Olympiad Problems

Find the common roots of the equation x²⁰⁰⁰+x²⁰⁰²+1 = 0 and x³+2x²+2x+1 = 0.

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x³+2x²+2x+1 = 0

=> x³+1+2x(x+1) = 0

=> (x+1)(x²-x+1)+2x(x+1) = 0

=> (x+1)(x²-x+1+2x) = 0

=> (x+1)(x²+x+1) =0

=> x = -1,w,w² where 'w' is the cube root of unity

Now putting these values in the expression x²⁰⁰⁰+x²⁰⁰²+1 we see that w and w² only reduce it to zero. So the common roots are w,w²

Find real x for which 1/[x] + 1/[2x] = {x} + 1/3, where [x] = greatest integer less than equal to x and {x} = x – {x}.

Let x = z + f, where z is the integer part and 0 <= f < 1

Now we have two cases (a) 0 <= f < 1/2 (b) 1/2 <= f < 1

Also note that the R.H.S of the given equation i.e., {x} + 1/3 > 0 so x has to be positive.

Case (a) 0 <= f < 1/2, [x] = z and 2x = 2z + 2f , [2x] = 2z since 0 <= 2f < 1

and {x} = x – [x] = z + f – z = f.

This reduces the equation to

1/z + 1/2z = f + 1/3 ........... (1)

=> 1/z + 1/2z >= 1/3 (Since f >= 0 )

=> 3/2z >= 1/3

=> 9 >= 2z

=> 2z – 9 <= 0

=> z = 1,2,3,4

for z = 1, {x} = f = 1 which is invalid in this case

for z = 2, using equation (1) {x} = f = 5/12 < 1/2

for z = 3, using equation (1) {x} = f = 1/6 < 1/2

for z = 4, using equation (1) {x} = f = 1/24 < 1/2

so possible values of x in this case is 2+5/12, 3+1/6 and 4+1/24

i.e., x = 29/12, 19/6 and 97/24

Case(b) Following the steps of case (a) show that in this case no solution exists.

Show that abc(a³-b³)(b³-c³)(c³-a³) is always divisible by 7 where a, b and c are non-negative integers.

Result: Cube of any integer leaves remainder 0, 1 or 6 on dividing by 7.

if any one of a,b or c is divisible by 7, then abc(a³-b³)(b³-c³)(c³-a³) is divisible by 7. So we assume that none of the numbers a,b or c is divisible by 7.

Now using the result stated above a³, b³,c³ leaves remainder 1 or 6 (due to the assumption) otherwise if 7 | a³ or 7 | b³ or 7 | c³ => 7 | a or 7 | b or 7 | c ( since 7 is prime) A contradiction to the assumption.

Now by Pigeon Hole Principle two of the numbers a³, b³,c³ have the same remainder. Hence their difference is divisible by 7, thus

7 | abc(a³-b³)(b³-c³)(c³-a³) in any case.

Find the remainder when 2 is divided by 1990

Result: If p is prime & p do not divide a then a^p is congruent a (modulo p) (Fermat's Theorem)

Note that 1990 = 199 X 10 and 199 is a prime.

Using Fermats theorem we have 2¹⁹⁹ is congruent 2 (modulo 199)

=> (2¹⁹⁹ )¹⁰ is congruent 2¹⁰ (modulo 199)

=> 2 is congruent 1024 (modulo 199)

i.e 199 | 2 - 1024

Now 2 and 1024 has the same last digit, which is 4 ( try to find it for 2 ). Therefore

10 | 2 - 1024

Now g.c.d (199, 10) = 1 therefore 1990 | 2 - 1024

So the remainder is 1024