The Answer for The Riddle

The Answer for The Riddle

Salam.


Sorry for delaying the answer for the post. No time.
Anyway, Lurhaf's answer is partially right. 15621, can be the number of coconuts in the situation, but not the minimum amount.

K, the solution. Bear with me a few seconds.

5 students and 1 monkey are in a deserted island, they all gather coconuts then goes sleep. The 1st student wakes-up and divides all the coconuts into 5 shares with 1 extra coconut that he gives to the monkey,he then hides his part of the share and goes back to sleep; The 2nd student wakes-up and divides all the left-over coconuts into 5 shares which also results into 1 extra coconut that she also gives to the monkey, she also hides her share; The 3rd, 4th, and 5th student also do this. What is the minimum number of coconuts possible in the given situation?

Jumlah kokonut setiap murid dapat pada akhir situasi = C
Jumlah kokonut sebelum pembahagian terakhir = 5xC+ 1 = 5C +1

Disebabkan 5C+1 adalah 4/5 dari jumlah sebelumnya, katakan X (murid ke 4 bahagi 5, tinggalkan 4),

4/5 dari Y4 = 5C+1
4/5 x Y = 5C+1
Y4 = (5C+1)x 5/4

Dengan ini, jika kita bergerak kebelakang,
(5C+1)5/4 = Y4
5/4 (Y4+1) = Y3
5/4 (Y3+1) = Y2
5/4 (Y2+1) = Y1
Y1+1 = Total kokonut.

So, there's the theory, now the practical. Again, bear with me.

[5C+1] (5/4) = 25C/4 + 5/4 ------> Y4
[25C+ 5/4 +1] (5/4) = 125C/16 + 25/16 + 5/4 = 125C/16 + 45/16 ------> Y3
[125C/16 + 45/16 + 1] (5/4) = 625C/64 + 225/64 + 5/4 = 625C/64 + 305/64 ------> Y2
[625C/64 + 305/64 + 1] (5/4) = 3125C/256 + 1525/256 + 5/4 = 3125C/256 + 1845/256 ------> Y1

3125C/256 + 1845/256, simplify it we could :
= 12 53/256 (C) + 7 53/256
= 12C + 7 + 53/256(C) + 53/256
= 12 C + 7 + [53 (C+1)]/256 -----> 2

The bold one is the clue to the answer.

53 tidak boleh membahagikan 256, logiknya kerana jika kita berbuat demikian akan ada pecahan. Sedangkan tidak wujud 1/8 kokonut atau 1/13 kokonut. Oleh itu, (C+1) harus membahagikan 256. Nilai C yang paling kecil untuk membolehkan [53 (C+1)]/256 = nombor bulat (tiada pecahan) adalah 255. 255+ 1 = 256 bla3.

Oleh yang demikian, kita masukkan nilai C kita pada equation 2.
12 (255) + 7 + [53(256)/256] = 3120

Lupa pula pada satu yang awal2 diberikan kepada monyet, 3120 +1 = 3121.

Kajian mendalam terhadap jawapan Lurhaf.
15621-1 = 15620
15620 = 12C + 7 + [53(C+1)/256]
(15620 - 7)256 = 12C(256) + 53(C+1)
(15620 - 7)256 = 3072C + 53(C+1)
3996928 = 3072C + 53C + 53
3996928 - 53 = 3125C
C = 1279# (tiada pecahan)

Jawapan Lurhaf betul, tetapi kurang tepat, kerana ia bukan jumlah minimum.

Wallahualam. Kudos Lurhaf.
Salam. God bless us all.