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.
(In reply to
solution by Charlie)
Really nice approach!
This is the same basic idea but without involving bases. If w1,w2, ..w5<=w are the weights, then:
w1+w2*w+w3*w^2+w4*w^3+w5*w^4=2800
Now:
w1+w2*w+w3*w^2+w4*w^3+w5*w^4<1+w+w^2+w^3+w^4=(w^5-1)/(w-1)
and
w1+w2*w+w3*w^2+w4*w^3+w5*w^4>=w+w^2+w^3+w^4+w^5=w(w^5-1)/(w-1)
therefore:
(w^5-1)/(w-1)<2800<=w(w^5-1)/(w-1)
Only w=5 and w=6 satisfy this condition. The rest is easy: say for w=5 we calculate the largest w5 for which w5*5^4<2800, w5=4, then w4 in the same way from the reminder and so on.
Edited on February 22, 2007, 3:32 am
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Posted by Art M
on 2007-02-22 03:30:32 |
re: solution
