2014/051) Find smallest positive a such that 5x^13+13x^5 + 9ax is divisible by 65 for all x

let f(x) = x(5x^12 + 13x^4 + 9a)

x can be factored out so 65 should divide (5x^12 + 13x^4 + 9a)


So taking mod 5 we have 13x^4 + 9a   mod  5 = 0

It is true for any x so take x co-prime to 5 so we have x^4 = 1

So we get 13 + 9a  mod 5 = 0

or 3 – a mod 5 = 0

or a = 3 mod 5

Similarly we have (5x^12 + 13x^4 + 9a) mod 13 = 0

 or 5 + 9a mod 13 = 0

or a = 11 mod 13

a = 3 mod 5 and a = 11 mod 13 can be solved by using Chinese  remainder theorem also but we see that
a = -2 mod 5 and a = -2 mod 13 so a = -2 mod 65 or 63

a = 63 + 65n and a = 63 is the lowest positive integer

you can find some discussion at http://mathhelpboards.com/challenge-questions-puzzles-28/divisibility-challenge-10574.html#post49059

2014/050) find the integral part of (2005)^3/(2003 * 2004) – (2003)^3 / (2004 * 2005)

To keep the arithmetic simple put

2004 = x to get

(x+1)^3/(x(x-1) - (x-1)^3/(x(x+1)

= ((x+1)^4 – (x-1)^4) / x(x+1)(x-1)

= 2 (4x^3 + 4x)/(x(x+1)(x-1)

= 8x(x^2+1)/x(x^2-1)

= 8(x^2 + 1)/(x^2-1)

= 8 + 2//(x^2-1)

Now the beauty is for any x > 2 the integral part is 8

Hence ans is 8.

Q2014/049) prove that ( sin 120/sin 40) = 1+ ( sin 20)/ sin 100)

we have RHS
= ( sin 100 + sin 20)/ sin 100
= ( 2 sin 60 cos 40)/ sin 80
= ( 2 sin 60 cos 40)/( 2 sin 40 cos 40)
= sin 60/ sin 40
= ( sin 120)/ sin 40
proved

2014/048) The polynomial g(x) is cubic. What is the largest value of a if f1(x)=x^2+(a−29)x−a and f2(x)=2x^2+(2a−43)x+a are both factors of g(x)?

f1(x) and f2(x) must have a common factor. Otherwise g(x) shall be product of  f1(x) and f2(x) and order 4

let f1(x) = (x-m)(x-p)

and let f2(x) = 2(x-m)(x-q)

comparing constant term

of  f1(x) = mp = - a and 2 mq = a we get p = - 2q or m = 0 => a = 0

then taking the product and comparing coefficient of x

we get m+p = 29-a ...(1)

m – q = (43-2a)/2

or 2m – p = 43 – 2a ... (2)

solving (1) and (2) 3 m = (72-3a) or m = 24 – a

so p = 2m + 2a – 43 = 48 – 43 = 5

now  - a = mp = 5(24-a) or 4a = 120 or a= 30

2014/047) For how many 2 digit numbers the sum of digits is greater than the product of digits?

(x+y) > xy
or xy - x - y < 0
x(y-1) < y

y = 0 or 1 for all x

x < y/(y-1)

or x < 1+ 1/(y-1)
so x has to be 1

so multiples of 10 or any one of digits sould be 1

that 10, 20, 30, 40,50,60,70,.80,90, 11,12,13,14,15,16,17,18,19, 21,31,41,51,61,71,81,91 that is 26 numbers

2014/046) solve for a :a + 2a^2 + 3a^3 + ... = 30

Because the series converges so we must have

|a| < 1

Now we are given
a + 2a^2 + 3a^3 + ... = 30 ... (1)

 Multiply (1) by a to get

a^2 +  2a^3 + 3a^4 + ... = 30a ... (2)

 Subtract (2) from (1) to get

a + a^2 + a^3 + ... = 30(1-a)  ... (1)

 Or a/(1-a) = 30(1-a)

Or 30(1-a)^2 = a

Or(30 – 61 a + 30a^2) = 0

Or (5-6a)(6-5a) = 0

hence a = 5/6 o 6/5

As |a| < 1 so a = 5/6 

2014/045) Solve for x: (ab)^2 =(bc)^ 4 =(ca)^ x =abc

we have ab = (abc)^(1/2)
bc = (abc)^(1/4)
ca = (abc)^(1/x)
multiply and get
(abc)^2 = (abc)^(1/2 + 1/4 + 1/x)

hence 1/2 + 1/4 + 1/x = 2 or x = 4/5

2014/044) Find all positive integers a and b such that a(a+2)(a+8)=3^b

all of a , a + 2 and a + 8 have to be power of 3( power 0 included)

a cannot be  power of 3 >= 1 or >= 3 then a+ 2 and a+ 8 are not divisible by 3

so check a =3^ 0 = 1

 that give a + 2 = 3 and a + 8 = 9 both power of 3

so we get a(a+2)(a+8)=3 3  

so a = 1 and b= 3 is the only solution

2014/043) For any polynomial P(x) show that P(a) - P(b) is divisible by a-b

 Proof:
Let p(x) =  t(n)x^n + t(n-1)x^{n-1} + ...  + t(0)
Then
 p(a) =  t(n)a^n + t(n-1)a^{n-1} + ...  + t(0)
 p(b) =  t(n)b^n + t(n-1)b^{n-1} + ...  + t(0)

So  p (a) – p(b) =  t(n)(a^n- b^n)  +t(n-1)(a^(n-1)-b^(n-1)  + ....  + t(1)(a-b)

As each of the a^k-b^k  is divisible by a- b so  p(a) – p(b) is divisible by a-b.

As a corollary


If p(x) has integer coefficients and P(0) and P(1) are odd it does not have any integer root.
This is so because P(even) – p(0) is even and P(odd) – p(1) is even so neither can be zero

2014/0042) show that x = 2 + 2^(1/3) + 2^(2/3) is a root of x^3 - 6x^2 + 6x -2 = 0

Let x = 2 + 2^(1/3) + 2^(2/3)
so x -  2 =  2^(1/3) + 2^(2/3)  ... (1)

cube both sides to get (x-2)^3 = 2 + 4 + 3 *2 * (2^(1/3) + 2^(2/3)) = 6 + 6 (x-2) ( from 1)
or x^3 - 6x^2 + 12 x - 8 = 6x - 6
or x^3 - 6x^2 + 6x - 2 = 0

so x is a root of above equation
proved