Question 201841
the formula is a = 4b + 26
b can be any value as long as it's a positive integer.
a can be any value that presumably must also be a positive integer?
a being divisible by all implies that there is no remainder from the division?
that's what i'm thinking, anyway.
under that assumption, we should be able to prove that all of the numbers can divide evenly into a except one of them.
we make a table of possible value for a starting from b = 1 and working our way up sequentially one increment at a time.
b,a
1,30
2,34
3,38
4,42
5,46
6,50
7,54
8,58
9,62
10,66
11,70
12,74
13,78
14,82
since 2 divides into 30, 2 is good.
since 5 divides into 30, 5 is good.
since 6 divides into 30, 6 is good
that leaves 4 and 7.
multiples of 4 are:
4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,.....
doesn't look like 4 divides evenly.  they keep missing each other by 2.
multiples of 7 are:
7,14,21,28,35,42,49,56,63,70,.....
when b = 11, a = 70
70 is divisible by 7 so 7 is good.
looks like 4 is the one that will not divide evenly into a.
when b = 4, a = 42 which is also divisible by 7 (7*6)
answer is selection b.) which is 4.