有 5 个标有 1 到 5 的袋子。给定袋子中的所有硬币重量相同。有些袋子有 10 克重的硬币，其他袋子有 11 克重的硬币。我从袋子 1 到袋子 5 中分别挑选 1、2、4、8、16 个硬币。它们的总重量为 323 克。那么有 11 克硬币的袋子标签的乘积是___。
(一) 15
(乙) 12
(三) 8
(四) 1答案：(乙)
解释：
There are 5 bags numbered 1 to 5.
We don't know how many bags contain 10 gm and
11 gm coins.
We only know that the total weights of coins is 323.
Now the idea here is to get 3 in the place of total
sum's unit digit.
Mark no 1 bag as having 11 gm coins.
Mark no 2 bag as having 10 gm coins.
Mark no 3 bag as having 11 gm coins.
Mark no 4 bag as having 11 gm coins.
Mark no 5 bag as having 10 gm coins.
Note: The above marking is done after getting false
results for some different permutations, the permutations
which were giving 3 in the unit place of the total sum.
Now, we have picked 1, 2, 4, 8, 16 coins respectively
from bags 1 to 5.
Hence total sum coming from each bag from 1 to 5 is 11,
20, 44, 88, 160 gm respectively.
For the above combination we are getting 3 as unit digit
in sum.
Lets find out the total sum, it's 11 + 20 + 44 + 88 + 160 = 323.
So it's coming right.
Now 11 gm coins containing bags are 1, 3 and 4.
Hence, the product is : 1 x 3 x 4 = 12. 替代解决方案：
有11g和10g硬币的袋子。从袋子 1、2、3、4、5 中取出的 1、2、4、8、16 个硬币的总重量为 323。
现在我们需要找到总和为 323g 的 10g 和 11g 硬币数量。
设 x 为重量 10g 的硬币数量，y 为重量 11g 的硬币数量
因此 10x + 11y = 323…………1
现在我们现在选择的硬币总数即 1+2+4+8+16=31。
因此 x + y = 31…………2
求解方程 1 和 2，我们得到 x=18，y=13。
因此袋子 2 和袋子 5(2+16=x=18) 有 10g 硬币，袋子 1、3 和袋子 4(1+4+8=y=13) 有 11g 硬币。
因此，具有 11 克硬币的袋子标签的乘积是 1 x 3 x 4 = 12。这个问题的测验