Math problem

[kingW3]

I guess he meant "how many". Obvioudly 2 satisfies condition, but how would you prove it's the only one? Or that there's more? Or infinity? It's actually quite simple, just think of it.

Anyway, why did you resort to trigonometry? This particular one could be done by perceptive 5th grader understanding how angles work in circle, by noticing one thing.

T_Bag
Should have told ya only first 8 grade math,because with sine it's pretty easy but all solutions count .
div.ide Well also it requires knowing that the sum of the opposite angles of the cyclic quadrilateral is equal to 180,which is done at 8th grade(actually our school didn't even mention it but ok) or you done it differently?
Anyway few problems incoming

[courierpower]

I just did a few problems for my bro's admission,so wanted to post some here xD
1.If p is a prime number and 3p+1 and 5p+1 are also prime number,how much are there p numbers that satisfy the conditions?
2.Triange ABC on which AB is equal to 6cm and the opposite angle of the side(the ∠ACB angle) is equal to 150°,What is the area of the circumscribed circle on the ABC triangle.
If anyone wants help or explanation of the problem(due of my English ) feel free to ask.

2 fits there, and no any other can it must be even number and only even prime is 2
btw i want to share all prime numbers (but not 2,3) can be written as 6n+1 or 6n-1 which makes it eaiser to find them in some other problems
and the triangle y t bags solution really fast u create secondary perpendicular triangle from the C corner and see their circumscribed circles are the same then 6/sin150=x/sin90 x=12 so area is 144pi

[T-Bag]

I guess he meant "how many". Obvioudly 2 satisfies condition, but how would you prove it's the only one? Or that there's more? Or infinity? It's actually quite simple, just think of it.

Anyway, why did you resort to trigonometry? This particular one could be done by perceptive 5th grader understanding how angles work in circle, by noticing one thing.

T_Bag
Should have told ya only first 8 grade math,because with sine it's pretty easy but all solutions count .Well also it requires knowing that the sum of the opposite angles of the cyclic quadrilateral is equal to 180,which is done at 8th grade(actually our school didn't even mention it but ok) or you done it different?

As I mentioned above, Im out of school for 8 years, so I just used stuff what I remember (poor me that I dont remember 8th grade )
and regarding first post about p I used Logical Statements (or w/e its called) to make sure its only number 2.

[kingW3]

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
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 ,and yeah i love circles

Spoiler for picture:

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

[div.ide]

Visual representation of simplest way to deal with circumscribed triangle:

Spoiler for Hiden:

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

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:

Spoiler for Hiden:

[kingW3]

Touche

[div.ide]

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

Spoiler for Hiden:

Case 1)

Spoiler for Hiden:

n=37
thus (n+1)/2 = 19 = 3k and  is natural -> no valid resuts

Case 2)

Spoiler for Hiden:

n+1 = 37
thus n/2 = 18 = 3k
k=6, p = 666 -> fits
n= 36
sum of digits = 9

Case 3)

Spoiler for Hiden:

n/2 = 37
n+1 = 75 = 3k
k = 15 -> out of range

Case 4)

Spoiler for Hiden:

n+1/2 = 37
you can guess from case 3 this won't fit either

So answer's 9, but it isn't that much of an elegant solution.

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

[div.ide]

Aha, just realised how could I make it much shorter.

n(n+1)/2 = 37 * 3k -> n(n+1) = 37 * 2 *3k

and since 0<k<10, it's obvious you'd have to group it as 37 * 6k because no other combination would result in 2 consecutive numbers (n*[n+1]), considering k<10

Now choosing between 6k = 36 and 6k = 38 (k being prime) is easy.

[kingW3]

Well,Good job
I shall start with a bit advanced problems
How much is the sum of all the solutions of the equation:
x + sqrt(x^2+16)=40/sqrt(x^2+16)

Value of a real parameter m for which the sum of the square of the solutions of equation x^2-mx+m-3=0 smallest value of m is