# Gold and Nickel

You have 15 identical coins – 2 of them made of pure gold and the other 13 made of nickel (covered with thin gold layer to mislead you). You also have a gold detector, with which you can detect if in any group of coins, there is at least one gold coin or not. How can you find the pure gold coins with only 7 uses of the detector?

**SOLUTION**

First, we note that if we have 1 gold ball only, then we need:

- 1 measurement in a group of 2 balls
- 2 measurements in a group of 4 balls
- 3 measurements in a group of 8 balls

Start by measuring **1**, **2**, **3**, **4**, **5**.

- If there are gold balls in the group, then measure
**6**,**7**,**8**,**9**,**10**,**11**.- If there are gold balls in the group, then measure
**5**,**6**,**7**.- If there are no gold balls among them, then there is a gold ball among
**1**,**2**,**3**,**4**, and a gold ball among**8**,**9**,**10**,**11**, so we can find the gold balls with the remaining 2 measurements. - If there are gold balls in
**5**,**6**,**7**, then measure**5**,**8**,**9**. If there are gold balls there, then**5**must be gold, and we can find the other gold ball among**6**,**7**,**8**,**9**,**10**,**11**with the remaining 3 measurements. If there is no gold ball among**5**,**8**,**9**, then there is a gold ball among**1**,**2**,**3**,**4**, and a gold ball among**6**,**7**, so again we can find them with only 3 measurements.

- If there are no gold balls among them, then there is a gold ball among
- If there are no gold balls in the group, then measure
**5**,**12**,**13**.- If there are no gold balls among them, then measure
**14**,**15**. If none of them is gold, then measure individually**1**,**2**, and**3**to find which are the 2 gold balls among**1**,**2**,**3**,**4**. Otherwise, there is a gold ball among**1**,**2**,**3**,**4**, and among**14**,**15**, and we can find them with the remaining 3 measurements. - If there are gold balls among
**5**,**12**,**13**, then measure**5**,**14**,**15**. If none of them is gold, then there is a gold ball among**1**,**2**,**3**,**4**, and a gold ball among**12**,**13**, so we can find them with 3 measurements. Otherwise,**5**is gold, and again we can find the other gold ball among**1**,**2**,**3**,**4**,**12**,**13**,**14**,**15**with 3 measurements.

- If there are no gold balls among them, then measure

- If there are gold balls in the group, then measure
- If there are no gold balls among
**1**,**2**,**3**,**4**,**5**, then we measure**6**,**7**,**8**.- If there are gold balls in the group, then measure
**9**,**10**,**11**,**12**,**13**.- If there are no gold balls among them, we measure individually
**6**,**7**,**8**,**14**. - If there is a gold ball among
**9**,**10**,**11**,**12**,**13**, then there is another one among**6**,**7**,**8**. We measure**8**,**9**. If none of them is gold, then we can find the gold among**6**,**7**, and the gold among**10**,**11**,**12**,**13**, with 3 measurements total. If there is a gold ball among**8**,**9**, then we measure**10**,**11**,**12**,**13**. If none of them is gold, then**9**is gold and we find the other gold ball among**6**,**7**,**8**with 2 more measurements. If there is a gold ball among**10**,**11**,**12**,**13**, then we can find it with 2 measurements. The other gold ball must be**8**.

- If there are no gold balls among them, we measure individually
- If there are no gold balls in the group, then measure
**9**,**10**.- If there are no gold balls among them, then measure individually
**11**,**12**,**13**,**14**. - If there are gold balls among
**9**,**10**, then measure**11**,**12**,**13**,**14**. If there is a gold ball among them, then there is another one among**9**,**10**, and we can find them both with 3 measurements. Otherwise, we measure**9**and**10**individually.

- If there are no gold balls among them, then measure individually

- If there are gold balls in the group, then measure

Discuss this puzzle in the forum.

**email**.

Split into 4 4 4 3

– Step 1, 2, 3: Scan the 3 groups of 4

Now you have 3 possible results:

*RESULT 1: detected = 0 => the 2 gold coins are in the group of 3. Use steps 4, 5 to scan each coin. Done

*RESULT 2: detected = 1. Split 4 coins into A=2, B=2. Split 3 coins into C=2, D=1

– Step 4: scan A:

+ If A is detected: use step 5 to scan C. If C is detected, use steps 6, 7 to scan a coin in A and in C. Done. If C is not detected, use steps 6, 7 to scan D and a coin in A. Done

+ If A is not detected: use step 5 to scan C. If C is detected, use steps 6, 7 to scan a coin in B and in C. Done. If C is not detected, use steps 6, 7 to scan D and a coin in B. Done

*RESULT 3: detected = 2. Split these 2 groups into sub-groups of 2. Then each group 2-2 you need 2 more steps, that is 7 steps in total. Done

Good analysis, but I feel there is an issue with RESULT 2, A detected, C not detected. Then, it is possible that there is a gold coin in group B.

Ahhh, thank you. That was my mistake. My solution was wrong

I feel this problem has no solution, or it can be solved with 8 steps

The puzzle ended up quite complex, but I added a solution which you can see above. I also made a forum post with some notes regarding the solving process; you might find it interesting. If you have any questions, feel free to respond there:)

https://www.puzzleprime.com/forums/topic/gold-and-nickel/

Solved.

1. As previously commented, separate coins into groups of 3. Scan to narrow down your coins to 6, being sure not to mix up each group of coins. Rearrange each group of 3 into two columns and three rows as you can see columns (1 & 2) and rows (A, B & C), G is Gold, N is Not Gold:

1 2

A: G N

B: N G

C: N N

2. Now, scan each ROW. Sort out the row that doesn’t have gold:

1 2

A: G N

B: N G

3. Then, in either row, swap the item in column 1 with the one in column two. Scan either COLUMN again to find your gold coins.

1 2

A: G N

B: G N

Step 1: 4 scans

Step 2: 2 scans

Step 3: 1 scan

TOTAL: 7 Scans

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This solution doesn’t appear to be guaranteed, at least as far as I can see. What if you arrange the six, and both gold coins are in column A. Then you scan 2x, eliminating Col B. But the switch and last scan will only leave you with one gold coin. It appears to me to take 8 scans to guarantee the 2 gold coins.

Or am I wrong?

You cannot find the gold coins with 100% accuracy with this method. You can use logical deduction to reduce the amount of scans you are doing (like the first step, there is no need to scan the last pile of 3 in any case whatsoever) but then your method gets rather case specific and would not work for all the arrangements. I do not have the answer to this either, I just don’t believe this is solved yet.

I’ve ALMOST got it. If you divide them into five groups of 3, you only have to scan four of those groups maximum to narrow it down to 6 coins. If none of the four groups has any gold coins, then you can easily solve which two of the remaining three coins are gold with 2 more scans. If two of the four have gold coins, scan 2 out of 3 from each group. If neither have gold, you’ve found your coins, and if only one has gold, you’ll find your last coin with one more scan. If both have gold, you take the remaining two pairs and pair them off with one from the opposite pair. One of the pairs now has two gold coins, which you can find with your last scan. If only one of the four has any gold coins, then the remaining two groups of 3, the scanned group with gold and the last, unscanned group, have to be separated and each coin is paired with one from the opposite group. You scan two of these pairs. If neither have gold, you’ve found your coins. If both have gold, you take the remaining two pairs and pair them off with one from the opposite pair. One of the pairs now has two gold coins, which you can find with your last scan. However, if only one of the two pairs has gold, I don’t know what to do. Because you don’t know whether that one pair has two gold coins or if the last, unscanned pair has one of the coins. Help! Am I way off or am I getting close and just not seeing something? Thanks!

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