Re: problem with EvenQ
- To: mathgroup at smc.vnet.net
- Subject: [mg105950] Re: [mg105923] problem with EvenQ
- From: "David Park" <djmpark at comcast.net>
- Date: Sun, 27 Dec 2009 02:26:49 -0500 (EST)
- References: <8534260.1261873319290.JavaMail.root@n11>
The trouble is that EvenQ[y] returns False unless y is manifestly an even integer. You could obtain a result in the following manner: FindInstance[2 y < 100 \[And] y == 2 x, {x, y}, Integers, 3] {{x -> 24, y -> 48}, {x -> -247, y -> -494}, {x -> -245, y -> -490}} Here is another example. Find some odd integers between 6 and 25. FindInstance[6 <= y <= 25 \[And] y == 2 x + 1, {x, y}, Integers, 4]; y /. % {23, 7, 21, 17} Here is the case using a RandomSeed. This gives different results each time you evaluate. FindInstance[6 <= y <= 25 \[And] y == 2 x + 1, {x, y}, Integers, 4, RandomSeed -> RandomInteger[{0, 100}]]; y /. % {13, 21, 11, 17} one example But if we ask for just a single instance then RandomSeed appears not to work. That looks like a bug to me. David Park djmpark at comcast.net http://home.comcast.net/~djmpark/ From: dvholten [mailto:info at dvholten.de] Hi folks, i puzzled all afternoon, but couldnt get this one solved: what is the proper way of using EvenQ[] within FindInstance[] ? I cant use it like FindInstance[ y*2 < 100 && EvenQ[y], {y}, Integers ] or even FindInstance[ EvenQ[y], {y}, Integers ] Actually, the expression used in FindInstance is much more complex, but i condensed the problem to be EvenQ[] - i expect some kind of special notation to help here. thanks dvh