杏吧原创

Solution to Enigma No. 1606

Answer: 53 聽聽聽聽聽聽

The winner is Tony Griffin of Etobicoke, Ontario, Canada.

Worked solution

The script states that 鈥渋f you could determine the total weight of the balls, you could work out the number of balls there are鈥.

However, if the total weight were 120 grams, there could be 2 balls @ 60g; 3 @ 40g; 4 @ 30g; 5 @ 24g; 6 @ 20g etc 鈥 so, in this case, the total weight fails to reveal a definitive number of balls.

But if the total weight were 121 grams, there are two solutions 鈥 but both are the same: 11 balls @ 11 grams, giving the same definitive solution either way round.

So the weight must be a squared prime number of grams (P2).

So possible numbers (or weight) of balls can only be:聽聽聽聽

P = 3, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, (67)

giving, respectively, total weights of:聽聽聽聽

P2 =9,49,121,169,289,361,529,841,961,1369,1681,1849,2209,2809,3481,3721,(4489)

Total weights above 4000 are impossible as there are no weights available to weigh them.

The weights for the balance are factors of 4000. So possible weight sizes are:

1,2,4,5,8,10,16, 20,25,32,40,50, 80,100,125,160,200,250, 400,500,800,1000 & 2000 grams.

Let鈥檚 consider the potential weighing process.聽聽聽聽聽

If, say, the weights were each 20g, these would produce a set of 200 鈥榮teps鈥 and every P2 would occupy its own step and be instantly identifiable 鈥 if the weights could be used. But without using them, no one individual P2 could be identified 鈥榳ith certainty鈥.

But if the weights were each 500g, the number of steps is reduced to 8, as many of the P2 would share steps as follows: 9-361; 529-961; 1369; 1681-1849; 2209; 2809; 3481; 3721. Even so, none of the individual P2, 1369, 2209, 2809, 3841 or 3721 can be selected 鈥榳ith certainty鈥.

However, if the weights were each 800g, the number of steps reduces to 5 as follows:

9-529; 841-1369; 1681-2209; 2809; 3481-3721. So with weights of 800g, one P2 (and only one), 2809, is identifiable 鈥榳ith certainty鈥.

And with 1000g and 2000g weights, no individual P2 is identifiable at all.聽聽聽聽聽

So 2809 is the only total weight which can be uniquely deduced from the information given. So there are 53 ball-bearings each weighing 53 grams 鈥 so the lock number is 53 too.

Number 53 opens the security lock.聽聽聽聽聽