tag:blogger.com,1999:blog-15628310.post1795863615201855304..comments2021-04-15T02:41:46.574-04:00Comments on Question of the day: Good Old Fashioned BankingAnonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-15628310.post-52646960177392008642012-04-03T16:25:24.664-04:002012-04-03T16:25:24.664-04:00Got it too, though in a weird binary equation way....Got it too, though in a weird binary equation way. 2x+10x+5y=100<br />=>12x+5y=100<br />=>12x=100-5y<br />=>12x=5(20-y)<br />=>20-12/5x=y<br />=>20-2.4x=y<br />now, we know that y is a whole no. therefore, 2.4x must be a whole no. The nearest number that makes the expression whole is if we assume x=5.<br />Therefore, using this value in the first equation, we get<br />12*5+5y=100<br />=>y=8<br />And the rest follows..Chaitalihttps://www.blogger.com/profile/04107875891164057947noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-64797757425649182912012-04-03T05:31:47.532-04:002012-04-03T05:31:47.532-04:00Anonymous is correct. It would be an interesting t...Anonymous is correct. It would be an interesting twist to the riddle, if you would change this sentence:<br /><br />"I need some two-dollar bills, ten times as many one-dollar bills"<br /><br />as follows:<br /><br />"I need some one-dollar bills, ten times as many two-dollar bills"<br /><br />This will make an interesting situation that there is no solution to the problem that includes $2 and $1 notes, but only 20 x $5 notes. Here is the reasoning for this changed scenario:<br /><br /><br />GET TO 100d!!!<br /><br />If we assume that John wants at least 1 x 1d bill, thus, (1) is not correct.<br />But since that is the only one that works out, John only gets 20 x 5d bills.<br />We assume that he gets whole bills only, thus, not 1.57 1d bills or something like that.<br /><br />REASONING: <br /><br />(2) - (5): remainder of 1d and 2d is not divisible by 5, so does not work out.<br /><br />(6): Amount of 1d and 2d is over 100d, thus that does not work out.<br /><br />We conclude that John gets only 20 x 5d bills and no 1d and 2d bills.<br /><br />(1)<br />0 x 1d = 0<br />0 x 2d = 0<br />20 x 5d = 100<br /><br />(2)<br />1 x 1d = 1<br />10 x 2d = 20<br />undefined x 5d = 79<br /><br />(3)<br />2 x 1d = 2<br />20 x 2d = 40<br />undefined x 5d = 58<br /><br />(4)<br />3 x 1d = 3<br />30 x 2d = 60<br />undefined x 5d = 37<br /><br />(5)<br />4 x 1d = 4<br />40 x 2d = 80<br />undefined x 5d = 16<br /><br />(6)<br />5 x 1d = 5<br />50 x 2d = 100<br />undefined x 5d = -5<br /><br />Good puzzle, thanks!<br /><br />KobusKobushttps://www.blogger.com/profile/15684206417271790460noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-24797402626023455582012-04-02T17:12:37.202-04:002012-04-02T17:12:37.202-04:005 two-dollar bills
50 one-dollar bills
8 five-doll...5 two-dollar bills<br />50 one-dollar bills<br />8 five-dollar billsAnonymousnoreply@blogger.com