2010/027) What is the square root of -i

THis can be done both with and with out De Moivre's formula as mentioned in

http://en.wikipedia.org/wiki/De_Moivre%2…

1st with it

-i = 0 + (-1) i = r cos t + r sin t

r^2 = 1

cos t = 0 and sin t = - 1 so t = 3pi/2

so -i = e^(3pi/2) i

so sqrt(i) = e^(3pi/4) i or e^5pi/4 taking (3pi/2 + 2npi/2) n = 0 and 1 n =2 3 gives 1st value

= cos 3pi/4 + i sin 3 pi/4 or cos 5pi/4 + i sin 5 pi/4

= - 1/sqrt(2) + 1/sqrt(2) i or 1/sqrt(2) - 1/sqrt(2) i

in case you are not familiar with De Moivre's formula then say

sqrt(-i) = (a+ib)

sqare both sides

-i = a ^2-b^2 + 2abi

so a^2-b^2 -= 0
and 2ab = -1
a^2-b^2 = 0 => a = +/-b so ab = -1/2 and we get a = 1/sqrt(2) b= - 1/sqrt(2)

or a = -1/sqrt(2) b= + 1/sqrt(2)

giving same results

2010/026) Strong induction problem

Suppose x is a real number, x does NOT equal 0, and (x + 1/x) is an integer

Prove for all n ≥ 1, x^n + 1/(x^n) is an integer.
x+ 1/x is in integer = n (given)

so (x+ 1/x)^2 = n^2

x^2 + 1/x^2 = n^2-2 is an integer
so true for 1 => true for 2

let it be true for upto k
now x^k+ 1/x^k

(x^k+ 1/x^k)(x+ 1/x)

= x^(k+1) + 1/x^(k-1) + x^(k-1) + 1/x^(k+1)

so x^(k+1) + 1/x^(k+1) = (x^k+1/x^k)(x+1/x) - (x^(k-1) + 1/x^(k-1))

if it is true for all n upto k so RHS is integer then LHS is integer so for k+ 1 so the induction step is proved

hence proved

2010/025) For which positive integer values of n occur EXACTLY two 0's at the end of 1^n + 2^n + 3^n + 4^n (base 10)

As it can have maximum 2 zeroes
Let us first find the period at which 1^n+2^n+3^n+ 4^n mod 100 repeats.

The period has to be < 100 as there are maximum number if 100 remainders



1^n mod 100 =1

As 3 is coprime to 100

3^40 = 1 because 40 is for n = 100 as per http://en.wikipedia.org/wiki/Euler%27s_t…

Because 2 and 4 are factors of 100 we need to find period n for which 2^n repeats to base 25. The for 25 is 20

That is 2^20 mod 25 = 1 and 4^20 mod 25 = 1

3 has a period factor of 40( 40 is maximum) and 2 and 4 have a period factor of 20 that is after that \it repeats

So 1^n,2^n, 3^n, 4^n has a period of 40 that is it repeats every 40 times for mod 100

For 1 to 40 we need to find for which the remainder 1^n+2^n+3^n+4^n is 0 mod 1

By a computer program it is found to be 3,5,7,11,15,19,23,25,27,31,35,39.

The values are 40k + 3,40k + 5, 40k + 7, 40k + 11, 40k + 15, 40k + 19, 40k + 23, 40k + 25, 40k + 27, 40k + 31, 40k + 35, 40k + 39 where k is an integer >= 0

2010/024) If the roots of [b-c]x*x+[c-a]x+[a-b]=0are equal then prove that 2b=a+c

roots are equal so discriminant is zero

so (c-a)^2 - 4(b-c)(a-b) = 0

or c^2-2 ac + a^2 - 4(ab-ac- b^2 + bc) = 0

or a^2+4b^2+c^2+2ac - 4ab - 4bc = 0

or(a+c-2b)^2 = 0

or a+c-2b = 0 or a + c = 2b

proved

2010/023) prove that a^4+b^4+c^4> or =abc(a+b+c)

we know

(a^2-b^2) ^2 >= 0

or a^4+b^4 >= 2a^b^2

similarly

b^4 + c^4 >= 2 b^2 c^2

c^4+ a^4 >= 2 c^2 a^2
adding all 3 above and deviding by 2

a^4+b^4+c^4 >= (a^2b^2+b^2c^2 + c^2 a^2) ... 1


now a^2+c^2 >= 2ac

multiply by b^2 on both sides

b^2(a^2+c^2) >= 2b^2ac -- 2

similarly
a^2(b^2+c^2) >= 2a^2bc ... 3

and c^2(a^2+b^2) >= 2c^2ab .. 4

adding (2) (3) and (4) we get

2(a^2b^2+b^2c^2+c^2a^2) >= 2abc(b+a+c)

or (a^2b^2+b^2c^2+c^2a^2) >= abc(a+b+c) ...5

from 1 and 5 we get a^4+b^4+c^4>=abc(a+b+c)

proved

2010/022) What is the largest number of 0's that can occur at the end of 1^n+2^n +3^+4^n

for every positive integer n

1^n + 2^n + 3^n + 4^n

n = 1 gives 10 and hence one zero
n =2 gives 41 so zero
n =3 gives 100 so 2 zeros

and it cannot end with > 2 zeros as it can be proved in steps as below

for it to end with k zeroes it need to be divisible by 10^k that is 2^k and 5^k. If we can show that is it not divisible by 8 then we are through.

for n > = 3, 2^n and 4^n are divisible by 8

now 1^n =1 for all n
3^n = 1 mod 8 for even n and 3 mod 8 for odd n

So 1^n + 3^n = 2 mod 8 for n even and 4 mod 8 odd n

So not divisible by 8 for any n and hence it cannot have >2 zeros

so ans is 2(for n = 3)

2010/021) Prove that there is no a for which a^2-3a -19 is divisible by 289

Proof:
As a first step as we see that 289 = 17^2.

Now a^2-3a-19 = (a-10)(a+7) + 51

The 2nd term that is 51 is divisible by 17 and for the 1st term that is product to be divisible by 17 either (a-10) or (a+7) is divisible by 17. but if one of them is divisible by 17 then the 2nd one is divisible by 17.

So 1st term is divisible by 289 and 2nd one is not divisible by 289 so sum is not divisible by 289. Or the 2nd term is divisible by 17 and 1st term is not divisible by 17 so sum is not divisible by 17.

So the expression is not divisible by 289.