Hard fraction (Help me)
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From: donno on 28 Jan 2010 08:44 Hello Everone I try to solve this problem 2 days. But can't solve. :P Please help me if you can solve.! In the picture a, b, c, d, e, f, g, h, i, j are different digit (0-9) http://img25.imageshack.us/img25/1208/hardfractionproblem.png ab mean a number have 2 digit = 10a + b find a, b, c, d, e, f, g, h, i, j at least 5 solutions. TIA
From: g.resta on 28 Jan 2010 10:38 On Jan 28, 2:44 pm, do...(a)abc.def wrote: > Hello Everone > > I try to solve this problem 2 days. > > But can't solve. :P > > Please help me if you can solve.! > > In the picture a, b, c, d, e, f, g, h, i, j are different digit (0-9) > > http://img25.imageshack.us/img25/1208/hardfractionproblem.png > > ab mean a number have 2 digit = 10a + b > > find a, b, c, d, e, f, g, h, i, j at least 5 solutions. Ok. This is a cryptarithm. You have AB/CD + EFG/GHI = 1 To solve this kind of problems you can use my on-line Alphametics creator / on-line Cryptarithms solver at http://www.iread.it/cryptarithms.php However, my program does not work on fractions, but only on integers so you have to manipulate a little your equation. By multiplicating everything by CD*GHI you obtain this equation that you can plug in my program: AB*HIJ + CD*EFG = CD*HIJ There are a lot of solutions. I've copied some below. For example, the first one leads to 36/81 + 540/972 = 1 bye, g. BJDGAHICEF ---------- 6210397854 7510294863 7820536914 9570246831 5201476938 5201396748 8261470935 8261390745 4861270953 4971560823 8502146937 0863752914
From: donno on 28 Jan 2010 11:07 On 28-Jan-2010, "g.resta(a)iit.cnr.it" <g.resta(a)iit.cnr.it> wrote: > Path: > border1.nntp.dca.giganews.com!nntp.giganews.com!nx01.iad01.newshosting.com!newshosting.com!198.186.194.249.MISMATCH!news-out.readnews.com!transit3.readnews.com!postnews.google.com!k35g2000yqb.googlegroups.com!not-for-mail > From: "g.resta(a)iit.cnr.it" <g.resta(a)iit.cnr.it> > Newsgroups: sci.math > Subject: Re: Hard fraction (Help me) > Date: Thu, 28 Jan 2010 07:38:40 -0800 (PST) > Organization: http://groups.google.com > Lines: 55 > Message-ID: > <5cbeae81-43f7-4303-a500-3bc49f220ec2(a)k35g2000yqb.googlegroups.com> > References: <TsudnbkU56xICfzWnZ2dnVY3go-dnZ2d(a)giganews.com> > NNTP-Posting-Host: 82.49.164.153 > Mime-Version: 1.0 > Content-Type: text/plain; charset=ISO-8859-1 > Content-Transfer-Encoding: quoted-printable > X-Trace: posting.google.com 1264693120 29351 127.0.0.1 (28 Jan 2010 > 15:38:40 GMT) > X-Complaints-To: groups-abuse(a)google.com > NNTP-Posting-Date: Thu, 28 Jan 2010 15:38:40 +0000 (UTC) > Complaints-To: groups-abuse(a)google.com > Injection-Info: k35g2000yqb.googlegroups.com; posting-host=82.49.164.153; > posting-account=J59fvAkAAADIKX4kGvgiDNbRO_xXLDqE > User-Agent: G2/1.0 > X-HTTP-UserAgent: Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US; > rv:1.9.2) > Gecko/20100115 Firefox/3.6 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Bytes: 2429 > Xref: number.nntp.dca.giganews.com sci.math:1274080 > > On Jan 28, 2:44�pm, do...(a)abc.def wrote: > > Hello Everone > > > > I try to solve this problem 2 days. > > > > But can't solve. :P > > > > Please help me if you can solve.! > > > > In the picture a, b, c, d, e, f, g, h, i, j are different digit (0-9) > > > > http://img25.imageshack.us/img25/1208/hardfractionproblem.png > > > > ab mean a number have 2 digit = 10a + b > > > > find a, b, c, d, e, f, g, h, i, j at least 5 solutions. > > Ok. This is a cryptarithm. > > You have AB/CD + EFG/GHI = 1 > > To solve this kind of problems you can use my > on-line Alphametics creator / on-line Cryptarithms solver at > > http://www.iread.it/cryptarithms.php > > However, my program does not work on fractions, but only > on integers so you have to manipulate a little your > equation. By multiplicating everything by CD*GHI you obtain > this equation that you can plug in my program: > > AB*HIJ + CD*EFG = CD*HIJ > > There are a lot of solutions. I've copied some below. > For example, the first one leads to > > 36/81 + 540/972 = 1 > > bye, > g. > > BJDGAHICEF > ---------- > 6210397854 > 7510294863 > 7820536914 > 9570246831 > 5201476938 > 5201396748 > 8261470935 > 8261390745 > 4861270953 > 4971560823 > 8502146937 > 0863752914 Thank you very much for many solution g.resta :)))))) I will bookmakrs your website .! However if anyone know how to solve this problem by other method. Please tell me.
