What is the "minimum" total number of nuts they could have started with?

Question

5 men and a monkey picked nuts all day, scooped the nuts into a single pile and went to bed.

During the night 1 man got up, divided the nuts into 5 even piles but there was 1 nut left over which he gave to the monkey. He kept 1 pile for himself and scooped the remaining 4 piles back together and returned to bed. Later on a second man got up, divided the nuts into 5 even piles and again there was 1 nut left over which he gave to the monkey. He also kept one pile for himself, scooped the remaining 4 piles together and went back to bed.

Similarly during the night each of the remaining men (3, 4 and 5) got up one by one and divided the nuts into 5 even piles, each time there was 1 nut left over which he gave to the monkey, kept 1 pile for himself and scooped the remaining 4 piles together and went back to bed.

In the morning they all got up, divided the nuts into 5 even piles, once again there was 1 nut left over which they gave to the monkey, took 1 pile each and went home.

Answer 1

Let x be the number of nuts to start. Then x≡1 (mod 5). Thus x=5y+1. After the first man left, he removed y+1 nuts (y to him, 1 to the monkey).

Therefore there wer 4y nuts left.

Likewise 4y=5z+1 so 4y≡1 (mod 5) or y≡4 (mod 5) or y=5(for the second)
4z=5s+1 (for the third)
4s=5t+1 (fourth)
4t=5u+1 (fifth)
In the morning: 4u=5v+1 So 4u≡1 (mod 5) or u≡4 (mod 5). Thus u=5a-1
4t=5(5a-1)+1=25a-4 thus 4t≡1 (mod 25) or t≡-1 (mod 25)
t=25b-1
4s=5(25b-1)+1, likewise s≡-1 mod(125)
z≡-1 (mod 625)
y≡-1 (mod 3125) so y=3125c-1
and x=5(3125c-1)+1=15625c-4. This is minimum (and positive) when c=1 so x=15621

Well, I have just checked my answer, and it works. It is also less than the number provided by the guy below me. But it still seems way too high. But I can't find a problem with the math. Hopefully I did it correctly, and the number is just higher than one would think.

I think what the two people below me are doing is this:
The first man wakes up splits them into 5 piles, gives the monkey 1 and leaves 1 pile (thus taking 4), instead of leaving 4 and taking 1.

Answer 2

The answer is -4 nuts. You may think I'm nuts for suggesting this answer, since you can't have a negative number of nuts, but the math works out. Here's how:

During the night Man A gets up, divides the -4 nuts into 5 piles of -1 nut each, for a total of -5 nuts. -4 - (-5) = 1 nut left over, which he gave to the monkey. He then took one pile of -1 nuts for himself, leaving behind -4 nuts, or the same as before Man A got up.

In turn, Men B, C, D, E each get up, divide the -4 nuts, give one to the monkey, then take -1 nut for themselves leaving behind -4 nuts.

In the morning, they divide the -4 nuts just as before and give one to the monkey, but this time each man walks off with -1 nut leaving 0 nuts left.

Of course -4 nuts can't exist, but note that if you add 5^6 or 15,625 nuts to the pile, it still works out, since at each step the 15,625 nuts are divided into 5 piles, so the number of them loses a factor of 5. So the number of nuts could be -4 + 15,625 or 15,621. But it could also be 31,246 or 46,871 nuts. In general, the number of nuts could be any positive number of the form 15,625N - 4 for some integer N. The problem statement calls for the minimum possible number of nuts so the answer is 15,621.

Answer 3

assuming the last pile = 6 nuts one per man and one for the monkey
write this in reverse and multiply using z as the unknown number

p1(pile) = 5z+1 =6
new z=6
p2 =(6X5) +1 = 5z +1 => 31
new z = 31
p3 = (31X5)+1 = 5z +1 = 156
p4 = (156X5)+1 =781
P5 = 3906
P6 = 19531
P7 (undivided pile) = 97656
please check though Im numerically dislsxic

Answer 4

Hmm... say there were "a" nuts at the beginning.

What you're saying is this:
(a-1) / 5 = b
(b-1) / 5 = c
(c-1) / 5 = d
(d-1) / 5 = e
(e-1) / 5 = 1, the minimum number for one pile.

so e = 6, d = 31, c = 156, b = 781, and a = 3906.

The minimum number is 3906.

EDIT: oh, whoops... I forgot the last one, in the morning. The answer in this case is (3906 * 5) + 1 = 19531.

Answer 5

if you take m as the minimum no nuts at the beginning

this event happens 6 times in all - one for each man and then once at the end

if you use the equation

4x ( (m-1)/5)^6 you will be on the right track

Answer 6

Man One started with (5a + 1) nuts, keeping (a) for himself, giving one to the monkey, and leaving (4a) for Man Two.
Man Two started with (5b + 1) nuts, keeping (b) for himself, giving one to the monkey, and leaving (4b) for Man Three.
Man Three started with (5c + 1) nuts, keeping (c) for himself, giving one to the monkey, and leaving (4c) for Man Four.
Man Four started with (5d + 1) nuts, keeping (d) for himself, giving one to the monkey, and leaving (4d) for Man Five.
Man Five started with (5e + 1) nuts, keeping (e) for himself, giving one to the monkey, and leaving (4e) for the morning.
In the morning, there were (5f + 1) nuts.

4a = 5b + 1 [The amount One left was the amount with which Two started.]
4b = 5c + 1
4c = 5d + 1
4d = 5e + 1
4e = 5f + 1

Working backward,
4e = 5f + 1
e = 5f/4 + 1/4

4d = 5e + 1
4d = 5(5f/4 + 1/4) + 1
4d = 25f/4 + 5/4 + 1
4d = 25f/4 + 9/4
d = 25f/16 + 9/16

4c = 5d + 1
4c = 5(25f/16 + 9/16) + 1
4c = 125f/16 + 45/16 + 1
4c = 125f/16 + 61/16
c = 125f/64 + 61/64

4b = 5c + 1
4b = 5(125f/64 + 61/64) + 1
4b = 625f/64 + 305/64 + 1
4b = 625f/64 + 369/64
b = 625f/256 + 369/256

4a = 5b + 1
4a = 5(625f/256 + 369/256) + 1
4a = 3125f/256 + 1845/256 + 1
4a = 3125f/256 + 2101/256
a = 3125f/1024 + 2101/1024

They started with 5a + 1 nuts.
= 5(3125f/1024 + 2101/1024) + 1
= 15625f/1024 + 10505/1024 + 1
= 15625f/1024 + 11529/1024
= (15 265/1024) f + 11 265/1024

Since this number must be an integer, the 265/1024's have to drop out of the equation... so let f = 1023. [Multiplying by one less than the denominator will make a -265/1024 to pair with the 265/1024. I know I could be using better terminology with modular arithmetic here, but I hope you understand what I mean.] Any whole number f <1023 will force "a" to be a mixed number.

15625f/1024 + 11529/1024
= 15625(1023)/1024 + 11529/1024
= 15625(1023)/1024 + 11529/1024
= 15621 nuts

Checking:
Man One starts with 15621, gives 1 to the monkey, keeps 3124 for himself, leaves 12496.
Man Two starts with 12496, gives 1 to the monkey, keeps 2499 for himself, leaves 9996.
Man Three starts with 9996, gives 1 to the monkey, keeps 1999 for himself, leaves 7996.
Man Four starts with 7996, gives 1 to the monkey, keeps 1599 for himself, leaves 6396.
Man Five starts with 6396, gives 1 to the monkey, keeps 1279 for himself, leaves 5116.
In the morning, they have 5116, the monkey gets 1, each of the rest get 1023.

They started with 15621 nuts.