence the initial assumption that there are two integers, both exactly divisible by 6, that sum to 53000, is false.
.
+33615 Assuming you mean that the two parts must be integers, then there is no way of doing this.
Assume the two integers are a and b.
a + b = 53000
Both a and b are exactly divisible by 6, so let a = 6m and b = 6n where m and n are also integers. Then
6m + 6n = 53000
m + n = 53000/6
Because m and n are integers the LHS must be an integer. However, the RHS, 53000/6, is not an integer (53000/6 ≈ 8833.333).
Hence the initial assumption that there are two integers, both exactly divisible by 6, that sum to 53000, is false.
.
