So stupid.
528 * 719 * 1110 * 139 * 177 * 195 * 235 * 295 * 313 * 373 * 412 * ... * 592 * 61 * ... * 113
2124*389
Note - all power bases in above fraction are prime.
TBH it doesn't look any simpler :D
Well you got that hyptoneuse is 12(x=12),and because the center of circumscribed circle is at the middle of the hypotenuse therefore r is equal to x/2 so r=6 and P=36pi,but nice way of doing it :) .About the p,i'll wait for div.ide he seemed like he wanted to say something about it :P
Anyway first
In a circle two normal chords(by normal i mean they form a angle of 90° degrees) cut each other to lines 3cm and 7cm.What is the distance from the center of circle and the dot where the two chords cut.
A picture cause my english :P ,and yeah i love circles :D
(http://imageshack.us/a/img29/9801/circlew.png)
2.If n is a natural number so that 1+2+3+...+n is a 3-digit number with same digits(111,222,333,what i mean by this).Then the sum of the digits on n is equal to?
3.When n is divided by 3 you get 2 as a leftover(n % 3= 2 modulo operator),when it's divided by 37 leftover is 22 how much is the leftover when dividing 111(n % 111)
Feel free to ask questions :P
Visual representation of simplest way to deal with circumscribed triangle:
(http://i.imgur.com/eHmMfuU.png)
No, I don't have to say much about p, because
Only 2 fits there, and no any other number can. It must be even number, and only even prime is 2 :D
If you fix english, it's basically the answer. Any odd p would result in even numbers after operations, and gneraly even numbers are not prime by definition. Thus, we need odd prime p, and there's only one odd prime number = 2, which incidentally satisfies our conditions.
Strings problem is another simple constrution:
(http://i.imgur.com/T1asNcB.png)
2.If n is a natural number so that 1+2+3+...+n is a 3-digit number with same digits(111,222,333,what i mean by this).Then the sum of the digits on n is equal to?
1+2+3+...+n = n(n+1)/2
111 = 3*37*1
222 = 3*37*2 etc.
let's call all those 3-digit numbers p
p = 3*37*k, k belongs to {1,2,...,9}
so, n(n+1)2 = 3*37*k // if these 2 numbers are equal, they must have same prime factors, thus
Case 1)n=37
thus (n+1)/2 = 19 = 3k and is natural -> no valid resuts
Case 2)
n+1 = 37
thus n/2 = 18 = 3k
k=6, p = 666 -> fits
n= 36
sum of digits = 9
Case 3)n/2 = 37
n+1 = 75 = 3k
k = 15 -> out of range
Case 4)n+1/2 = 37
you can guess from case 3 this won't fit either
3.When n is divided by 3 you get 2 as a leftover(n % 3= 2 modulo operator),when it's divided by 37 leftover is 22 how much is the leftover when dividing 111(n % 111)
using the fact(s) that 3|111 and 37|111
1) 2 (mod 3) = 3k + 2 (mod 111), k (belongs to) {0,1,2,...,36}
2) 22 (mod 37) = 37k + 22, (mod 111), k (belongs to) {0,1,2}
so, 2) gives us 3 potential numbers: 22, 59, 81
22 = 3*7 + 1
59 = 3*19 + 2
81 = 3*27 + 0