DIY Challenge 3: 10 bars!
How many times will you need to use the weighing scale?
What students worked out7
Student 4
We need 2 or 3 weighings to do this. First we weigh ingots 12345 and 678910. If 678910 ingots are heavier, we weigh ingotsa 67 and 89. If both weigh the same, 10 is the real gold bar. If 67 is heavier than 89, we weigh 6 and 7. If 6 is heavier than 7, 6 is the real ingot.
If my answer is confusing, I can explain in the next class.
Student 2
it kinda is
Student 7
WE CAN DO IT IN THREE WHEIGHINGS
Student 2
first, we weigh 1,2,3,4,5 v.s 6,7,8,9,10
if 1,2,3,4,5 is heavier then we weigh
1,2 v.s 3,4
if one side is heavier eg:1,2
then we weigh 1 v.s 2
if they are equal then the answer is 5Student 1
Basically we first weigh 1,2,3,4,5vs6,7,8,9,10 for example 12345 are heavier we measure 1,3vs2,4 if equal then 5 is on side heavier the measure those to e.g 1,3 are heavier we do 1vs3.
Student 1
🙁
Student 3
for 10 bars minimum 3 weighing required
group 10 bars into 4 groups
123,456,78,9and101. 123vs456
2. 78vs9 and 10
3. weigh group separately which is heavierso minimum 3 weighing required without luck
same strategy can work for even 11 and 12 bars