Question 252666: If a and b are different positive integers, and 5a + 2b = 34 is an integer,
what is the sum of all possible values of a?
A) 6 B) 10 C) 12 D) 14 E) 16
Found 2 solutions by richwmiller, palanisamy:
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! 5a+2b=34
Answers could be a=6 and b=2
2b must be even but b can be even or odd.
so if 2b must be even then 5a must be even so a must be even since 5 is odd and 34 is even.
If you subtract an even number from an even number, you get an even number.
a must be less than 7 since 5*7=35 and both a and b must be positive.
But the answers are sums of a not a itself
10 is possible because 5*4+2*7=34
6+4=10
5*2+2*12=34
so we could have 6,4 and 2 for a total of 12
since 6, 4 and 2 are possible values for a 14 can't be since it would need 2 twice.
16 has a similar problem with 4
since we already have a sum of 12 a sum of 16 would need 4 again.
so the highest sum possible for a would be 12 using 2,4 and 6. We could also have a =0 and b would equal 17 but the sum of the a's would not change.
Answer by palanisamy(496)  (Show Source):
You can put this solution on YOUR website! Given, a and b are different positive integers,
and 5a + 2b = 34 is an integer,
5a = 34-2b
a = (34-2b)/5
In order that (34-2b)/5 tobe a positive integer, 34-2b must be a multiple of 5.
The possible values of b are 2,7,12 and 17.
If b = 2, then a = (34-4)/5 = 30/5 = 6
If b = 7, then a = (34-14)/5 = 20/5 = 4
If b = 12, then a = (34-24)/5 = 10/5 = 2
If b = 17, then a = (34-34)/5 = 0
Therefore the sum of the values of a = 6+4+2 = 12
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