
nuggets
The old answer was 43, my answer is 44, here is why:
1)First let's solve this: Find the smallest integer N(N>0), where
N=6X+9Y+20Z
and X, Y ,Z are nonnegative integers.
The solution would be X=1, Y=Z=0, and N=6;
2)Next we need to make sure N+1, N+2, N+3, N+4, N+5
can be expressed as 6X+9Y+20Z as well. (another pack of 6 will take care of N+6). Consider the following:
1 = 2*6+920
2 = 202*9
3 = 96
4 = 4*620
5 = 2069
In other words, to buy one more nugget, we need to add 2 packs of 6, one pack of 9, and remove 1 pack of 20; To buy 2 more nuggets, we would add 1 pack of 20, and remove 2 packs of 9, and so on.
Thus Y and Z can not be 0, instead, we must have
Y=2 and Z=1 ==>
N=6X+9Y+20Z=6*1+9*2+20*1=44
Pan, Wenyu
Monday, December 16, 2002
actually i believe the answer should be 41, since,
20 + 9 + 6 + 6 = 41,
i'm assuming the question is looking for the smallest number N, equal to or greater than which any number of nuggets can be bought.
since 43 nuggets can be bought (20 + 6*4 + 9) and 42 (6*7), AND 41, that should be the correct answer
ubaid dhiyan
Monday, December 16, 2002
ubaid,
How can you buy 43? 20+6*4+9 = 53
Pan, Wenyu
Monday, December 16, 2002
Recent Topics
Fog Creek Home
