Can we relate this?
⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ a + b = 8 a b + c + d = 2 3 a d + b c = 2 8 c d = 1 2
Let the above system of equations have n sets of real solutions. Let these solutions be ( a 1 , b 1 , c 1 , d 1 ) , … , ( a n , b n , c n , d n ) . Find the value of
n + i = 1 ∑ n ( a i + b i + c i + d i ) .
The answer is 66.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
@Sharky Kesa how do you got to know about that we can transform an equation like this which you have made. generally after seeing this question one would square ,subtract, divide to get the values. please help over this
Log in to reply
Doing tonnes of expansion problems can give you the upper hand when something weird like this question appears. This question was based simply off the observation above.
Log in to reply
Same way.
Also solution upvoted(+1).
One more thing, from which book you study and practice olympiad questions?
Log in to reply
@Priyanshu Mishra – Too many to count. I have a wide variety of them from which I solve.
Log in to reply
@Sharky Kesa – OK.
But tell me name of only one geometry book you use.
Log in to reply
@Priyanshu Mishra – Geometry Revisited by Coxeter and Grietzer.
Log in to reply
Log in to reply
@Priyanshu Mishra – Number Theory? Problem Solving Tactics - AMT
Log in to reply
@Sharky Kesa – Full form of AMT?
Log in to reply
@Priyanshu Mishra – Australian Mathematics Trust. Only available in Australia. Sorry.
Log in to reply
@Sharky Kesa – Thanks for the information.
You can also use the book Number theory-structures, examples by Titu Andreescu . It is a very well written book on N.T.
Substituting b = 8 − a and d = 2 3 − c − a ( 8 − a ) into the third equation, we obtain ( a − 4 ) [ 2 c − ( a 2 − 4 a + 7 ) ] = 0 If a = 4 , we deduce that b = 4 and c + d = 7 , c d = 1 2 . We obtain the solutions ( 4 , 4 , 3 , 4 ) and ( 4 , 4 , 4 , 3 ) .
Otherwise we have c = 2 1 ( a 2 − 4 a + 7 ) and hence d = 2 1 ( a 2 − 1 2 a + 3 9 ) . Thus 0 = 4 c d − 4 8 = ( a − 3 ) 2 ( a − 5 ) 2 leading to the solutions ( 3 , 5 , 2 , 6 ) and ( 5 , 3 , 6 , 2 ) . These four solutions yield the answer 6 6 .
How did u get the equation in the 2nd line
Log in to reply
As I said, I substituted expressions for b and d in terms of a and c into the equation a d + b c = 2 8 , and tidied up.
Did almost the same way as you did, sir.I found it quite difficult to factorise.How did you do it so easily,sir?
Notice the following
( x 2 + a x + c ) ( x 2 + b x + d ) = x 4 + ( a + b ) x 3 + ( a b + c + d ) x 2 + ( a d + b c ) x + c d
Substituting these values, we get
P ( x ) = x 4 + 8 x 3 + 2 3 x 2 + 2 8 x + 1 2 = ( x + 1 ) ( x + 2 ) 2 ( x + 3 ) = ( x 2 + 4 x + 4 ) ( x 2 + 4 x + 3 ) = ( x 2 + 3 x + 2 ) ( x 2 + 5 x + 6 )
From this, we get the possible solution cases are ( 4 , 4 , 3 , 4 ) , ( 4 , 4 , 4 , 3 ) , ( 3 , 5 , 2 , 6 ) , ( 5 , 3 , 6 , 2 ) , so the answer is 66.