Minus came up with ten possible combinations. Here they are:
2 blue, 4 green 2 black 4 blue 8 green |
1 black, 1 blue, 2 green 4 green, 2 blue 1 black, 2 blue 6 green, 1 blue |
1 black, 1 blue, 2 green 1 blue, 6 green 1 black, 4 green 4 blue |
2 blue, 1 black 1 blue, 6 green 2 blue, 4 green 4 blue |
1 black, 2 green, 1 blue 6 green, 1 blue 4 green, 1 black 4 blue |
2 green, 1 blue, 1 black 4 green, 1 black 4 green, 2 blue 2 green, 3 blue |
1 black, 2 blue 3 blue, 2 green 4 blue 8 green |
3 blue, 2 green 1 black, 4 green 8 green 2 black |
2 green, 1 black, 1 blue 4 green, 1 black 1 black, 2 blue 8 green |
4 green, 1 black 2 green, 3 blue 8 green 2 black |
vloney [vloney@northcom.net] started Minus off with a bit of a "palate-cleanser". We're not sure if Iva salivated, but Minus sure did!
What made Iva salivate was a tantalizing menu of delectable bejewelled dishes, 4 necklaces with the following ingredients: one with 8 green gemstones, one with 4 blue gemstones, one with 2 black gemstones, and one with 2 blue gemstones and 4 green gemstones.
Our feature chef this week is Daniel Sheehan of Christ Church Grammar School in Western Australia. Daniel cooked up a storm and provided Minus with the solid reasoning (sorta like seasoning?) behind his tasty dish.
We know that 2 greens = 1 blue and 2 blues = 1 black. 1 necklace is worth 4 blues. There are 4 necklaces and 20 gems in total.I know that a necklace can't have an odd number of green gems because if you do you would be left with a remainder of half a blue in the end.
The most 1 necklace can have is 8 green gems because 8 green gems are equal to 4 blues. That leaves us with 12 gems remaining with 3 necklaces to go.
I can use 4 green gems because 4 is even. That means I can use either 1 black gem or 2 blue ones. Since I am running out of gems I'll use 1 black gem with the 4 green gems. I now have 2 necklaces to go with 7 gems remaining.
I'll now use 2 green gems for this next necklace. This means I can use either 1 blue and 1 black or 3 blues. If I use 1 black and 1 blue the last necklace would have to have 1 black and 2 blues which would work. If I use 3 blues the last necklace would have to have 2 blacks which would also work. Therefore there are at least 2 solutions to this problem.
Solution 1 and 2: Necklace 1 has 8 green gems. Necklace 2 has 4 greens and 1 black gem Solution 1 only: Necklace 3 has 2 greens, 1 black and 1 blue gem Necklace 4 has 1 black and 2 blue gems Solution 2 only: Necklace 3 has 2 greens and 3 blue gems Necklace 4 has 2 black gems
Fifth graders, L.K. Benjamin, Lindsay Maidment, Nicholas Bradley and Tracy Wu, at South Scotland School in Laurinburg. NC, came up with this scrumptious mix of combinations.
1)8 green; 2) 1 black, 2 greens, 1 blue; 3) 2 blue, 1 black; 4) 1 black, 4 green
1) 4 greens, 2 blues 2) 4 greens, 1 black; 3) 2 greens, 3 blues; 4) 2 greens, 1 black, 1 blue
1) 2 blue, 1 black; 2) 4 greens, 1 black; 3) 4 blues 4) 8 greens
Minus got a horrible case of indigestion, chomping through all of this week's solutions. The big fella has gone to bed early, but don't worry... If your solution isn't featured on this week's menu, we'll plastic-wrap them and serve them to Minus as leftovers. Nothing goes to waste around here!
Special thanks go to:
Sean Bleibtrey, Dan McNerny, Joe Artuso, and Darcy WhitenightJulie Weaver, Andrea Brighton, Miranda Peer, and David Fritz
Jessica Oesterling, Tiffany Stickel, Jennifer Leu, and Tania Shaffer
John Kostelnik, Gary Vanderwater, Arthur Rodgers, Brian Gesalman, and John Lawrence
Jim Britz, Bria Bennie, Jason Pishney, and Jesse Harr
Kwok Wah [kwokwah@cadvision.com]
Jim Kearns [jkearns@wiu.k12.pa.us]
Dane Coalson
from Pulaski Middle SchoolCaroline Warburton
from Pulaski Middle SchoolBrenna Blevins
from Pulaski Middle SchoolJessica Brown
from Pulaski Middle SchoolDonald Norris
from Pulaski Middle SchoolSushant Saukar
from Chowgule College, Goa, IndiaAdam Harley & Thomas Pickersgill
from Kent Street Senior High School in Perth, Western Australia