From: Robert Israel on 28 Jan 2010 12:53 "g.resta(a)iit.cnr.it" <g.resta(a)iit.cnr.it> writes: > On Jan 28, 2:44=A0pm, do...(a)abc.def wrote: > > Hello Everone > > > > I try to solve this problem 2 days. > > > > But can't solve. :P > > > > Please help me if you can solve.! > > > > In the picture a, b, c, d, e, f, g, h, i, j are different digit (0-9) > > > > http://img25.imageshack.us/img25/1208/hardfractionproblem.png > > > > ab mean a number have 2 digit =3D 10a + b > > > > find a, b, c, d, e, f, g, h, i, j at least 5 solutions. > > Ok. This is a cryptarithm. > > You have AB/CD + EFG/GHI =3D 1 > > To solve this kind of problems you can use my > on-line Alphametics creator / on-line Cryptarithms solver at > > http://www.iread.it/cryptarithms.php > > However, my program does not work on fractions, but only > on integers so you have to manipulate a little your > equation. By multiplicating everything by CD*GHI you obtain > this equation that you can plug in my program: > > AB*HIJ + CD*EFG =3D CD*HIJ > > There are a lot of solutions. I've copied some below. > For example, the first one leads to > > 36/81 + 540/972 =3D 1 > > bye, > g. > > BJDGAHICEF > ---------- > 6210397854 > 7510294863 > 7820536914 > 9570246831 > 5201476938 > 5201396748 > 8261470935 > 8261390745 > 4861270953 > 4971560823 > 8502146937 > 0863752914 Or you could use my alphametic applet at <http://www.math.ubc.ca/~israel/applet/metic/metic.html> which allows input in the form AB/CD + EFG/HIJ = 1 and produces the output 10/45 + 287/369 = 1 10/45 + 728/936 = 1 10/96 + 473/528 = 1 70/96 + 143/528 = 1 10/28 + 369/574 = 1 31/62 + 485/970 = 1 21/96 + 375/480 = 1 32/80 + 417/695 = 1 12/60 + 748/935 = 1 12/54 + 609/783 = 1 12/96 + 735/840 = 1 12/96 + 357/408 = 1 42/87 + 315/609 = 1 32/48 + 169/507 = 1 13/52 + 678/904 = 1 13/26 + 485/970 = 1 34/51 + 269/807 = 1 24/63 + 507/819 = 1 24/96 + 531/708 = 1 54/87 + 231/609 = 1 74/89 + 105/623 = 1 35/70 + 481/962 = 1 45/90 + 381/762 = 1 15/30 + 486/972 = 1 45/90 + 186/372 = 1 35/70 + 148/296 = 1 45/90 + 138/276 = 1 45/61 + 208/793 = 1 36/81 + 540/972 = 1 36/81 + 405/729 = 1 16/32 + 485/970 = 1 46/92 + 185/370 = 1 56/84 + 307/921 = 1 56/84 + 109/327 = 1 27/81 + 630/945 = 1 27/81 + 306/459 = 1 57/92 + 140/368 = 1 27/54 + 309/618 = 1 17/89 + 504/623 = 1 18/90 + 372/465 = 1 18/90 + 276/345 = 1 38/61 + 207/549 = 1 38/95 + 426/710 = 1 38/76 + 451/902 = 1 48/96 + 351/702 = 1 38/76 + 145/290 = 1 48/96 + 135/270 = 1 39/51 + 204/867 = 1 39/65 + 284/710 = 1 29/87 + 310/465 = 1 19/57 + 308/462 = 1 19/58 + 273/406 = 1 29/58 + 307/614 = 1 -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: donno on 28 Jan 2010 16:55
> Path: > border1.nntp.dca.giganews.com!nntp.giganews.com!news.glorb.com!news2.glorb.com!news.acm.uiuc.edu!not-for-mail > From: Robert Israel <israel(a)math.MyUniversitysInitials.ca> > Newsgroups: sci.math > Subject: Re: Hard fraction (Help me) > Date: Thu, 28 Jan 2010 11:53:16 -0600 > Organization: Usenet @ UIUC - http://www.acm.uiuc.edu/news/ > Lines: 120 > Sender: rbisrael(a)outside.world > Message-ID: <rbisrael.20100128174621$1987(a)news.acm.uiuc.edu> > References: <TsudnbkU56xICfzWnZ2dnVY3go-dnZ2d(a)giganews.com> > <5cbeae81-43f7-4303-a500-3bc49f220ec2(a)k35g2000yqb.googlegroups.com> > NNTP-Posting-Host: www1.acm.uiuc.edu > X-Trace: news.acm.uiuc.edu 1264701196 12194 172.22.35.81 (28 Jan 2010 > 17:53:16 GMT) > X-Complaints-To: news(a)acm.uiuc.edu > NNTP-Posting-Date: Thu, 28 Jan 2010 17:53:16 +0000 (UTC) > X-Local-Trace: rbisrael, 174.6.131.74, Thu Jan 28 11:53:15 2010 > X-Newsreader: News::Web v0.51 > Bytes: 3826 > Xref: number.nntp.dca.giganews.com sci.math:1274106 > > "g.resta(a)iit.cnr.it" <g.resta(a)iit.cnr.it> writes: > > > On Jan 28, 2:44=A0pm, do...