Candles....(From CMC test)

[bristy1588]

Noushin has $n$ candles. On the first day she lights 1 candle for 1 hour, on the second day she lights 2 candles for 2 hours.... Like this on the n-th day she lights $n$ candles for $n$ hours. The candles chosen according to one's wish. If after the n-th day all the candles had been lighted for the same number of hours, find $n$.

[confused]

candles can be lit maximum :n(n+1)/2 =m hr.m must be an intreger.so n must be odd.

[bristy1588]

Confused, i meant on day 2, 2 candles were lit each for 2 hrs.. and so on

[nafistiham]

যেকোনো বেজোড় সংখ্যাই এমন হতে পারে
ধরা যাক,$n=2k+1$
$1+2+3+...........+2k+(2k+1)$
এখন, $1+2k=2+2k-1=3+2k-3=.............$
সুতরাং, এভাবে মোম ব্যাবহার করা যেতে পারে ।

[bristy1588]

Tiham,
Lets see the case when n=3. The 3 candles are A,B and C. Now. We know on the first day 1 candle was lit, candle A, therefore, After day 1, Candle A was lit for 1 hour and the candle B,C for 0 hours, Now let on day 2, candles B and C were lit, each for 2 hours. So after day 2 in total, Candle A was lit for 1 hour, Candle B for 2 hours and Candle C for 3 hours. At day 3, All the 3 candles were lit for 3 hours each. In total after day 3, A was lit for 4 hours, B for 5 hours and C for 5 hours. However, all candles were supposed to be lit for the same number of hours.
I think u did not understand the question.

[Labib]

Cracked it...

Let's count how many hours in total were the candles lit...
$1$st day, we lit the candles $1*1=1^2$ hours.
$2$nd day, we lit the candles $2*2=2^2$ hours.
$\cdots\cdots$
$\cdots\cdots$
$n$th day we lit the candles $n*n=n^2$ hours.

So, in total we lit the candles for ...

$1^2+2^2+\cdots+n^2=\frac{(n(n+1)(2n+1)}6$ hours.
Thus each candle was lit $\frac{(n+1)(2n+1)}6$ hours.

Now let's calculate backwards...

If $n>1$,

After the $n$th day, each candle was lit $\frac{(n+1)(2n+1)}6$ hours.
After the $(n-1)$th day, each candle was lit $[\frac{(n+1)(2n+1)}6-n]$ hours.
After the $(n-2)$th day, $(n-1)$ candles were lit $[\frac{(n+1)(2n+1)}6-n-(n-1)]$ hours. Other $1$ candle was lit $[\frac{(n+1)(2n+1)}6-n]$ hours.
After the $(n-3)$th day, $(n-2)$ candles were lit $[\frac{(n+1)(2n+1)}6-n-(n-1)-(n-2)]$ hours. Other $2$ candles were lit $[\frac{(n+1)(2n+1)}6-n]$ and $[\frac{(n+1)(2n+1)}6-n-(n-1)]$ hours respectively.
$\cdots\cdots$
$\cdots\cdots$
On the $(n-n-1)=1$st day, $(n-n-2)=2$ candles were lit $[\frac{(n+1)(2n+1)}6-\frac{n(n+1)}2+1]$ hours. Other $(n-2)$ candles were lit $[\frac{(n+1)(2n+1)}6-n],[\frac{(n+1)(2n+1)}6-n-(n-1)],\cdots, [\frac{(n+1)(2n+1)}6-\frac{n(n+1)}2+3]$ hours respectively.

Certainly,
$[\frac{(n+1)(2n+1)}6-n]$ is the highest hour a candle was lit after the First day.
So,
$[\frac{(n+1)(2n+1)}6-n]=1$.
$\Rightarrow n=-1,\frac 52$.
So, $n>1$ doesn't have a positive integer solution for $n$.

But, when $n=1$, After the $(n-1)=0$th day, each candle was lit $[\frac{(n+1)(2n+1)}6-n]=0$ hours.
Which 'obviously' satisfies the conditions.
Therefore, $n=1$ is the only solution for $n$.

[bristy1588]

Labib,
Please tell me where I m getting u wrong:
Let me denote the total number of hours that each candle was lit for after $n$ days as $K$. And let the candle which was not lit on the $(n-1)$ is Candle A.
On the $(n-2)$ day, We have 2 cases:
1. Candle A was lit:
In this case, On the $(n-3)$ day, Candle A was lit:$(k-n-(n-2))$. 2 candle was lit for $(k-n-(n-2))$ hrs and $(n-3)$ candles were lit for$(k-n-(n-1)-(n-2))$ hrs.
2. Candle A was NOT lit:
In this case, On the $(n-3)$ day, Candle A was lit for $(k-n)$ hrs. 1 candle was lit for $(k-n-(n-1))$ hrs and $(n-3)$ candles were lit for $(k-n-(n-1)-(n-2))$ hours

In Either Way, I do not get what u proposed to happen on the $(n-3)$ day. Could u please explain?

[Labib]

Fixed the bug... It should be okay... I think...

[bristy1588]

Labib,
Why didnt u consider the case 1. i.e When Candle$A$ was lit. How do u prove that Candle A was never lit on day $(n-2)$ ?

[Labib]

Then I think the solution is wrong...
Will try to think on it tomorrow.