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After our previous misadventure with The Number Kidnapper, Derek, Alice, and I got two new friends to replace Peter - Timothy and Katie.

We were walking along again in the Town Square when we were kidnapped again!

Some time later, we woke up.

"Hello again! Some of you may remember me." said a familiar voice.

"Oh, not again." groaned Derek.

"It's me! The Number Kidnapper!" he said. "After you escaped due to the help of those 'friends' of yours, I began to think of how to finally get rid of you for good!"

"Well? What is it?" I said.

"Again, each of you have a number. If you want to leave, you have to find it! But this time, it won't be as easy."

He then had us hold hands in this fashion:

Me

Alice - Katie

Timothy - Peter

"Now then. The sum of Alice and Katie's numbers is one half that of Timothy and Peter's numbers, and twice that of ASCIIThenANSI's. Can you guess your number?" he asked.

"Easy!" said Alice. "30."

"Wrong!" he shouted. "Guards? Take her away!"

"But- but that was what it was last time!" Alice stammered.

"I said it wouldn't be as easy, now would it? By the way, your number was 4." he said.

Then we held hands like this:

Me - Timothy

Katie - Peter

"Now the sum of Katie and Peter's numbers is one third that of the sum of ASCIIThenANSI's and Timothy's numbers. Do you think you can guess your numbers?"

Once again, I need your help! I need everyone's numbers so we can all escape!

What are the numbers of me and my friends?

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  • $\begingroup$ -1 for having more than one solution even though the wording didn't give that impression. Credit to Hugh Meyers for proving this in his answer. $\endgroup$
    – Rosie F
    Sep 1, 2016 at 10:30
  • $\begingroup$ So Derek is gone now, and Peter came back? "Oh, not again." groaned Derek. and left :D $\endgroup$
    – Gintas K
    Sep 1, 2016 at 12:00

1 Answer 1

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Unless there is another condition or I have made a mistake,

Your life hangs on a coin toss! Both of these appear to work:
You 4, Katie 4, Timothy 14, Peter 2 and
You 8, Katie 12, Timothy 31 and Peter 1.
Surely, I am missing something because this seems like simple arithmetic.

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  • $\begingroup$ You're not missing something - although I didn't consider there might be multiple solutions. $\endgroup$
    – ASCIIThenANSI
    Mar 21, 2016 at 12:19
  • $\begingroup$ @ASCIIThenANSI You gave us 4 unknowns and 3 simultaneous equations, so no surprise here. If you can give us one more relation, which can't be completely derived from the ones we already have, then the solution will be unique. $\endgroup$
    – Reti43
    Mar 22, 2016 at 17:03
  • $\begingroup$ @Reti43 I was surprised that there were only two solutions. Limiting it to integers almost did the trick. $\endgroup$
    – Hugh Meyers
    Mar 22, 2016 at 17:15
  • $\begingroup$ I just tried solving by hand, and came up with $Y = \frac27, K = \frac{-24}7, T = \frac{-23}7, P = \frac{31}7$. I suspect there are many more solutions. $\endgroup$
    – GentlePurpleRain
    Mar 22, 2016 at 18:40
  • $\begingroup$ I was assuming he meant "natural number" but it isn't actually stated. Certainly there are infinite solutions if you allow rationals. $\endgroup$
    – Hugh Meyers
    Mar 22, 2016 at 19:16

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