BDMO Secondary National 2021 #2

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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BDMO Secondary National 2021 #2

Unread post by Mehrab4226 » Sat Apr 10, 2021 12:38 pm

\(x\) আর \(y\) এমন ধনাত্মক পূর্ণসংখ্যা যেন \(2(x+y) = \gcd(x, y) + \text{lcm}(x, y)\) হয়। \(\text{lcm}(x, y)/\gcd(x, y)\)-এর মান বের করো।

Let \(x\) and \(y\) be positive integers such that \(2(x+y) = \gcd(x, y) + \text{lcm}(x, y)\). Find \(\text{lcm}(x, y)/\gcd(x, y)\).
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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Re: BDMO Secondary National 2021 #2

Unread post by Pro_GRMR » Sat Apr 10, 2021 5:29 pm

Mehrab4226 wrote:
Sat Apr 10, 2021 12:38 pm
Let $x$ and $y$ be positive integers such that $2(x+y)=G.C.D(x,y)+L.C.M(x,y)$. Find $L.C.M(x,y)/G.C.D(x,y)$.
Assume, $gcd(x,y)=d$.
So, $x = ad$ and $y = bd$ Such that a, b are coprime. Also, $lcm(x, y) = abd$.
We get:
$2(x+y)=G.C.D(x,y)+L.C.M(x,y)$. Find $L.C.M(x,y)/G.C.D(x,y)$
$\Longrightarrow 2(ad+bd) = d + abd$
$\Longrightarrow 2d(a+b) = d(1 + ab)$
$\Longrightarrow 2a+2b = 1 + ab$
$\Longrightarrow ab - 2a - 2b + 1 = 0$
$\Longrightarrow ab - 2a - 2b + 4 = 3$
$\Longrightarrow a(b - 2) - 2(b - 2) = 3$
$\Longrightarrow (a - 2)(b - 2)= 3$
We see that $3$ has only two factors $1$ and $3$
Thus, $a - 2 = 1$ or $a - 2 = 3$
$\Longrightarrow a = 3, 5$
$\Longrightarrow (a,b) = (3, 5), (5,3)$
$\Longrightarrow ab = 15$
Notice that,
$L.C.M(x,y)/G.C.D(x,y) = abd/d = ab = \boxed{15}$
"When you change the way you look at things, the things you look at change." - Max Planck

J. M. Akibur Rahman
Joined:Sun Apr 11, 2021 7:33 pm

Re: BDMO Secondary National 2021 #2

Unread post by J. M. Akibur Rahman » Sun Apr 11, 2021 11:31 pm

I have a different approach to this.
After your 4th line, we can do as follows:
or, 2(a+b)=1+ab
Here, L.H.S. is even. So R.H.S. should also be even. That is possible only if a and b both are odd.
So let's assume,
and b=2q-1
where p and q and positive integers.
So we get:
or, 2(2p+2q-2)=4pq-2p-2q+2
or, 2p+2q-2=2pq-p-q+1
or, 3p+3q-3=2pq
or, 3(p+q-1)=2pq
Here, L.H.S. is divisible by 3. Hence R.H.S. should also be divisible by 3. In that case, one of the p and q must be divisible by 3.
Let's assume,
q is divisible by 3.
So we can write q=3n
where n is a positive integer.
We get:
or, p+3n-1=2pn
or, p-1=2pn-3n
or, p-1=n(2p-3)
or, n=(p-1)/(2p-3)
As n and p are positive integers,
for p=1, we get n=0 which is not acceptable.
for p=2, we get n=1 which is acceptable.
for other possible values of p, n becomes a proper fraction which is not acceptable.
So p=2 and n=1

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