Question 201841: if a = 4b + 26 and b is a positive integer. the a could be divisible by all of the following except
a)2 b) 4 c) 5 d)6 e) 7
Found 2 solutions by Theo, jim_thompson5910:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 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.
Answer by jim_thompson5910(35256)  (Show Source):
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