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Re: If a and b are both integers, how many possible solutions are there to [#permalink]
22 May 2011, 21:10

1

This post was BOOKMARKED

50 = 3a + 2b Since the sum of 3a and 2b is to be even (50) either both are odd or both are even. since 2b cant be odd, 3a and 2b should both be even. for 3a to be even, a can be : 2,4,6,8, and 0. however, 7 > |–a|

This means that -7<a<7 If a is positive/zero: it can have values 0,2,4 and 6. cant be 8. 4 solutions.

But we have to consider negative values for a too. Similar to above negative value of 3a will mean subtracting a number from 2a. for subtracting from an even number (2a) to result in 50, the product 3a has to end with an even digit. so a can be -6,-4,-2.

e.g. if a = -4, 3a= -12 and b= 31

we have 7 solutions uptil now, considering all possible values that 2 can take within this bracket of -7<a<7 and a and b being integers.

Note: We dont have to think of values of b, since a is the one with constraints. if b was -ve, a will have to be positive and a much higher valuse than 7, so these are not to be considerd. Answer is 7 possible solutions.

Re: If a and b are both integers, how many possible solutions are there to [#permalink]
23 May 2011, 17:35

Expert's post

2

This post was BOOKMARKED

u0422811 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

Remember that |–a| = |a| e.g. |–3| = 3 = |3| If |a| < 7, this implies that the absolute value of a is less than 7. So a could range from -6 to 6 (since a can only be an integer). Those are 13 values. Now look at this: 50 = 3a + 2b 50 - 3a = 2b (50 - 3a)/2 = b Since b has to be integer too, 50 - 3a must be divisible by 2. Since 50 is even, a should be even too (Even - Even = Even). From -6 to 6, there are 7 even numbers and 6 odd numbers. So there are 7 possible solutions (a = -6, b = 34; a = -4, b = 31 etc) _________________

Re: If a and b are both integers, how many possible solutions are there to [#permalink]
19 Jan 2015, 03:19

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Re: If a and b are both integers, how many possible solutions are there to [#permalink]
19 Jan 2015, 12:19

Expert's post

Hi All,

In this question, you can use Number properties to your advantage (as some of the others posters have pointed out). If you got "stuck" on this question, then you should note that the answer choices are relatively small and the math is basic arithmetic. This means that you can use "brute force" to get to the answer. You can easily determine the solutions by taking some notes and doing some basic math.

We're told: 1) A and B are both INTEGERS 2) 3A + 2B = 50 3) 7 > |-A|

We're asked for the total number of POSSIBLE solutions to the equation. From the answers, we know that there are at least 4 and at most 8. Let's find them....

Keep things simple at first...and take notes so that you can make deductions... IF.... A = 0, then B = 25

IF.... A = 1, then B = 23.5 This is NOT a possible solution (since B is NOT an integer). This tells us that A CANNOT be ODD, since that causes B to become a non-integer. From here on, we WON'T WASTE TIME testing ODD numbers for A....

IF.... A = 2, then B = 22

IF.... A = 4, then B = 19 Notice the pattern emerging: when A increases by 2, B decreases by 3.....

IF.... A = 6, then B = 16

We're not allowed to go any higher (the inequality limits us here). So far, we have 4 answers. The question NEVER stated that A and B couldn't be negative though....and that 'absolute value' IS there for a REASON.....

IF.... A = -2, then B = 28

IF... A = -4, then B = 31

IF... A = -6, then B = 34

Now, we're not allowed to go any lower (again, the inequality limits us here). We now have 3 additional answers.

Harvard asks you to write a post interview reflection (PIR) within 24 hours of your interview. Many have said that there is little you can do in this...