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Find the number of ordered pairs \(a, b\) of integers such that\(\frac{a + 2}{a + 5} = \frac{b}{4}.\)

 Apr 18, 2024
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Multiply both sides of the equation by (a+5):

 

a+2=b⋅4a+5​

 

Multiply both sides by 4:

 

4(a+2)=b(a+5)

 

Expand the left side:

 

4a+8=b(a+5)

 

This equation represents the following:

 

4a+8 is a multiple of b (since it's equal to b times (a+5))

 

a+5 is a divisor of 4a+8 (since the right side is a multiple of the left side)

 

Since we're looking for integer solutions for (a,b), let's analyze the divisibility conditions for both sides:

 

For 4a+8 to be divisible by b, b must be a divisor of 8. The divisors of 8 are 1, 2, 4, and 8.

 

For a+5 to be a divisor of 4a+8, we need to check divisibility for each possible value of b (divisors of 8):

 

If b=1, then a+5 must divide 4a + 8. In this case, the only integer solution for a is -5 (since 4a + 8 = -a + 8, which is divisible by -1 when a = -5).

 

If b=2, then a+5 must divide 4a + 8. This doesn't have any integer solutions for a, because the left side (always even) cannot be equal to the right

side (always odd).

 

If b=4, then a+5 must divide 4a + 8. Here, a = -3 is the only integer solution (since 4a + 8 = a + 8, which is divisible by 4 when a = -3).

 

If b=8, then a+5 must divide 4a + 8. The only integer solution for a is -1 (since 4a + 8 = 3a + 8, which is divisible by 8 when a = -1).

 

So we found possible integer solutions for (a,b) as:

 

(-5, 1)

 

(-3, 4)

 

(-1, 8)

 

There are a total of 3​ such ordered pairs.

 Apr 18, 2024

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