An algebra problem by Isaac Wright
If x + y = 3 6 and x 2 + y 2 = 1 0 9 8 , what is the value of x y ?
The answer is 99.
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4 solutions
Simple approach
So you post solutions to level zero problems also!
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Yes. Any Problem?
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Nope, well, It has an advantage of up votes!
I hope you don't take this comment seriously :(
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@Swapnil Das – I am not in the set of people who die for upvotes. eg: Sravanth Chebrolu.
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@Nihar Mahajan – No, no I meant you deserve upvotes, sorry if the comment hurts:(
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@Swapnil Das – Everyone deserves upvotes for his/her own solution.
Good solution
x + y = 3 6 x 2 + y 2 = 1 0 9 8 x 2 + y 2 = 1 0 9 8
x 2 + 2 x y + y 2 = 1 0 9 8 + 2 x y
( x + y ) 2 = 1 0 9 8 + 2 x y
3 6 2 = 1 0 9 8 + 2 x y
1 2 9 6 = 1 0 9 8 + 2 x y
1 9 8 = 2 x y
x y = 9 9
If you start with 35= x and 1= y , it matches the first rule but not the second ( 3 5 2 + 1 2 =1226.) You try the same for 34 and 2. It doesn't work. 3 and 33 however do work (33+3=36, 3 3 2 + 3 2 =1098, 3 3 × 3 = 99 ) This is one of many approaches, and the most simple for a non-expert in algebra
Such a solution is acceptable because it was proven and understandable. Trial and error is always an acceptable approach. Great problem Isaac, and good solution!
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Trial and Error is acceptable to a degree. If there are a limited number of cases to consider, then trial and error definitely is an acceptable way to find the answer. However, in his solution, he did not go through every single positive integer ordered pair ( x , y ) with x + y = 3 6 , he only kept on going until he found numbers that worked. Without showing that none of the other numbers can possibly be fit the second condition (or there are more ordered pairs, but with the same product x y ), his solution cannot be considered complete.
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This was actually my first ever solution so I didn't actually think about it this way. Thanks for the helpful tips!
How do we know x and y are positive integers? They could be irrational.
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I agree. How about fractional numbers? (Though that might not be possible, But still.)
What do you mean? It say positive integers???
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@Isaac Wright Don't try to act over smart. We all read that the question initially stated "Find the value of xy".
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@Mehul Arora – :) Sorry, just messin
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@Isaac Wright – You don't actually have to state that x and y are positive integers. Your solution however didn't show why they should be positive integers,
@Isaac Wright This is bashing. Such a solution is not acceptable. For a proper solution, See @Nihar Mahajan 's Solution.
(x+y) = 36
(x+y)^2 = 36^2 = 1296
(x+y)^2 = x2 + y2 +2xy
x2+y2 = 1098
so : 1098+ 2xy = 1296
xy= [ (1296-1098)/2]
xy= 99 :)
( x + y ) 2 = x 2 + y 2 + 2 x y ⇒ ( x + y ) 2 − ( x 2 + y 2 ) = 2 x y ⇒ 3 6 2 − 1 0 9 8 = 2 x y ⇒ 1 9 8 = 2 x y ⇒ x y = 9 9