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StackVille Bank & Trust - Robbed!

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The Robbery

The bank is being robbed. There are millions in cash in the vault. The lone robber takes the manager to the safe and demands he opens it.

The manager was scared, as he did not know the three numbers to the vault. He did know, however, where the bank president kept clues to the combination. He opened a drawer, pulled out a piece of paper, and read the clues:

The Combination

  1. If the second number was not a multiple of 8, then it was a number from 80 through 89.

  2. If 1st number was not a multiple of 3, then it was a number from 60 through 69.

  3. If the third number was a multiple of 3, then it was a number from 50 through 59.

  4. If the first number was a multiple of 2, then it was a number from 50 through 59

  5. If the second number was a multiple of 6, then it was a number from 40 through 49.

  6. If 2nd number was not a multiple of 7, then it was a number from 60 through 69.

  7. If the first number was not a multiple of 4, then it was a number from 70 through 79.

  8. If 3rd number was not a multiple of 4, then it was a number from 60 through 69.

9.If the third number was not a multiple of 6, then it was a number from 70 through 79.

The vault safe numbers range from 50-79.

The Get-A-Way

The robber looked at the manager, smiled and said, "Thanks!"

He opened the vault, grabbed the cash, and made a smooth escape. The robber figured out the vault's combo - can you?

What is the vault combination?

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  • $\begingroup$ I get that there are actually two choices for the second number: rot13(sbegl-avar cyhf frira) and rot13(svsgl-avar cyhf svir). $\endgroup$ – shoover Apr 15 at 23:41
  • $\begingroup$ @shoover that first number does work! However I think we have to assume that the number is in the ranges. That or there’s been a mistake $\endgroup$ – Beastly Gerbil Apr 15 at 23:46
  • $\begingroup$ Hi guys, I triple checked my puzzle for typos and errors. Everything looks good. There are no mistakes, and a solution is just around the corner. Good luck. Thanks for looking. John $\endgroup$ – John S. Apr 15 at 23:55
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Rearrange:

    1. If 1st number was not a multiple of 3, then it was a number from 60 through 69.
    1. If the first number was a multiple of 2, then it was a number from 50 through 59
    1. If the first number was not a multiple of 4, then it was a number from 70 through 79.
    1. If the second number was not a multiple of 8, then it was a number from 80 through 89.
    1. If the second number was a multiple of 6, then it was a number from 40 through 49.
    1. If 2nd number was not a multiple of 7, then it was a number from 60 through 69.
    1. If the third number was a multiple of 3, then it was a number from 50 through 59.
    1. If 3rd number was not a multiple of 4, then it was a number from 60 through 69.
    1. If the third number was not a multiple of 6, then it was a number from 70 through 79.

The answer is:

75 - (56 or 64) - 76

As:

If first number is a multiple of 2 but not a multiple of 4 (e.g. 2, 6, 10, ...), then it is both between 50-59 and between 70-79. This cannot be true, so one of the following is true

* First number is multiple of 4 and is between 50-59: 52, 56
* First number is not a multiple of 2 and is between 70-79: 71, 73, 75, 77, 79

If first number is not a multiple of 3, then it is between 60-69, which is not true, so first number is a multiple of 3.

=> First number is 75

If second number is a multiple of 6, then it is between 40-49: 48
Then it is not a multiple of 7, so it must be between 60-69, which cannot be true, so second number is not a multiple of 6.

Second number cannot be simultaneously between 80-89 and between 60-69, so it cannot be simultaneously not a multiple of 7 and not a multiple of 8.
Option 1: Not a multiple of 7, but a multiple of 8: then it is between 60-69, so 64 (which is not a multiple of 6)
Option 2: A multiple of 7, but not a multiple of 8: then it is between 80-89, so 84, but this is a multiple of 6, so this option is false.
Option 3: A multiple of 7 and a multiple of 8: 56, which is not a multiple of 6

=> Second number is 56 or 64

If third number is a multiple of 3 but not a multiple of 6 (3, 9, 15, ...), then it is both between 50-59 and between 70-79. This cannot be true so one of the following is true:
* Third number is multiple of 6 and is between 50-59: 54
* Third number is not a multiple of 3 and is between 70-79: 70, 71, 73, 74, 76, 77, 79

If third number is not a multiple of 4, then is between 60-69, which is not true, so third number is a multiple of 4.

