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There are 6561 balls out of them 1 is heavy Find the min no of times the balls have to be weighed for finding out the heavy ball?
6561 if u notice is 3^8...n can be divided by 3(prime factor) only...we keep on dividing it into sets of 3 and in 1 weighing itself you can determine which set is heavy. Divide that set into further 3 lots...keep doing that..
6561
2187 2187 2187 1st weighing
729 729 729
243 243 243
81 81 81
27 27 27
9 9 9
3 3 3
1 1 1
ANswer is 8 attempts
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To find the number of balls in a ball pit containing 0.032 moles of balls, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) particles per mole. Multiplying 0.032 moles by Avogadro's number gives: [ 0.032 , \text{moles} \times 6.022 \times 10^{23} , \text{balls/mole} \approx 1.93 \times 10^{22} , \text{balls} ] Therefore, there are approximately (1.93 \times 10^{22}) balls in the ball pit.
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There are 6561 balls out of them 1 is heavyFind the min no of times the balls have to be weighed for finding out the haevy ball?
6561 2187 2187 2187 1st stage 729 729 729 2 stage 243 243 243 3 stage 81 81 81 4 stage 27 27 27 5 stage 9 9 9 6 stage 3 3 3 7 stage 1 1 1 8 stage
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What would 0.032 moles of balls in a ball pit be explain math?
To find the number of balls in a ball pit containing 0.032 moles of balls, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) particles per mole. Multiplying 0.032 moles by Avogadro's number gives: [ 0.032 , \text{moles} \times 6.022 \times 10^{23} , \text{balls/mole} \approx 1.93 \times 10^{22} , \text{balls} ] Therefore, there are approximately (1.93 \times 10^{22}) balls in the ball pit.
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