1. ## Fun-Quiz

In a fun-quiz there were n people including the host of the show, viz. A,B,C,D,E,F.......etc. As per the convention of the party everyone has got chocolate in the following manner A-1,B-2,C-3,D-4.......chocolates respectively. Before anyone has eaten a bit of chocolate, due to some urgent call the host of the show left the show with his chocolates. Later on the rest attendants recollected their chocolates in a box and then redistributed all the chocolates among themselves and thus everyone has received 13 chocolates.

1) What could be the minimum number of persons who attended the party ?

2)As per the previous question, maximum how many chocolates were there to be received by all of them,initially ?

3)Is it possible to determine the total number of people, initially who have attended the meeting?

4)Is it possible to determine who is the host of the show i.e A or B or C or................etc. ?

2. ## Re: Fun-Quiz

1) Minimum number of people who attended the party: 24 (including the host)
2) 24 people would have received 24x25/2 = 300 chocolates. However, you asked for the maximum number of chocolates - it's possible that 25 or 26 people attended. If 26 attended then they would have received 26x27/2 = 351 chocolates.
3) No - could be 24, 25, or 26 - we don't know which.
4) The host was A if 24 people attended, or M if 25 attended, or Z if 26 attended.

3. ## Re: Fun-Quiz

So there were a total of 1+ 2+ 3+ ...+ n= n(n+1)/2 chocolates distributed. If the host got m chocolates, there would be n(n+1)/2- m chocolates left to be distributed among n- 1 people. Since each of those people got 13 chocolates we must have 13(n- 1)= n(n+1)/2- m. Multiplying that out, we get $n^2- 25n+ 26= 2m$. Of course, m, and so 2m is a positive number so $n^2- 25n+ 26> 0$.
Solve that equation- or graph it- to determine where n is positive.

4. ## Re: Fun-Quiz

Originally Posted by HallsofIvy
Solve that equation- or graph it- to determine where n is positive.
Another condition: the value of $n^2-25n+26$ must be less than or equal to 2n.

5. ## Re: Fun-Quiz

Originally Posted by HallsofIvy
So there were a total of 1+ 2+ 3+ ...+ n= n(n+1)/2 chocolates distributed. If the host got m chocolates, there would be n(n+1)/2- m chocolates left to be distributed among n- 1 people. Since each of those people got 13 chocolates we must have 13(n- 1)= n(n+1)/2- m. Multiplying that out, we get $n^2- 25n+ 26= 2m$. Of course, m, and so 2m is a positive number so $n^2- 25n+ 26> 0$.
Solve that equation- or graph it- to determine where n is positive.
Curious ! The discriminant for n2-25n + 26 >0 is not a perfect square. Then how do we derive the value of n ?

6. ## Re: Fun-Quiz

Originally Posted by ebaines
Another condition: the value of $n^2-25n+26$ must be less than or equal to 2n.
Hi ebaines,

Why should the value of n2-25n + 26 must be less than or equal to 2n ?

7. ## Re: Fun-Quiz

Originally Posted by SPOCK
Curious ! The discriminant for n2-25n + 26 >0 is not a perfect square. Then how do we derive the value of n ?
That means that n2- 25n 26 is not equal to 0 for any integer. But you only need to find an integer so that it is greater then 0, not equal to 0.

8. ## Re: Fun-Quiz

Originally Posted by HallsofIvy
That means that n2- 25n+ 26 is not equal to 0 for any integer. But you only need to find an integer so that it is greater then 0, not equal to 0.
Thanks! HallsofIvy. So the least such value of an integer is 24 for which n2-25n+26 is greater than zero. But one thing I don't understand why ebaines mentioned the condition that n2-25n+26 must be less than or equal to 2n ? Please explain .

9. ## Re: Fun-Quiz

Since n^2-25n+26 is two times the index number for the host, that number must be less than or equal to the number of people in the room. For example if there are 26 people, n^2-25n+26 = 52, so the host would be the 52/2 = 26th person in the room. Buit if n=30 then n^2-25n+26 = 176, and the host is number 176/2=88, which is impossible - you can't be the 88th person in a group of 30.

10. ## Re: Fun-Quiz

Hmmm...why can't it be this simple:

1 + 2 + 3 + .... + n = 13n

So:
n(n + 1) / 2 = 13n
n^2 + n = 26n
n + 1 = 26
n = 25

Over and out!

11. ## Re: Fun-Quiz

Originally Posted by ebaines
Since n^2-25n+26 is two times the index number for the host, that number must be less than or equal to the number of people in the room. For example if there are 26 people, n^2-25n+26 = 52, so the host would be the 52/2 = 26th person in the room. Buit if n=30 then n^2-25n+26 = 176, and the host is number 176/2=88, which is impossible - you can't be the 88th person in a group of 30.
Thanks! and yes the number of people cannot be more than 26 because the number of chocolates left after the host took away his chocolates must be divisible by 13 and (n-1). But if we take n>26 then the number of persons after dividing by 13 will be more than (n-1), which is impossible. So the condition n2-25n + 26 < 2n ensures that n is less than or equal to 26. Am I right ?

12. ## Re: Fun-Quiz

Originally Posted by Wilmer
Hmmm...why can't it be this simple:

1 + 2 + 3 + .... + n = 13n

So:
n(n + 1) / 2 = 13n
n^2 + n = 26n
n + 1 = 26
n = 25

Over and out!
You're ignoring the fact that one person (the host) takes his chocolates with him and doesn't share in the 13 per person redistribution. So you're answer misses the other solutions of N=24 and N=26. For example if N=24, and the host is the first person, the remaining 23 participants get a total of 2 + 3 + 4 + ..+24 = 299 chocolates which they can redistribute 13 apiece.

13. ## Re: Fun-Quiz

OK; should have read the FULL problem!

A similar one:

At a birthday party for n kids, n(n+1)/2 chocolates were available,
which the host intended to distribute evenly to the kids.

However, EBaines (the janitor) stole and ate 7 of the chocolates!

After the host realised this, he distributed the chocolates in this manner:
g chocolates to each of the girls
(g - 1) chocolates to each of the boys

The number of girls was twice the number of boys.
How many kids?

14. ## Re: Fun-Quiz

21 children.

Let b = number of boys. The number of girls is 2b, the total number of children is 3b, and the original number of chocolates is 3b(3b+1)/2. Let k = number of chocolates the host originally intended to give each child; hence k3b = 3b(3b+1)/2, or k=(3b+1)/2. In order for k to be an integer b must be odd.

After the 7 chocolates are taken the remaining are distributed, so

$\frac {(3b+1)3b} 2 - 7 = 2bg +b(g-1)$

Rearrange:

$g = \frac {9b+5-\frac {14} b} 6$

In order for g to be an integer, b must divide into 14. The only odd numbers that satisfy that requirement are 1 and 7. But if there's only 1 child there aren't enough chocolates for me to steal 7, so that isn't a good answer. Hence there are 7 boys and 14 girls.

Right on!