Question

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Question: {Refer the pic attached}
It's a challenging one though..

For easiness, u can assume that (a,b)>0

[100 POINTS]

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Answer 1

Answer:-

Refer the attachment

Answer 2

$\mathfrak{\huge{\blue{\underline{\underline{AnswEr :}}}}}$

The six ordered pairs are :

❖ (1009, 2018)

❖ (2018, 1009)

❖ (1009 · 337, 674) =(350143, 674)

❖ (1009 · 1346, 673) = (1358114, 673)

❖ (674, 1009 · 337) = (674, 350143)

❖ (673, 1009 · 1346) = (673, 1358114)

$\mathfrak{\huge{\blue{\underline{\underline{ExplanaTion :}}}}}$

First rewrite the equation as

2 · 1009(a + b) = 3ab, and note that 1009

is prime, so at least one of a and b must be divisible by 1009.

If both a and b are divisible by 1009, say with a = 1009q, b = 1009r, then we have 2(q + r) = 3qr.

But qr ≥ q + r for integers q, r ≥ 2, so at least one of q, r is 1. This leads to the solutions :

q = 1, r = 2 and r = 1, q = 2, corresponding to the ordered pairs (a, b) = (1009, 2018)

and (a, b) = (2018, 1009).

In the remaining case, just one of a and b is divisible by 1009, say a = 1009q. This gives 2 · 1009(1009q + b) = 3 · 1009qb, which can be rewritten as

2 · 1009q = (3q − 2)b.

Because the prime 1009 does not divide b, it must divide 3q − 2; say 3q − 2 = 1009k. Then 1009k + 2 = 3 · 336k + k + 2 is divisible by 3, so

k ≡ 1 (mod 3). For k = 1, we get q = 337, a = 1009 · 337, b = 2q = 674. For k = 4,

we get q = 1346, a = 1009 · 1346, b = q/2 = 673.

We now show there is no solution

with k > 4. Assuming there is one, the corresponding value of q is greater than 1346,

and so the corresponding

$\huge{b = \frac{2q}{3q - 2} .1009}$

is less than 673. Because b is an integer, it follows that b ≤ 672, which implies

$\huge{ \frac{1}{b} ≥ \frac{1}{672} > \frac{3}{2018} }$

Contradicting,

$\huge{ \frac{1}{a} + \frac{1}{b} = \frac{3}{2018} }$

Finally, along with the two ordered pairs (a, b) for which a is divisible by 1009 and b is not, we get two more ordered pairs by interchanging a and b.

$<marquee> Thanks ❤$