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The ratio of GCD and LCM of two integers are 1:36 and sum of the integers is 5460. What is the difference between these two integers?

This problem was posted earlier. You can find it through this link here below:
http://www.matholympiad.org.bd/forum/vi ... =13&t=2919

thnx.................

Is ther any problem here?
We assume two integers: (a,b) where, a=x×y and, b=y×z, such that: gcd(x,z)=1 and y is the GCD of these two integers. So, the sum is: (a+b)=(x×y)+(y×z)=y(x+z)=5460 and, differnece: (a-b)=(x×y)-(y×z)=y(x-z)
So, gcd(a,b)=y and, lcm=(a,b)=x×y×z. Then,
lcm(a,b)/gcd(a,b) = 36/1 [reverse]
(x×y×z)/y = 36
x×z = 36 = 1×36 = 4×9.
If, (x×z)=1×36, then, (x+z)=37, y=5460/37. But, 5460 is congruent to 21 modulo 37. So, (x+z) isn't equal to 37.
Then, (x×z)=4×9, then, (x+z)=13, y=5460/13=420.
So, (a-b) = y(x-z) = 420 (9-4) = 420×5 = 2100.

Your solution is correct. But please latex your posts. It's mandatoey in the BdMO forum.

Tasnood wrote:Is ther any problem here?
We assume two integers: (a,b) where, a=x×y and, b=y×z, such that: gcd(x,z)=1 and y is the GCD of these two integers. So, the sum is: (a+b)=(x×y)+(y×z)=y(x+z)=5460 and, differnece: (a-b)=(x×y)-(y×z)=y(x-z)
So, gcd(a,b)=y and, lcm=(a,b)=x×y×z. Then,
lcm(a,b)/gcd(a,b) = 36/1 [reverse]
(x×y×z)/y = 36
x×z = 36 = 1×36 = 4×9.
If, (x×z)=1×36, then, (x+z)=37, y=5460/37. But, 5460 is congruent to 21 modulo 37. So, (x+z) isn't equal to 37.
Then, (x×z)=4×9, then, (x+z)=13, y=5460/13=420.
So, (a-b) = y(x-z) = 420 (9-4) = 420×5 = 2100.

Please, use LaTex to write your solution. If you are new in BdMO forum, take a look: viewtopic.php?f=25&t=2 .