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)