Thanks to Bilbao for providing this cool math problem!
I have 3 friends at home whom I want to surprise with a magic trick: Albert, Brian, Charles.
I place 3 objects on a table in front of them: an apple, a biscuit, and a carrot. I also lay on the table a plate with 24 hazelnuts.
I give Albert 1 hazelnut from the plate. I give 2 to Brian and 3 to Charles.
I tell them that I will leave the room, and while I’m out, they must choose among the objects and keep one each without my knowledge. I also tell them that:
The person who takes the apple must take as many hazelnuts as I gave him in the beginning.
The person who takes the biscuit must take twice the amount of hazelnuts I gave him in the beginning.
The person who takes the carrot must take four times the hazelnuts I gave him in the begining.
Finally, I tell them to eat all the hazelnuts they have with them. I go out of the room. Upon coming back in, I find 6 hazelnuts on the plate.
Who took the carrot?
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Albert took the carrot.
Albert took the carrot
Brain took the apple
Charles took the biscuit
ALBERT !!:
24 - 6 = 18 remaining
18 - 6 = 12 hezelnuts to work with
12 = 4(1) + 1(2) + 2(3) …
So, the answer is ALBERT the guy with one hazelnut.
1 + 2 + 3 = 6
Remainder = 6
Therefore, 24 - 6 - 6 = 12 (The Unknown amount taken while away)
Charles get carrot, out, because 3 x 4 = 12 (too many)
Brian get carrot, out, because 2 x 4 = 8 (Charles apple and Albert biscuit is 13, the opposite will give 15)
Therefore, left Albert getting the carrot!
Brian the apple and Charles the biscuit.
Verify, 1×4 + 2×1 + 3×2 = 12
Albert took the carrot, Charles took the biscuit and Brian took the apple.
Albert took the carrot,
Brian the apple,
Charles the biscuit.
Albert
24 hazelnuts in the beginning – 6 he gives to the boys = 18
He finds 6 hazelnuts on the table => 12 have been taken
4A + 1B + 2C = 12
Taking all possibilities into consideration we find out that A = 1, B = 2 and C = 3.
As the person who chose the carrot takes 4 times the hazelnuts he had in the first place, and Albert takes one hazelnut (4×1=4) he’s the one who has the carrot.
Makes more sense in my head though.
Albert.
Albert took the carrot
Brian took the apple
Charles took the biscuit
There were 24 hazlenuts initially, he hands out 6 (leaving 18), and then 6 are left when returns, so it means 12 have been taken when he is out of the room.
If he who had 3 originally has to take twice the amount = 6
If he who had 2 originally has to take the same amount = 2
If he who had 1 originally has to take four times amount = 4
This will give the 12 that have been taken
As we are asked for who took the carrot (and therefore he who had 1 initially and takes 4), it must be Albert.
Albert took the carrot.
Brian took the apple, and Charlie took the biscuit.
A=1+(4*1)=5
B=2+(1*2)=4
C=3+(2*3)=9
A+B+C=18
24-18=6 hazelnuts still on the plate
Albert took the carrot.
We start out with 24 hazelnuts. Albert receives 1, Brian receives 2, and Charles receives 3, which leaves 18 hazelnuts up for grabs.
Each one of the three objects on the table represents a multiplier that is applied to each person’s original hazelnut allotment. Let us represent these multipliers as follows: apple = a = 1, biscuit = b = 2, and carrot = c = 4.
Let us represent the three original hazelnut allotments as h1, h2, and h3. So h1 + h2 + h3 = 6.
We are told that there are 6 hazelnuts left after the instructions are carried out. So 18 - a*h1 - b*h2 - c*h3 = 6. Subsequently:
- a*h1 - b*h2 - c*h3 = -12, and a*h1 + b*h2 + c*h3 = 12.
If we substitute the numeric multipliers for their variables we have:
h1 + 2*h2 + 4*h3 = 12.
We know that h1, h2, and h3 are the numbers 1, 2, and 3 in some arrangement. So we must find a way to multiply the elements of [1,2,3] and [1,2,4] so that each number of each set maps to only one number of the other set, and the sum of their products is 12.
Now, you could do this the long way with equations and substitutions, or matrices if you like, but a little trial and error will yield results much more quickly, in my opinion.
Ultimately, we see that 1*4 + 2*1 + 3*2 = 12. So the person who got 1 hazelnut took 4 (which means he took the carrot), the person who got 2 hazelnuts took 2 (which means he took the apple), and the person who got 3 hazelnuts took 6 (which means he took the biscuit).
We know that Albert was given 1 hazelnut to start, so Albert took the carrot.
Albert took the carrot.
After dishing out the initial hazelnuts (1+2+3) 18 remain. If, upon returning to the room, 6 are still in the dish, then 12 must have been taken.
The only way this could happen is if Albert took the carrot (1×4), Brian took the apple (1×2), and Charles took the biscuit (2×3).
Albert Brian Charles
(1) (2) (3)
apple (x1) 1 2 3
biscuit (x2) 2 4 6
carrot (x4) 4 8 12
To have 6 nuts left on the plate, the dudes have to take a total of 12 hazelnuts in addition to the 6 they get initially.
