superprismatic Posted November 7, 2011 Report Share Posted November 7, 2011 Suppose I have 2520 coins, each with a value of 1. I can easily make N piles of coins with each pile having the same value, for N=2,3,4,5,6,7,8,9,10. You may make a set of coins of any positive integer denominations you would like as long as the total value of all the coins is 2520. What is the minimum number of coins you could make this way such that you could split the coins into N equal- valued piles for each integer N from 2 to 10? For example, I can make 2269 coins which can satisfy the piles criteria as stated above: 2268 coins each having a value of 1, and one coin having a value of 252. But 2269 coins is unlikely to be minimal. By the way, 252 is the largest value a coin can be, otherwise 10 equal piles can't be made. I don't know the answer to this puzzle. I figured that I'd put it out on the Den and see what a bunch of good, imaginative puzzle-solvers can do with it! Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 7, 2011 Report Share Posted November 7, 2011 I think we only have to be concerned with 6-10. Since 2-5 can be made by combining the previous. Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 8, 2011 Report Share Posted November 8, 2011 (edited) 36 coins Smallest count for 10 stacks 10 coins of 252 then smallest count for 9 stacks would be 18 252 x 9 28 x 9 Each stack has to have at least two coins to also make 10 stacks... Keep dividing a 252 coin into N pieces until you get to 6 10+(9-1)+(8-1)+(7-1)+(6-1)=36 coins total Example: 252 35 35 28 28 17 10 8 7 252 45 35 28 28 28 4 252 60 45 35 28 252 60 45 35 28 252 60 45 35 28 252 60 45 35 28 Edited November 8, 2011 by curr3nt Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 8, 2011 Report Share Posted November 8, 2011 (edited) Im not sure if i understood the problem statement correctly.... 50 coins. Coin value x No of coins 84 x 10 60 x 10 45 x 10 35 x 10 28 x 10 Using above coins, all piles values for n=2 to 10 can be obtained. Edited November 8, 2011 by 14.swapnil.14 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 8, 2011 Report Share Posted November 8, 2011 to clear up any confusion (it took me a few read throughs as well to understand), the goal is to be able to get 10 piles of 252, 8 piles of 315, 7 piles of 360, etc. Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted November 8, 2011 Author Report Share Posted November 8, 2011 to clear up any confusion (it took me a few read throughs as well to understand), the goal is to be able to get 10 piles of 252, 8 piles of 315, 7 piles of 360, etc. Yes, you correctly understand what I meant. Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 8, 2011 Report Share Posted November 8, 2011 Haven't been able to reduce my coin count further... yet... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 8, 2011 Report Share Posted November 8, 2011 14swapnil14 solution definately works, but it seems high. I'm unsure about current's method, 8 doesn't devide 252. And even if it did, it may not leave the right currancy. [ Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 8, 2011 Report Share Posted November 8, 2011 36 coins 252 x 6 60 x 4 45 x 5 35 x 7 28 x 9 17 x 1 10 x 1 8 x 1 7 x 1 4 x 1 10 stacks 252 252 252 252 252 252 60+60+60+60+8+4 45+45+45+45+45+17+10 35+35+35+35+35+35+35+7 28+28+28+28+28+28+28+28+28 9 stacks 252+28 252+28 252+28 252+28 252+28 252+28 60+60+60+60+28+8+4 45+45+45+45+45+28+17+10 35+35+35+35+35+35+35+28+7 8 stacks 252+35+28 252+35+28 252+35+28 252+35+28 252+35+28 252+35+28 60+60+60+60+35+28+8+4 45+45+45+45+45+28+28+17+10+7 7 stacks 252+45+35+28 252+45+35+28 252+45+35+28 252+45+35+28 252+45+35+28 252+35+28+28+17 60+60+60+60+35+28+28+10+8+7+4 6 stacks 252+60+45+35+28 252+60+45+35+28 252+60+45+35+28 252+60+45+35+28 252+45+35+28+28+28+4 252+35+35+28+28+17+10+8+7 Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 8, 2011 Report Share Posted November 8, 2011 (edited) I get slightly different numbers but the same count of 36 coins. Minimum for 6 stacks = 6 coins 420 420 420 420 420 420 Change to 7 stacks = 12 coins 360 360 360 360 360 360 60 x6 Change to 8 stacks = 19 coins 315 315 315 315 315 315 60 x5 + 15 45 x7 Change to 9 stacks = 27 coins 280 280 280 280 280 280 60 x4 + 25 + 15 45 x6 + 10 35 x8 Change to 10 stacks = 36 coins 252 252 252 252 252 252 60 x3 + 32 + 25 + 15 45 x5 + 17 + 10 35 x7 + 7 28 x9 These coins can then be restacked back to 6. Combine 2 stacks from 10 