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 Unusual Equation Problem (Posted on 2006-11-28)
Let us denote by [x] the greatest integer ≤ x.

How many positive integers p satisfy [p/95]=[p/97]?

How many positive integers q satisfy [q/2005]=[q/2007]?

 See The Solution Submitted by K Sengupta No Rating

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 no programming Comment 2 of 2 |
I obtained the same answers as Charlie:
2,303 for p, and 1,006,008 for q

Imagine a modified problem of r, where GI(r/5)=GI(r/7), and temporarily allow zero in addition to the positive integers.
GI function       0                    1                         2
left side         0,1,2,3,4           5,6,7,8,9              10,11,12,13,14
right side  0,1,2,3,4,5,6   7,8,9,10,11,12,13  14,15,16,17,18,19,20

If the GI function is 0, there are 5 possible values of r
If the GI function is 1, there are 3 possible values of r
If the GI function is 2, there is 1 possible value of r

So there can be 1 + 3 + 5 - 1 = 8 possible r's (subtract 1 for the value r=0 which is disallowed since we want only positive integers)

In general, 1+3+5+...+n  where n is odd is    ((n+1)/2)^2
So for p, the answer is ((95+1)/2)^2 - 1 = 2,303
So for q, the answer is ((2005+1)/2)^2 - 1 = 1,006,008
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I just reread Charlie's solution, and now I realize his sol'n did not rely on programming either; just for verification.

Edited on November 28, 2006, 4:10 pm
 Posted by Larry on 2006-11-28 16:07:15

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