\(x\) 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 \(\text{lcm}(x, y)/\gcd(x, y)\).
BDMO Secondary National 2021 #2
 Mehrab4226
 Posts:230
 Joined:Sat Jan 11, 2020 1:38 pm
 Location:Dhaka, Bangladesh
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
Henri Poincaré
Re: BDMO Secondary National 2021 #2
Assume, $gcd(x,y)=d$.Mehrab4226 wrote: ↑Sat Apr 10, 2021 12:38 pmLet $x$ and $y$ be positive integers such that $2(x+y)=G.C.D(x,y)+L.C.M(x,y)$. Find $L.C.M(x,y)/G.C.D(x,y)$.
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

 Posts:1
 Joined:Sun Apr 11, 2021 7:33 pm
Re: BDMO Secondary National 2021 #2
I have a different approach to this.
After your 4th line, we can do as follows:
2a+2b=1+ab
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,
a=2p1
and b=2q1
where p and q and positive integers.
So we get:
2(2p1+2q1)=1+(2p1)(2q1)
or, 2(2p+2q2)=4pq2p2q+2
or, 2p+2q2=2pqpq+1
or, 3p+3q3=2pq
or, 3(p+q1)=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:
3(p+3n1)=2p×3n
or, p+3n1=2pn
or, p1=2pn3n
or, p1=n(2p3)
or, n=(p1)/(2p3)
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
Now,
q=3n=3×1=3
And,
a=2p1=2×21=3
Again,
b=2q1=2×31=5
Hence,
L.C.M(x,y)/G.C.D(x,y)=abd/d=ab=3×5=15
After your 4th line, we can do as follows:
2a+2b=1+ab
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,
a=2p1
and b=2q1
where p and q and positive integers.
So we get:
2(2p1+2q1)=1+(2p1)(2q1)
or, 2(2p+2q2)=4pq2p2q+2
or, 2p+2q2=2pqpq+1
or, 3p+3q3=2pq
or, 3(p+q1)=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:
3(p+3n1)=2p×3n
or, p+3n1=2pn
or, p1=2pn3n
or, p1=n(2p3)
or, n=(p1)/(2p3)
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
Now,
q=3n=3×1=3
And,
a=2p1=2×21=3
Again,
b=2q1=2×31=5
Hence,
L.C.M(x,y)/G.C.D(x,y)=abd/d=ab=3×5=15