2014/010) Solve this system of congruences, 2x +3y ≡ 22 mod27, and x +4y ≡ 7 mod27

2x + 3y ≡ 22 mod 27  … (1)
x + 4y ≡ 7 mod 27 … (2)

multiply the second by 2
2x + 8y ≡ 14 mod 27  … (3)

subtract the first from the last one
5y ≡ -8 mod 27

as GCD(5,27) = 1, which divides any number, the congruence has one solution between 0 and 26, namely y = 20

substitute in the second equation
x + 80 ≡ 7 mod 27
x   ≡ - 73  mod 27 or 8

hence general solution is
(x = 8 + 27h, y = 20 + 27k)

2014/009)Given ab=1 bc=2 cd=3 de=4 ea=5 find a,b,c,d,e

Solution 2014/009)

Multiply all 5 to get (abcde)^2 = 120
Or abcde = +/-120^(1/2)
Devide by product of ab and cd to get e = = +/-120(1/2)/ 3= = +/-(40/3)(1/2) = +/-2(10/3)(1/2)

Similarly you can find the rest. It should be noted that all should be positive or all should be –ve.

Q2014/008) Solve in positive integers for 577(bcd+b+c)=520(abcd+ab+ac+cd+1)

577(bcd+b+c)=520a(bcd+b+c)+520(cd+1)

(577−520a)(bcd+b+c)=+520(cd+1)

First conclusion:  a=1

57(bcd+b+c)=+520(cd+1)

57b(cd+1)+57c=+520(cd+1)

57c=(520−57b)(cd+1) →b<10

Further

57c=(520−57b)(cd+1) 
=> 520 - 57b < 57
or b >= 9
so b = 9
hence

57c=7(cd+1)

c(57−7d)=7 →d=8 →c=7
refer to http://mathhelpboards.com/challenge-questions-puzzles-28/solve-positive-integers-8202.html#post37808

Q2014/007) show that the five roots of the quintic

a5 x^5+ a4x^4+a3x^3+a2x^2+a1x+a0=0 are not all real if 2a2^4<5a5 a3.="" nbsp="" p="">

I shall prove it by contradiction.
Without loss of generality let a5 = 1
 Let the roots be y1,y2, y3,y4,y5
 Then a4^2 = (y1+y2+y3+y4+y5)^2 = y1^2 + y2^2 + y3^2 + y4^2 + y5^2 + 2a3 as a3 consists of product of 2 elements that are separate

Or 2a4^2 =4a3 + 2y1^2 + 2y2^2 + 2y3^2 + 2y4^2 + 2y5^2) 

= 4a3 + ½(4y1^2 + 4y2^2 + 4y3^2 + 4y4^2 + 4y5^2) 

= 4a3 + ½(sum(ym-yn)^2+ 2ymyn)

>= 4a3 + a3 as sum (ym-yn)^2 >=0 and sum ymyn= a3

 >= 5a3

So all the roots are real => 2a4^2 >= 5a3

So under given condition all roots cannot be real

Q2014/006) Find positive integers n such that the average of 1, 2, 3, 4, ..., n is a perfect square.

average of 1,2, … n is = (n+1)/ 2 = p^2 so n = 2 p^2 - 1, p = 1 => n = 1 p = 2 => n = 7

Q2014/005) If z= -2+ 2√3i , then find z ^(2n) + [2^(2n)][z^n]+ 2^(4n)?

We have  z= -2+ 2√3i = 4 cis 120 or 4 w where w is cube root of 1

So z ^(2n) + [2^(2n)][z^n]+ 2^(4n)

= 4^2n w^2n  + 4^2n w^n + 4^2n

= 4^2n ( w^2n + w ^n + 1)

= 16^n( w^2n + w ^n + 1)

Now there are 2 cases

1)      n is multiple of 3 :  so w^n = w^2n = 1 hence sum = 3 * 16^n

2)      n is not multiple of 3 : sum = 0

Q2014/004) p(x) = a2012 x^2012 + a2011 x^2011 + … + a1 x + a0 a2012 etc., are just coefficients. and p(x) = 1/x for integer x = 1,2,3,…,2013 What is p(2014)

We have xp(x) – 1 = 0 at 1,2,…. 2013 and of order 2013

so xp(x) = 1 – m(x- 1) ..(x-2013)
at x =0 we get 1 – m(-1)(-2)(-2013) or 1 + m * 2013! = 0
m = -1/2013!
at x = 2014
2014p(2014) = 1 + 1/2013!(2013) !
= 1 + 1 = 2
or p(2014) = 2/2014 = 1/1007

Q2014/003) How many positive integers are there , for: a^2=9555^2+c^2

9555=3∗5∗7^2∗13
So 9555^2=3^2∗5^2∗7^4∗13^2
This has (2+1)(2+1)(4+1)(2+1) or 135 factors out of which one is 9555 and 67 are below 9555 and 67 are above 9555
(a + c) > 9555 and (a-c) < 9555 is a set of solution
So number of solutions 67
For a-c  we need to take3x∗5y∗7z∗13m (x,y,z,m to be chosen based on limit for example x between 0 and 3 such that a +c > 9555) and for a-c it is 9555^2/(a+c)

Q2014/002) Let x,y,z be real numbers such that 9x−10y+z=8 and x+8y−9z=10.

Evaluate x^2−2y^2+z^2.

Solution

 We are given:

(1) 9x−10y+z=8

(2) x+8y−9z=10

To eliminate z, multiply (1) by 9 and add to (2) to get:

82x−82y=82

Which implies:

x=y+1

Substituting for x in (1), we get:

9(y+1)−10y+z=8 or z−y=−1 so y−z=1

putting this in (2), we find no contradiction.

Thus, we have:

(3) x−y=1=> x = y+ 1

(4) y−z=1=> z = y- 1

x^2−2y^2+z^2= (y+1) ^2 + (y-1)^2 – 2y ^2 = 2 after expanding and simplifying

Q2014/001)Given that 1^2+2^2+3^2+.....+k^2=k*(k+1)*(2k+1)/6,what is the is the smallest integer k such that 1^2+2^2+3^2+.....+k^2 is divisible by 100.

k*(k+1)*(2k+1)/6 must be divisible by 100

or k*(k+1)*(2k+1) must be divisible by 600

600 = 25 * 8 * 3

One number is always divisible by 3 so we need not check for it.  only one that is either k or k + 1 is by 2 ( hence 8) and one by 5 ( hence 25)

So 2k + 1 > = 25 or k > = 12

If k is even then multiple of 8: so candidates 16,24

If k is odd then  multiple of 8 – 1 that is 15, 23

So we need to try 15,16,23,24 so on and checking one by one we get  24 as answer