So:
Brian(1*2nuts)+Charles(2*3nuts)+Albert(4*1nut)=12nuts
Which tells us that Albert has the carrot.
Albert took the carrot
Charles took the biscuit
Brian took the apple
Albert
Let the one who took the apple have started off with a hazelnuts,
the biscuit b hazelnuts and the carrot c hazelnuts: a + b + c = 6 (x)
After taking the items,further hazelnuts taken: a + 2b + 4c = 12 (y)
Subtracting (x) from (y) b + 3c = 6
The only possible solution is that b = 3, c = 1 and a = 2
Therefore, Albert took the carrot and ate 5 hazelnuts, Brian the apple and ate 4 hazelnuts, Charles the biscuit and ate 9 hazelnuts
There were 24 nuts to begin with, you gave away 1+2+3=6 right away. Of the remaining 18 nuts, 6 were left. Therefore, the 3 people took 12 nuts among them. Charles must take 6 to make numbers even, so he took the biscuit. So, Albert took the carrot and 4 nuts, and Brian took the apple and 2 nuts.
Albert.
Supposing Brian took the apple, he would have to pick 2 hazelnuts (1 time 2 = 2) and so he would now have 4 hazelnuts (2 + 2 = 4)
Supposing Charles took the biscuit, he would have to pick 6 hazelnuts (2 times 3 = 6) ans so he would now have 9 hazelnuts (6 + 3 = 9)
Supposing Albert took the carrot, he would have to pick 4 hazelnuts (4 times 1 = 4) ans so he would now have 5 hazelnuts (4 + 1 = 5)
They would have eaten, in total 18 hazelnuts (4 + 9 + 5 = 18), leaving 6 hazelnuts on the plate (24 - 18 = 6).
ALbert took the carrot. Little glutton.
Albert took the carrot. Brian took the apple. Charles took the biscuit.
First he handed out 6 nuts to start with, and there were 6 nuts left when he returned. So that leaves 12 nuts out of the total 24.
A couple of ways to tackle this problem.
The “brute force” method:
6 possible ways to split the 3 items between the 3 friends.
Of these 6 ways, only one way has the nuts sum up to 12.
The “logical trial” method.
If Charles got the carrot, he would take 12 nuts leaving nothing for the other 2. So Charles is out.
If Brian got the carrot, he would get 8 nuts, leaving 4 for the other 2 which isn’t enough. So Brian is out.
If Albert got the carrot, he would take 4 nuts, leaving 8 for the other 2. If Brian got the apple, he would get 2 nuts, leaving 6 for Charles. Charles takes the biscuit, and thus takes the last 6 nuts.
So either way, you see Albert got the carrot.
Well, if it was breakfast time, and it was a sausage biscuit, I’d have to take the biscuit!
Albert took the carrot.
There were 24 hazlenuts and 6 are left so 18 were eaten.
Brian took the apple so took as many as his 2 = 1×2 + 2 = 4 he eats.
Charles took the biscuit so took twice as many as his 3 = 2×3 + 3 = 9 he eats.
Albert took the carrot so took four times as many as his 1 = 4×1 + 1 = 5 he eats.
Total eaten = 4 + 9 + 5 = 18 (leaving 6 from 24).
Chares has the cattor because
chares 3x carrot 4=12
bran 2x biscuit 2= 4
albert 1x apple 2= 2
_____
18
+ 6 left on the plate
___
24 hazelnuts that were on the plate when he started out!!!
7th grader at WWMS
Albert - Carrot
Brian - Apple
Charles - Biscuit
Albert,
Albert:carrot; Brian:apple; Charles: biscuit
so Albert: 1+4=5; Brian: 2+2=4; Charles: 3+6=9; total taken=18
left=6
Albert is the bunny
Bunnies like carrots
So Albert took the carrot
Brian took the apple
Charles took the biscuit
Hazlenuts:
Albert:1+4=5
Brian:2+2=4
Charles:3+6=9
5+4+9+6=24
Albert took it. After you give the nuts out, you only have 18 left. Then you minus the 6 remaining. From there, you have 12 left. Therefore, Albert takes the carrot, Charles takes the bisscuit and Brian takes the apple.
albert
Albert took the carrot.
Albert took the carrot. 12 extra hazelnuts were eaten after you left.
If albert took the carrot, he would take 4 more (1×4). Brian took the apple thereby taking 2 more (2×1). And Charles took the biscuit so he took 6 more (3×2). 4 + 2 + 6 + 6 (leftover) + 6 (original amount given) = 24
Albert takes the carrot.
I think that the one that took the carrot was Albert:
24 H - 1H(A) - 2H(B) - 3H(C)= 18 H
6H left= they eat 12 H betwwen them 4H (A)+ 2H (B) +6H(C)= 12(H)
this meets the criteria so Albert took 4 times what he has already taken, so he was the one that took the carrot, cool.
Albert took the carrot.
Brian took the apple.
Charles took the biscuit.
Therefore:
A = 4
B = 2
C = 6
Total = 12.
24 - 6 (given at the start) - 12 (taken) = 6 hazelnuts.
Very good, Albert is the answer. You can always see some nice explanations from some of our regulars MichaelC, MFox, Shawn, Migrated,Suineg,Oneiric, Fuzzy, Falwan. Some new visitors too with good explanations, as above.