Penalty fee

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Posts: 15
Joined: Sun Oct 21, 2018 11:06 am UTC

Penalty fee

Postby Jayraj » Fri Nov 16, 2018 6:20 pm UTC

One day, 3 drivers with their lorry stopped at a checkpost to pass the weight limit check.

The officer in charge presents them with a riddle and if they answer correctly within the first 2 attempts,

they will be allowed to pass through with no penalties even if they are over the weight limit.

The Officer hands each of them a notice showing their weight difference with that of the weight limit, down to the nearest Kilogram.

Officer: "Funnily enough, the numbers I gave you are consecutive, and if I add up the difference between each of your

load and the limit, and to that I subtract the multiplication of those same numbers, I get zero."

"My question is: can you guess which other two numbers are in your friend's hands?"

Using their first attempt, the three drivers answer at the same time:

"Yes", "No" and "No".

Can you save them all a potential penalty fee and use their last attempt to guess the numbers ?

Posts: 136
Joined: Tue Feb 03, 2009 2:05 pm UTC

Re: Penalty fee

Postby Yat » Tue Nov 20, 2018 9:31 am UTC

Calling x the smallest number, the others are x+1 and x+2, so the sum of all three is 3x+3 and the product is x*(x+1)*(x+2) = x^3+3x²+2x.
So we need to solve x^3+3x²+2x-(3x+3) = 0, which is x^3+3x²-x-3 = 0. 1 and -1 are trivial solutions, we easlily get to this:
The three values of x that give a difference of 0 are -3, -1 and 1.
Therefore, the numbers on the papers can either be
-3, -2 and -1,
-1, 0 and 1,
or 1, 2 and 3.
the only two numbers that are common to two solutions are -1 and 1, so the two guys that say no have them and therefore, the first guy has 0.

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