A not-so-hard quickie

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AlanMiller
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A not-so-hard quickie

Post by AlanMiller »

Two intelligent, honest students are sitting together at lunch one day when their math teacher hands them each a card. “Your cards each have an integer on them,” the teacher tells them. “The product of the two numbers is either 12, 15 or 18. The first to correctly guess the number on the other’s card wins.”

The first student looks at her card and says, “I don’t know what your number is.”

The second student looks at her card and says, “I don’t know what your number is, either.”

The first student then says, “Now I know your number.”

What number is on the loser’s card?

Alan

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HansV
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Re: A not-so-hard quickie

Post by HansV »

Spoiler
The loser has 6 on her card.
Regards,
Hans

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Claude
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Re: A not-so-hard quickie

Post by Claude »

I agree with Hans. Many many years ago, we had this in a math test, back in the days when people used to be able to calculate without the aid of an electronic calculator or computer.
Cheers, Claude.

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AlanMiller
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Re: A not-so-hard quickie

Post by AlanMiller »

Well done lads. As Claude says, it's the kind of thing you learned to do in your head ... before the age of graphical calculators and such.

Alan

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BobH
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Re: A not-so-hard quickie

Post by BobH »

I believe you, but could you please 'show your work'?
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HansV
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Re: A not-so-hard quickie

Post by HansV »

This was my reasoning:
Spoiler
We know that the numbers on the cards are whole numbers, that their product is 12, 15 or 18, and that neither student can immediately guess the number on the other's card. We'll successively rule out numbers.

12 = 1x12 or 2x6 or 3x4.
If the cards had been 1 and 12, the student with 12 would have known that the other has 1.
If the cards has been 3 and 4, the student with 4 would have known that the product is 12 since 15 and 18 cannot be divided by 4. So the other had 3.
This leaves 2 and 6.

15 = 1x15 or 3x5.
If the cards had been 1 and 15, the student with 15 would have known that the other has 1.
If the cards has been 3 and 5, the student with 5 would have known that the product is 15 since 12 and 18 cannot be divided by 5. So the other had 3.
This rules out 15 as product.

18 = 1x18 or 2x9 or 3x6.
If the cards had been 1 and 18, the student with 18 would have known that the other has 1.
If the cards has been 2 and 9, the student with 9 would have known that the product is 18 since 12 and 15 cannot be divided by 9. So the other had 2.
This leaves 3 and 6.

So the possibilities are 2 and 6 or 3 and 6.
If the first student had 6, she couldn't know whether the second student had 2 or 3, but the second student could have derived from that that the first student must have 6.
So the second student must have had 6. In the end, we still don't know whether the first student had 2 or 3, but that doesn't matter. We weren't asked to solve that.
Regards,
Hans

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StuartR
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Re: A not-so-hard quickie

Post by StuartR »

That's exactly the logic I used too Hans. I'd attach a photo of the whiteboard next to my computer, but I've wiped it now.
StuartR


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HansV
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Re: A not-so-hard quickie

Post by HansV »

Great minds...
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Hans

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BobH
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Re: A not-so-hard quickie

Post by BobH »

Thank you, Hans!

I was on the right track in identifying the factors for each result; however, my aged mental acuity blurred before I got to the end of it.
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