Trollathon #1.3 : A Sum and A Function

Level 2

Let A = 2 + 3 + . . . + 23 A = 2 + 3 + ... + 23

Answer = A p ( p ( p ( p ( 2 ) + 1 ) ) ) A - p(p(p(p(2) + 1)))

Hint : p ( 1 ) = 2 , p ( 3 ) = 5 , p ( 5 ) = 11 p(1) = 2, p(3) = 5, p(5) = 11 .


The answer is 41.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Zi Song Yeoh
Apr 2, 2014

Firstly, one notes that p ( n ) p(n) denotes the nth prime. So, p ( p ( p ( p ( 2 ) + 1 ) ) ) = 59 p(p(p(p(2) + 1))) = 59 . The more difficult part is the sum. One obviously would think it is the sum of all integers from 2 to 23, but you discovered that it is wrong after submitting. One notes that this problem is about primes and suspect that it is the sum of all primes from 2 to 23, which gives 100 100 . So, the answer is 100 59 = 41 100 - 59 = \boxed{41} .

Bad question, I considered p ( x ) p(x) as the minimum value of the closest prime to 2 x 2x , so

p ( 2 ) + 1 = 3 + 1 = 4 p ( p ( 2 ) + 1 ) = p ( 4 ) = 7 p ( p ( p ( 2 ) + 1 ) ) = p ( 7 ) = 13 p ( p ( p ( p ( 2 ) + 1 ) ) ) = p ( 13 ) = 23 \begin{aligned} p(2) + 1 & = & 3 + 1 = 4 \\ p(p(2) + 1) & = & p(4) = 7 \\ p(p(p(2) + 1)) & = & p(7) = 13 \\ p(p(p(p(2) + 1))) & = & p(13) = 23 \\ \end{aligned}

So the answer can also be 100 23 = 77 100 - 23 = \boxed{77}

Pi Han Goh - 7 years, 2 months ago

Log in to reply

LOL

Zi Song Yeoh - 7 years, 2 months ago

this problem really tolled me up

Kishlaya Jaiswal - 7 years, 2 months ago

Log in to reply

Shouldn't it be trolled me up?

Ameya Salankar - 7 years, 2 months ago

Log in to reply

LOL I missed that essential 'r'

Kishlaya Jaiswal - 7 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...