There are five bags of coins such that:
a) the weight of any coin (in ounces) is a positive integer less than or equal to w
b) all coins in a given bag weigh the same
c) the weight of a coin in bag #1 is different from the weight of a coin in bag #2
d) when one coin from bag #1, w coins from bag #2, w2 coins from bag #3, w3 coins from bag #4, and w4 coins from bag #5 are combined and weighed, the result is 2800 ounces
Determine analytically, the weight of a coin in each of the five bags.
Let y(k) represent the weight of a coin in bag #k.
Then, we must have:
y(1) + w * y(2) + w^2 * y(3) + w^3 * y(4) + w^4 * y(5) = 2800
Since the weight of each coin is at least 1, it follows that:
1+w+w^2+w^3+w^4 < 2800 ......... (i)
Also, the weight of a coin is at most w, so that:
w+w^2+w^3+w^4+w^5 > 2800 ......(ii)
If w=4, then from(ii), we observe that:
lhs =1364 <2800 = rhs. This leads to a contradiction.
If w=7, then from (i), we observe that:
lhs =2801 > 2800 = rhs. This leads to a contradiction.
Therefore, w= 5, or 6.
If w=5, then we have:
y(1) + 5 * y(2) + 5^2 * y(3) + 5^3 * y(4) + 5^4 * y(5) = 2800
=> y(1)== 0 (mod 5), so that: y(1)=5
Then, y(2) + 5 * y(3) + 5^2 * y(4) + 5^3 * y(4) = 559
=> y(2) == 4(mod 5), so that: y(2) = 4
Then, y(3) + 5 * y(4) + 5^2 * y(5) = 111
=> y(3) == 1 (mod 5), so that: y(3) =1
Then, y(4) + 5 * y(5) = 22
=> y(4) == 2(mod 5), so that: y(4) = 2
Then, y(5) = (22-2)/5 = 4
Accordingly, we have:
{y(1), y(2), y(3), y(4), y(5)} = (5,4,1,2,4)
If w=6, then we have:
y(1) + 6*y(2) + 6^2 * y(3) + 6^3 * y(4) + 6^4 *y(5) = 2800
=> y(1) == 4(mod 6), so that: y(2)=4
So, y(2) + 5 * y(3) + 5^2 * y(4) + 5^3 * y(5) = 466
=> y(2) == 4 (mod 6), so that: y(2) = 4
However, y(1) = y(2) = 4 violates the given conditions.
Accordingly, this case is NOT feasible.
Consequently, we must have:
y(1)=1, y(2)= 4, y(3)= 1, y(4)=2, and y(5)=4 as the only possible solution.
Edited on July 29, 2022, 2:07 am
Puzzle Solution
