71Didn't thought of such beauty of this sum.. !! Thanks!
Q:=> The no of ordered pair (x,y) satisfying \left[\frac{x}{2} \right]+\left[\frac{2x}{3} \right]+\left[\frac{y}{4} \right]+\left[\frac{4y}{5} \right] = \frac{7x}{6}+\frac{21y}{20}
where 0<x,y<30 and [ . ] is " GIF ".
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6 Answers
17 x6 = x2 + 2 x3
So , 7 x6 - [ x2 ] - [ 2 x3 ] = { x2 } + { 2 x3 }
Similarly observing that , 21 y20 = y4 + 4 y5
We may write down the given equation as -
{ x2 } + { 2 x3 } + { y4 } + { 4 y5 } = 0
Note that { b } ≥ 0 for all " b " , so each of these 4 terms must be zero .
Since , { a } = 0 if and only if , " a " is an integer , hence for each of these terms to be zero , " x " must be a multiple of " 6 " , and " y " has to be a multiple of " 20 " .
Accordingly , the total no. of ordered pairs of ( x , y ) are 4 .
71
1waise though i think ricky left this part "PURPOSELY"...coz it does not matter much for this problem but still for sake of completeness eliminate the cases like
1.{x/2}+{2x/3}>1
{y/4}+{4y/5}<1
2.
{x/2}+{2x/3}>1
{y/4}+{4y/5}>1
3.{x/2}+{2x/3}<1
{y/4}+{4y/5}>1
Clearly ALL OF THE ABOVE CASES GET REJECTED IF WE OBSERVE THE GIVEN EQUATION CAREFULLY!
and now the problem is COMPLETELY CRACKED!