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by rahul.s » Sun Jan 31, 2010 2:39 am
50 = 3a + 2b

7 > |-a|

If a and b are both integers, how many possible solutions are there to the system above?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

OA: D
Last edited by rahul.s on Sun Jan 31, 2010 2:55 am, edited 1 time in total.
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by ajith » Sun Jan 31, 2010 2:50 am
rahul.s wrote:50 = 3a + 2b

7 > |-a|

If a and b are both integers, how many possible solutions are there to the system above?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

OA: D
The options for a are {-6,-5,....0,.....4,5,6}

a cant be odd since if a is odd b cannot be an integer (3a+2b =50)

so a can be {-6,-4, -2,0,2,4,6}

So 7 possible solutions
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by sars72 » Sun Jan 31, 2010 2:55 am
i assume the second statement is 7 > mod(a) where mod stands for modulus

so, we have 50 = 3a+2b and -7<a<7

-> 2b = (50-3a) -> b = (50-3a)/ 2

Since b has to be an integer, the result of (50-3a) should be even i.e. divisible by 2

So, that leaves us with the possibile values of a as -6,-4,-2,0,2,4,6

Therefore there are 7 possible values of a and thus 7 pussible values for b

So, the answer is 7
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by rahul.s » Sun Jan 31, 2010 7:52 am
sars72 wrote:i assume the second statement is 7 > mod(a) where mod stands for modulus
yup. it's modulus(a).
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