Determine the total number of real y with 1 ≤ y ≤ 2007 satisfying this equation:

[y/2] + [y/3] + [y/8] + [y/9] = 77y/72

where [p] denotes the greatest integer ≤ p.

Hi!

Let notice for the beginning that 77/72 = 1/2+1/3+1/8+1/9

So because [x]<=x the equality can be true only when

[y/2] = y/2 ; [y/3]=y/3 ; [y/8]= y/8 ; [y/9]=y/9;

So y = multiple of 2 and 3 and 8 and 9.

But if y is multiple of 8, of course is multiple of 2,

and

if y is multiple of 9 of course is multiple of 3

so :

y is multiple of 8*9 = 72

So the number are the multiple of 72, mean :

y={72,144,216,288,....,1944}

The total number of solution is from 1*72 to 27*72

mean 27 numbers!

 

 

Posted by Chesca Ciprian on 2007-11-04 13:50:29