What is the smallest value of a + b, if a and b are positive integers such that..?

Question

What is the smallest value of a + b, if a and b are positive integers such that

1/(2a) + 1/(3a) + 1/(3a) = 1/(b² - 2b)

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Hi. I've asked this question already, but I was still confused by the answers given.

I simplified the expression to
7/(6a) = 1/(b² - 2b)

Then a = [(7b²-14b)/6], so 6a = (7b²-14b)

I don't know how to continue from here. Thanks. Or perhaps someone has a better way of solving this?

Answer 1

　

You correctly solved for a ----> a = (7b²−14b)/6



This is a quadratic expression, that has minimum value when b = 1

So for b > 1, as b increases, then so does a.



However, we have the following restrictions on integers a and b:

a > 0, b > 0

a = (7b²−14b)/6 > 0 ----> b < 0 or b > 2 ----> b > 2



Since a increases as b increases (for b > 4/7, and therefore for b > 2), we must find smallest integer value of b > 2 such that a = (7b²−14b)/6 is an integer.



Since a is an integer, 6 divides 7b²−14b = 7b(b−2) ----> 6 divides b(b−2)

----> 2 divides b(b−2) and 3 divides b(b−2)

b and b−2 are both even or both odd, so they must be both even to be divisible by 2

One of b or b−2 is divisible by 3. So either b = 6 or b−2 = 6 ----> smallest value of b = 6



b = 6

a = 7(6)(6−2)/6 = 28



Smallest value of a + b = 28 + 6 = 34

Answer 2

1/(2a) + 1/(3a) + 1/(3a) = 1/(b²-2b)



1/(2a) + 1/(3a) + 1/(3a) = 7/(6a) > 0, so b²-2b > 0

b > 2

b²-2b = b(b-2) = 3, 8, 15, 24, 35, 48, 63, ...



7/(6a) = 1/(b(b-2))

a = 7(b(b-2))/6, so b is divisible by 6.

b = 6

a = 28

a+b = 34

Answer 3

7/(6a)=1/(b^2-2b)=>6a/7=b(b-2).

taking b=6, 6a/7=24=>a=28.

These are the 1st +ve integers for a

& b satisfying the equation.

Thus the smallest a+b=28+6=34

Answer 4

It seems obvious to me, that "a" has to be a multiple of 7. I would try a = 7, 14, 21, 28, etc. For each value of "a", figure the corresponding values of "b". I get the first integer solution at a = 28, b = 6.

Answer 5

6a = (7b²-14b) Here a and b are both positive where b>2

a=0 where b=2, and a is negative where 0
a = (7b²-14b)/6 so (b²-2b) must be divisible by 6 for the integer a.

Checking b=6, a=28



da/db = (14/6)(b-1), which is always increasing for b>2,

so the smallest value for a+b (where a and b are positive integers) is 28+6 = 34 ← ← ← 𝐴𝑛𝑠𝑤𝑒𝑟

Answer 6

1/(2a) + 1/(3a) + 1/(3a) = 1/(b² - 2b)

3/(6a) + 2/(6a) + 2/(6a) = 1/(b² - 2b)

7/(6a) = 1/(b² - 2b)

7(b² - 2b) = 6a

a = 7/6(b² - 2b)

a and b must be a positive integer, (b² - 2b) must be positive, so b must be 3 or greater.

(b² - 2b) must also be a multiple of 6 to keep 7/6 of it an integer.

plug and chug, b = 3 then (b² - 2b) = 3, b = 4 then (b² - 2b) = 8, b = 5 then (b² - 2b) = 15, b = 6 then (b² - 2b) = 24. Look like b = 6 and a = 7/6(24) = 28



a = 28, b = 6



but you better check with others to make sure.