maths questions to challenge you!!!?

Question

1) Find all integers n>1 for which n^4+4^n is a prime.

2) Are there any integers x and y such that x^2-5y^2=2. Show it

3) How many integers are there such that n such that 2^n+27 is divisible by 7?

4) Find the least postive integer that leaves remainder of 1, 1, 4, 3 when divided by 2, 3, 5, and 4 respectively.

5) Solve the folowing congruences: x=5(mod 7), x=7(mod12) , x=3(mod 25).

6) Last one now. Solve x=5(mod6), x=4(mod 11), x=3(mod 17)

Answer 1

Question 3.
2^0 + 27 = 28 = 4 * 7
2^3 + 27 = 35 = 5 * 7
2^6 + 27 = 91 = 13 * 7
2^9 + 27 = 539 = 77 * 7
2^12 + 27 = 4123 = 589 * 7

if a = n/3
2^n + 27
2^(3a) + 27
8^a + 27

if 8^a + 27 = 7(5 + 8^1 + 8^2 + ... 8^(a-1)) then 8^a + 27 is evenly divisable by 7, and 2^n + 27 is divisible by 7 where n = 0, 3, 6, 9 ...

8^a + 27 = 7(5 + 8^1 + 8^2 + ... 8^(a-1))
(8^(a+1) + 27) / 7 = 5 + 8^1 + 8^2 + ... 8^a)
(8^(a+1) + 27) / 7 - 5 = 8^1 + 8^2 + ... 8^a)

prove using math induction.
(8^(k+1) + 27) / 7 - 5 = 8^1 + 8^2 + 8^3 + ... 8^k
If k = 1 then (8^2 + 27) / 7 - 5 = 8 -> TRUE
Solve for k+1
(8^(k+1) + 27) / 7 - 5 + 8^(k+1) = 8^1 + 8^2 +... 8^k + 8^(k+1)
8^(k+1) / 7 + 27 / 7 - 5 + 8^(k+1) = .....
(1/7)8^(k+1) + 27 / 7 - 5 + 8^(k+1) = .....
(1/7)8^(k+1) + 27 / 7 - 5 + 8^(k+1) = .....
(8/7)8^(k+1) + 27 / 7 - 5 = .....
(1/7)8^(k+2) + 27 / 7 - 5 = .....
8^(k+2) / 7 + 27 / 7 - 5 = .....
(8^(k+2) + 27) / 7 - 5 = 8^1 + 8^2 +... 8^k + 8^(k+1)
TRUE.

Answer 2

1) No integers rather than 1
2)No real x,y because x^2=5y^2+2 and there is no perfect square
Let A be an integer gives remainder m when it is divided by 5
hence the integer A can be written as
5N+m where N is a the integer portion of division A/5
If we squared A we get A^2=(5N+m)^2
=25n^2+10Nm+m^2
If we took the remainder of A^2 Which can be any perfect square
A^2 mode 5=(25n^2+10Nm+m^2) mode 5
=((25N^2 mode 5) +(10Nm mode 5)+(m^2 mode 5)) mode 5
=0+0+m^2 mode 5
m can be 0,1,2,3, or 2
(m^2 mode 5) can be 0, 1 or 4 only
any perfect square (5N+m)^2 can’t give remainder 2 when divided by five.
3)n=0,3,6,9,12,.....................
4)19
5)103
6)785
It is not perfect answer but it is the best here!

Answer 3

1) n=1
2) there is no answer or may be i dont know how to find
3) n=3
4) 19
5) x=7, x=12, x=25
6) x=6, x=11, x=17

Answer 4

when I was taking number theory, I found that struggling my way thru the homework let me do well on the midterms. Other approaches such as asking oter people to do my homeowrk, would make me feel confident, but when test-time came, .. uhh ... blankness of thought came too

Answer 5

1ans - for n=1 e get the prime no.plz follow mathematical induction
2ans- there are no integers whose square ends
with two or three
3ans)- substitute for n=1,2,3,4...
then e get series for n =3,6,9,12,....
therfoe it is 3(1+2+3+4+.........+infinity) no. of
integers
4 ans)-19

Answer 6

I know the answer of question 4

19

Answer 7

nice,Ok,this is the first level,lets go to 2nd one

Answer 8

omfg... this is tough..
i totally forgot all about it..
lol.......