(a)abc.def wrote: > > > Hello Everone > > > > > > I try to solve this problem 2 days. > > > > > > But can't solve. :P > > > > > > Please help me if you can solve.! > > > > > > In the picture a, b, c, d, e, f, g, h, i, j are different digit (0-9) > > > > > > http://img25.imageshack.us/img25/1208/hardfractionproblem.png > > > > > > ab mean a number have 2 digit =3D 10a + b > > > > > > find a, b, c, d, e, f, g, h, i, j at least 5 solutions. > > > > Ok. This is a cryptarithm. > > > > You have AB/CD + EFG/GHI =3D 1 > > > > To solve this kind of problems you can use my > > on-line Alphametics creator / on-line Cryptarithms solver at > > > > http://www.iread.it/cryptarithms.php > > > > However, my program does not work on fractions, but only > > on integers so you have to manipulate a little your > > equation. By multiplicating everything by CD*GHI you obtain > > this equation that you can plug in my program: > > > > AB*HIJ + CD*EFG =3D CD*HIJ > > > > There are a lot of solutions. I've copied some below. > > For example, the first one leads to > > > > 36/81 + 540/972 =3D 1 > > > > bye, > > g. > > > > BJDGAHICEF > > ---------- > > 6210397854 > > 7510294863 > > 7820536914 > > 9570246831 > > 5201476938 > > 5201396748 > > 8261470935 > > 8261390745 > > 4861270953 > > 4971560823 > > 8502146937 > > 0863752914 > > Or you could use my alphametic applet at > <http://www.math.ubc.ca/~israel/applet/metic/metic.html> > which allows input in the form AB/CD + EFG/HIJ = 1 > and produces the output > > 10/45 + 287/369 = 1 > 10/45 + 728/936 = 1 > 10/96 + 473/528 = 1 > 70/96 + 143/528 = 1 > 10/28 + 369/574 = 1 > 31/62 + 485/970 = 1 > 21/96 + 375/480 = 1 > 32/80 + 417/695 = 1 > 12/60 + 748/935 = 1 > 12/54 + 609/783 = 1 > 12/96 + 735/840 = 1 > 12/96 + 357/408 = 1 > 42/87 + 315/609 = 1 > 32/48 + 169/507 = 1 > 13/52 + 678/904 = 1 > 13/26 + 485/970 = 1 > 34/51 + 269/807 = 1 > 24/63 + 507/819 = 1 > 24/96 + 531/708 = 1 > 54/87 + 231/609 = 1 > 74/89 + 105/623 = 1 > 35/70 + 481/962 = 1 > 45/90 + 381/762 = 1 > 15/30 + 486/972 = 1 > 45/90 + 186/372 = 1 > 35/70 + 148/296 = 1 > 45/90 + 138/276 = 1 > 45/61 + 208/793 = 1 > 36/81 + 540/972 = 1 > 36/81 + 405/729 = 1 > 16/32 + 485/970 = 1 > 46/92 + 185/370 = 1 > 56/84 + 307/921 = 1 > 56/84 + 109/327 = 1 > 27/81 + 630/945 = 1 > 27/81 + 306/459 = 1 > 57/92 + 140/368 = 1 > 27/54 + 309/618 = 1 > 17/89 + 504/623 = 1 > 18/90 + 372/465 = 1 > 18/90 + 276/345 = 1 > 38/61 + 207/549 = 1 > 38/95 + 426/710 = 1 > 38/76 + 451/902 = 1 > 48/96 + 351/702 = 1 > 38/76 + 145/290 = 1 > 48/96 + 135/270 = 1 > 39/51 + 204/867 = 1 > 39/65 + 284/710 = 1 > 29/87 + 310/465 = 1 > 19/57 + 308/462 = 1 > 19/58 + 273/406 = 1 > 29/58 + 307/614 = 1 > -- > Robert Israel israel(a)math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada great website !!! Robert. :))) I will bookmarks your web. Thank you very much. However I still need another solution. If anyone can use any logic of mathematic. :P |