tag:blogger.com,1999:blog-6234783319349274920.post5275916664951978545..comments2024-03-27T21:58:46.972-04:00Comments on FOLIO OLIO: SUNDAY #4589Ralph Henryhttp://www.blogger.com/profile/05079364726250352589noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6234783319349274920.post-72555161139382279142021-07-18T12:39:48.397-04:002021-07-18T12:39:48.397-04:00There are eleven of each denomination of coin:
11...There are eleven of each denomination of coin:<br /><br />11 x 50p = 5.50<br />11 x 10p = 1.10<br />11 x 5p = 0.55<br />11 x 2p = 0.22<br /><br />Total = 7.37<br /><br />The clue here was that the 'target' figure ends in a 7. If this were achieved by 7 x 1p = 0.07 and remembering that the highest denomination is 50p, we would not have enough to reach 7.37, since 7 x 50p = £3.50, and the target is more than twice that.<br /><br />The only other way to arrive at a 7 is to combine an odd number of 5ps and 2ps to produce figures that end with a 5 and a 2 respectively. <br /><br />Since with 6 x 12 = 12p we would run into the same problem as above i.e. not enough coins, so 11 x 2p = 0.22p is the next possible choice. Adding 11 x 5p and 11 x 50p leaves us 1.10p short of our target, which conveniently is 11 x 10p.Robinhttps://www.blogger.com/profile/03891037740276969786noreply@blogger.com