[Guest]

A father bought KFC special custom made bucket chicken and wanted to distribute the chicken pieces to his sons. To his eldest he gave one piece plus 1/7 of the remaining pieces; to his second eldest, two pieces plus 1/7 of the remaining pieces ;to the third eldest three pieces plus 1/7 of remainig pieces and so forth.

All the pieces were distributed among his sons without remainder.

how many sons and how many pieces were there in the KFC special custom made bucket chicken??

Edited May 8, 2011 by TrUnX

[Guest]

0 and 0 or

36chicken things and 6 kids

[Guest]

i think you forgot to say that the last kid gets all the left over ones

[Guest]

i dint,it was implied in the question

explain the steps!!

Edited May 8, 2011 by TrUnX

[Aaryan]

36 chicken pieces and 6 kids

I used guess and check, like my eight-year-old brother told me to.

[Guest]

sory my second post made no sense

when there are 2 peolpe left to take the first person should have 7 peiceces beforing taking their 1/7 because the last personmust take 1/7 of nothing or else he wont be last. So the 1/7 of the second to last person should be 1 given that first takes one second takes 2. That means there must be 6 for the last person 7-1.If the last person takes 6 he must be the sixth son, so 6 kids 6pieces each 36 pieces

guessing and checking is easier

[Aaryan]

Impressive!My thoughts exactly!

after I posted my answer, I went back and figured it out algebraically too.

Now people will think I'm lying.

Edited May 8, 2011 by Aaryan

[Guest]

7 sons and 35 pieces.

1. each son gets 1/7 of the remaining pieces. so one can assume that there is at least 7 pieces. 2. if each son gets 1/7 pieces then one can assume there are 7 sons. 3. the first son gets 1 piece, the second 2, the third 3, etc. this means: a. 4th son gets 4, 5th son gets 5, 6th son gets 6, and 7th son gets 7 b. each son also gets 1 of the 7 remaining pieces. Therefore.... 4. first has 2, second has 3, third has 4, fourth has 5, fifth has 6, sixth has 7, and seventh has 8 (original pieces plus their 1 of the remaining) 5a. 2+3+4+5+6+7+8=35 OR 5b. 1+2+3+4+5+6+7=28...28+7=35

[Guest]

sory my second post made no sense

when there are 2 peolpe left to take the first person should have 7 peiceces beforing taking their 1/7 because the last personmust take 1/7 of nothing or else he wont be last. So the 1/7 of the second to last person should be 1 given that first takes one second takes 2. That means there must be 6 for the last person 7-1.If the last person takes 6 he must be the sixth son, so 6 kids 6pieces each 36 pieces

guessing and checking is easier

1. the question doesn't allow for the chicken pieces to be distributed equally so saying "6 kids 6 pieces each" is incorrect.

2. if each son gets 1/7 of the remaining then there would only be 1 for the last son from the remaining, leading one to believe that there are 7 sons as each gets 1/7.

3. the question doesn't state that the sons only get 1/7...they get some pieces AS WELL AS 1/7. you have to take into account all of the other pieces which are separate from the 1/7 remaining.

Edited May 8, 2011 by missepicfail

[Guest]

After the eldest takes 1, the number of pieces must be a multiple of 7. (Time 1)

After the eldest takes his 1/7th and the second eldest takes 2, the number of pieces must again be a multiple of 7. (Time 2)

Therefore, to go from Time 1 to Time 2, we move from a multiple of 7 to another multiple of 7. The easiest way to do this is to take away 7 pieces.

Then the eldest's 1/7th plus the second eldest's 2 pieces must equal 7.

Therefore, the eldest's 1/7th must equal 5, since 5 + 2 = 7.

If 1/7th is 5, 7/7ths is 35.

Add the 1 that the eldest took, and you find a total of 36 pieces.

Work through the problem, and you find 6 sons.

[Guest]

I solved it with algebra. After the first son gets all his pieces, there are N pieces left. The second son gets 2 pieces leaving N-2, and then gets 1/7 of that, therefore N-2 is divisible by 7 and so is that number plus 7, or N+5. N+5 must be the number that the 1st son gets 1/7 of. Since there are N left after taking 1/7, 5 must equal that 1/7th and so N+5=35. The son was given 1 piece before that, so the total number of pieces of chicken is 36. This works out to 6 children recieving 6 pieces each. They recieve each 1,2,3,4,5,6, and also 1/7 remainder which is 5,4,3,2,1,0.

I realized that in order for the chicken to always be divible by 7, the number a son gets plus the 1/7 that the prior son had must total 7. And since each son recieved one more than the prior before dividing by 7, that the multiple of 7 had to be decreasing by exactly 1 as well in order to keep the pieces divisible by 7.

Someone else had the exact same idea already.

Edited May 9, 2011 by Nana7

[Guest]

[spoiler=great work googon ]the answer is 6 sons and 36 pieces....

i think this was easy one for you, next time will find a difficult one

[Guest]

The answer is

36 pieces and 6 sons.. (square of the no. of sons)

[Guest]

I believe that

the answer to all similar questions for 1/n division would be "n-1" sons and square of (n-1) pieces...

Edited May 9, 2011 by Silver Surfer

[Guest]

the answer is total 36 pieces and 6 sons..

the details are like this..

1. use common hit and trial method to get this as above.

2. go like this

total pieces are X.

the equation will go like this..

(1+1/7(x-1))+ (2+ 1/7 (x-3))+ ....n n is no of persons...

(1+2+3+...n)+ 1/7(nx- (1+3+6+10...n)) After simplifying it

this should be integer and devisible by 7.

so the term (1+3+6+10......) should be devisible by seven...

by expanding the table it comes 1+3+6+10+15 and at this stage the sum is 35 and is divisible by seven.

the pieces will be 35+1(the eldest got), and the kids will be 6 (the last one get only his share not 1/7 of remaining cos 6+ 1/7(35-35)).

CHEEEERRRRSSSSS

[Guest]

Aaryan, I really like your solution, but how can you know that to get from one multiple of 7 to the other you use 7 instead of 14 without guessing and checking.

[Guest]

1st son takes a piece and takes 1/7 of 35 =5, 2nd son takes 2 pieces and takes 1/7 of 28 = 4, 3rd son takes 3 pieces and takes 1/7 of 21 = 3, 4th son takes 4 pieces and takes 1/7 of 14 = 2, 5th son takes 5 pieces and takes 1/7 of 7 = 1, and the last son takes the remaining 6 pieces.

then they all eat 6 pieces each, live a long life, grow old, and die