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Going Greatest With 2007 (Posted on 2007-11-04) Difficulty: 2 of 5
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.

See The Solution Submitted by K Sengupta    
Rating: 2.7500 (4 votes)

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Solution Solution | Comment 2 of 4 |

The total number of real y is 27.

lcd(1/2, 1/3, 1/8, 1/9) = 72.
And, in fact  1/2 + 1/3 + 1/8 + 1/9 = 77/72; therefore y must be a multiple of 72.  Using the floor function, |_ 2007/72 _|, we find the value 27, thus (72 x 27) is the largest value of y.

Where 1 <= y <= 2007,
y = {(72 x 1); (72 x 2); (72 x 3); ....; (72 x 27)}


  Posted by Dej Mar on 2007-11-04 17:16:22
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