to get the 5 stacks. Combine 2 stacks from 8 to get the 4 stacks. Combine 3 stacks from 9 to get the 3 stacks. Combine 4 stacks from 8 to get the 2 stacks. Edited November 8, 2011 by curr3nt Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 8, 2011 Report Share Posted November 8, 2011 take just 6 and 7... Each stack of 6 has to be at least two coins since one coin would be too large to work in a 7 stack. So the lowest count of coins would have to be 12. with just 7 and 8...each stack of 7 would have to be at least two coins for the same reason as above. So the lowest count of coins would have to be 14. with just 8 and 9...each stack of 8 would have to be at least two coins. Lowest count of coins is 16. with just 9 and 10...lowest count of coins 18 Each increase in stack count increases the coin count by at least the previous number of stacks because you have to take at least one coin out of a stack to make the value smaller. 6 stacks becomes 12 for 7 stacks. Then take one coin from each stack to make a 8th stack for 19 coins. Take one coin from each stack for a 9th stack for 28 coins and take one more coin from each stack for a 10th and 36 coins. Only way to reduce the coin count would be to take the 6 stack out of the equation... Can it be done since it shares factors with 8 and 9? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 8, 2011 Report Share Posted November 8, 2011 Not sure where the flaw in this logic is, but see next post for breakdown in fewer coins. take just 6 and 7... Each stack of 6 has to be at least two coins since one coin would be too large to work in a 7 stack. So the lowest count of coins would have to be 12. with just 7 and 8...each stack of 7 would have to be at least two coins for the same reason as above. So the lowest count of coins would have to be 14. with just 8 and 9...each stack of 8 would have to be at least two coins. Lowest count of coins is 16. with just 9 and 10...lowest count of coins 18 Each increase in stack count increases the coin count by at least the previous number of stacks because you have to take at least one coin out of a stack to make the value smaller. 6 stacks becomes 12 for 7 stacks. Then take one coin from each stack to make a 8th stack for 19 coins. Take one coin from each stack for a 9th stack for 28 coins and take one more coin from each stack for a 10th and 36 coins. Only way to reduce the coin count would be to take the 6 stack out of the equation... Can it be done since it shares factors with 8 and 9? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 8, 2011 Report Share Posted November 8, 2011 (edited) 31 coins: 252 252 252 252 252 252 112 112 105 105 105 60 60 60 56 56 28 28 28 20 20 20 7 7 7 3 3 3 1 1 1 6 Stacks: 252 112 56 252 112 56 252 105 60 3 252 105 60 3 252 105 60 3 252 28 28 28 20 20 20 7 7 7 1 1 1 7 Stacks: 252 105 3 252 105 3 252 105 3 252 60 28 20 252 60 28 20 252 60 28 20 112 112 56 56 7 7 7 1 1 1 8 Stacks 252 60 3 252 60 3 252 60 3 252 56 7 252 28 28 7 252 20 20 20 1 1 112 112 56 28 7 1 105 105 105 9 Stacks: 252 28 252 28 252 28 252 20 7 1 252 20 7 1 252 20 7 1 112 105 60 3 112 105 60 3 105 60 56 56 3 10 Stacks: 252 252 252 252 252 252 112 112 20 7 1 105 60 56 28 3 105 60 56 28 3 105 60 28 20 20 7 7 3 1 1 Seems like it could go lower though, stay tuned. Edited November 8, 2011 by MoMoney Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 8, 2011 Report Share Posted November 8, 2011 nice MoMoney Quote Link to comment Share on other sites More sharing options...
Question
superprismatic
Suppose I have 2520 coins, each with a value of 1.
I can easily make N piles of coins with each pile
having the same value, for N=2,3,4,5,6,7,8,9,10.
You may make a set of coins of any positive integer
denominations you would like as long as the total
value of all the coins is 2520. What is the
minimum number of coins you could make this way
such that you could split the coins into N equal-
valued piles for each integer N from 2 to 10?
For example, I can make 2269 coins which can
satisfy the piles criteria as stated above:
2268 coins each having a value of 1, and one
coin having a value of 252. But 2269 coins
is unlikely to be minimal. By the way, 252
is the largest value a coin can be, otherwise
10 equal piles can't be made.
I don't know the answer to this puzzle. I figured
that I'd put it out on the Den and see what a bunch
of good, imaginative puzzle-solvers can do with it!
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