Probability of Divisibility. (1 Viewer)

deswa1 · Sep 14, 2012

Is it 2/3?

Fus Ro Dah · Sep 14, 2012

RealiseNothing said:
$\\ $A random positive integer $ n $ is selected for the expression $ \\\\ S_n = \sqrt{\frac{\sum_{k=1}^{n} k^3 }{9}} \\\\ $ Find the probability that $ S(n) $ is an integer$.$

$\\ S_n = \sqrt{\frac{\sum_{k=1}^{n} k^3 }{9}} = \frac{\sqrt {\sum_{k=1}^{n} k^3} }{3} $ but one cool property of the Triangular Numbers is that if we define the nth Triangular Number to be $ T_n $, then $ \sum_{k=1}^{n} k^3 = T_n ^2 $ so what we actually have there is $ S_n = \frac{T_n}{3}. $Another cool thing about the Triangular Numbers is that assuming base 10, the digital root of a Triangular Number is either 1, 3, 6 or 9. The digital root sequence is 1 3 6 1 6 3 1 9 9 and combined with the fact that if the digital root is divisible by 3, then the number is divisible by 3, we have the probability of $ S_n $ being an integer is $ \frac{2}{3}$

deswa1 · Sep 14, 2012

Fus Ro Dah said:
$\\ S_n = \sqrt{\frac{\sum_{k=1}^{n} k^3 }{9}} = \frac{\sqrt {\sum_{k=1}^{n} k^3} }{3} $ but one cool property of the Triangular Numbers is that if we define the nth Triangular Number to be $ T_n $, then $ \sum_{k=1}^{n} k^3 = T_n ^2 $ so what we actually have there is $ S_n = \frac{T_n}{3}. $Another cool thing about the Triangular Numbers is that assuming base 10, the digital root of a Triangular Number is either 1, 3, 6 or 9 and combined with the fact that if the digital root is divisible by 3, then the number is divisible by 3, we have the probability of $ S_n $ being an integer is $ \frac{3}{4}$

Crap lol- I got all that (except for the digital root- not sure what that means) but I ended up with 2/3 (originally typo in previous post but I edited it...). Where did I make the mistake? (I'll try and post my working- give me a sec)

Fus Ro Dah · Sep 14, 2012

deswa1 said:
Crap lol- I got all that (except for the digital root- not sure what that means) but I ended up with 2/3 (soz- typo in previous post). Where did I make the mistake? (I'll try and post my working- give me a sec)

I made a mistake because I miscounted the sequence. The answer is 2/3.

deswa1 · Sep 14, 2012

Fus Ro Dah said:
I made a mistake because I miscounted the sequence. The answer is 2/3.

Oh ok- cool. That was a nice question- thanks Realise

And I didn't know that the sum of cubes was the triangular numbers squared- I wrote out the first ten terms of the cubes and noticed that haha so I learnt something new as well..

seanieg89 · Sep 14, 2012

Fus Ro Dah said:
$\\ S_n = \sqrt{\frac{\sum_{k=1}^{n} k^3 }{9}} = \frac{\sqrt {\sum_{k=1}^{n} k^3} }{3} $ but one cool property of the Triangular Numbers is that if we define the nth Triangular Number to be $ T_n $, then $ \sum_{k=1}^{n} k^3 = T_n ^2 $ so what we actually have there is $ S_n = \frac{T_n}{3}. $Another cool thing about the Triangular Numbers is that assuming base 10, the digital root of a Triangular Number is either 1, 3, 6 or 9. The digital root sequence is 1 3 6 1 6 3 1 9 9 and combined with the fact that if the digital root is divisible by 3, then the number is divisible by 3, we have the probability of $ S_n $ being an integer is $ \frac{2}{3}$

Pretty sure the answer is 2/3. The question is equivalent to finding the probability that n(n+1) is divisible by 3. This occurs precisely when n is congruent to 0 or 2 mod 3.

EDIT. You noticed this already, cool.

Probability of Divisibility. (1 Viewer)

RealiseNothing

what is that?It is Cowpea

deswa1

Well-Known Member

Fus Ro Dah

Member

deswa1

Well-Known Member

Fus Ro Dah

Member

deswa1

Well-Known Member

seanieg89

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)