Problem form American Invitational Mathematics Examination: Can you solve it?

A very beautiful problem with elegant solution. Suitable for students preparing for IIT and other entrance exams. Even for school students this problem is suited who wants to learn something different. As the problem doesn't require more than knowledge of multiplication of variables and solving simultaneous linear equations, a 10th grade students can also understand it!

Circle: Geometry

Given below is a fine problem based on simple properties of circles ( tangent and chord ). Try yourself before looking at the solution!

Here is the solution!

Olympiad Problems

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

#IIT #CMI #ISI

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

Pigeon Hole Principle and Divisibility

Here is a good problem Pigeon Hole Principle and Divisibility of integers. These type of problems are important for  Olympiads, Indian Statistical Institute (ISI) and  Chennai Mathematical Institute (CMI) .
Get the pdf file here. Please leave your comment.