Suppose that a and b are integers such that 3b=8−2a.How many of the first six positive integers must be divisors of 2b+12?
+6252 3b=8−2a2b+12=23(8−2a)+12=16−4a+363=52−4a3If I understand the question were are looking for the values of a∈{1,2,3,4,5,6} such that 52−4a3∈Z
52−4a3=16+4−4a3 so we want values of a∋4−4a3∈Z4−4a={0,−4,−8,−12,−16,−20} and the two of those divisible by 3 are 0 and -12corresponding to a=1, a=4 so there are 2 values of a that satisfy the original statement
.+6252 3b=8−2a2b+12=23(8−2a)+12=16−4a+363=52−4a3If I understand the question were are looking for the values of a∈{1,2,3,4,5,6} such that 52−4a3∈Z
52−4a3=16+4−4a3 so we want values of a∋4−4a3∈Z4−4a={0,−4,−8,−12,−16,−20} and the two of those divisible by 3 are 0 and -12corresponding to a=1, a=4 so there are 2 values of a that satisfy the original statement