=> Third number is 76

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  • 2
    $\begingroup$ Fixed spoilers :) $\endgroup$ – Beastly Gerbil Apr 16 at 0:02
  • $\begingroup$ Thanks for fixing those darned spoilers! $\endgroup$ – shoover Apr 16 at 17:17
  • $\begingroup$ Hello Everyone, my name is Heather I am John's partner. Unfortunately he has been in the hospital since yesterday, and tested positive for Covid. Doctors are very confident he will be home soon. He wanted me to get on here and take care of some of his affairs, and tending to this site was one of my tasks. A literal quote he wrote down for this website, "tell my students and the users on 'puzzling' I will be home soon and we will be tearing through math problems like always, soon" $\endgroup$ – John S. Apr 17 at 2:00
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The code is

75-64-76


Looking at the individual numbers:

For the first number:

If it's not a multiple of 3, then its 60-69
If it's a multiple of 2, then it's 50-59
If it's not a multiple of 4, then it's 70-79

So:

If it's 50-59, it's a multiple of 2, 3 and 4; the number must be a multiple of 12. No multiples of 12 fall in the range 50-59.

If it's 60-69, it's not a multiple of 3, not a multiple of 2 and a multiple of 4. As all multiples of 4 are also multiples of 2, this cannot be.

If it's 70-79, then it's not a multiple of 4, not a multiple of 2 and a multiple of 3. Multiples of 3 in this range are 72, 75 and 78. 75 fits, so the first number is 75.

For the second number:

If it's not a multiple of 8, then it's 80-89
If it's a multiple of 6, then it's 40-49
If it's not a multiple of 7, then it's 60-69

So:

If it's 40-49, it's a multiple of 8, a multiple of 7 and a multiple of 6. No such number exists.

If it's 60-69, it's not a multiple of 7, not a multiple of 6 and a multiple of 8. Multiples of 8 in this range are 64, which fits. So the second number is 64.

However

As @Shoover and @Magma points out, 56 also works, as do an infinite amount of numbers that are both a multiple of both 7 and 8, and not a multiple of 6. The OP has clarified in the comments under this answer that the number must be in the ranges given, meaning it is 64.

Finally, for the third number (initially typed a statement wrong):

If it's a multiple of 3, then it's 50-59
If it's not multiple of 4, then it's 60-69
If it's not a multiple of 6, then it's 70-79

So:

If it’s 50-59, then it’s a multiple of 3, 4 and 6. Multiples of 6 in this range are 54, which is not divisible by 4, so it’s not that.

If it’s 60-69, then it’s not a multiple of 3 or 4, and a multiple of 6. All multiples of 6 are also multiples of 3 so this cannot be.

If it's 70-79, then it's a not a multiple of 6 that's a multiple of 4 and not a multiple of 3. Multiples of 4 in this range are 72, 76 - only 76 fits here, so it’s 76.

Which gives us

75-64 - 76

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    $\begingroup$ There are infinitely many numbers that are multiples of 7 and 8 but not of 6, any of those numbers could be the second number. $\endgroup$ – Magma Apr 15 at 23:59
  • $\begingroup$ @Magma true, I have a feeling there’s a mistake in the puzzle. That or we just have to assume the number is in the ranges given (but it doesn’t say that anywhere) $\endgroup$ – Beastly Gerbil Apr 16 at 0:06
  • $\begingroup$ I think this ** SPOILER ** for the first number alone will help. SVEFG AHZORE EHYR #4: ryvzvangrf nyy zhygvcyrf bs 2 rkprcg gubfr sebz 50 gb 59. Ryvzvangrq gurersber ner 60, 62, 64, 66, 68 naq 70, 72, 74, 76 naq 78. EHYR #2: vaqvpngrf gung vs gur ahzore jnf abg n zhygvcyr bs 3, vg jnf n ahzore sebz 60 guebhtu 69. Guvf ryvzvangrf 50, 52, 53, 55, 56, 58, 59 naq 71, 73, 77 naq 79. Ehyr #7: vaqvpngrf gung vs gur ahzore jnf abg n zhygvcyr bs 4, gura vg jnf n ahzore sebz 70 guebhtu 79. Guvf ryvzvangrf 51, 54, 57 naq 61, 63, 65, 67 naq 69. Gur erznvavat ahzore, 75, fngvfsvrf nyy nobir $\endgroup$ – John S. Apr 16 at 0:12
  • $\begingroup$ @JohnS. It’s the second number that’s the problem, the first number is easy enough to find $\endgroup$ – Beastly Gerbil Apr 16 at 0:14
  • $\begingroup$ Yes, absolutely, the combination number must follow the rule, for instance: If the first number was not a multiple of 4, then it was a number from 70 through 79 - then the number has to be within this range. Same for the rest. Edit for clarification if needed, but it seems pretty clear I believe. $\endgroup$ – John S. Apr 16 at 0:14

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