MIOTP WEEK 3 SOLUTIONS

1) Find the sum of the digits of N

N=9+99+999+...+99....99                                                 (320   9's)

Solution 

(10-1)+(100-1)+(1000-1)............

+(100..00 -1)

  (320  zeros)

=111......110 - 320

   

2)Let a, b, and c be distinct nonzero real numbers such that

 

Find |abc|

Solution 

From the given conditions it follows that

Multiplying the above equations gives (abc)²= 1, from which the desired

result follows.

|abc|=1

3)

Solution.

n³+100=(n+10)(n²+an+b)+c

            =n³+(10+a)n²+n(b+10a)+c

           

Equating coefficients yields

10+a=0

b+10=0

10b+c=100

Solving this system yields a = −10, b = 100, and c = −900. Therefore, by the

Euclidean Algorithm, we get

n+10=gcd(n³+100,n+10)

         =gcd(-900,n+10)

         =gcd(900,n+10)

n+10=900

n=890 

                       LEVEL 2

1)

Consider the sequence a₁, a₂, . . . defined by

aₙ = 2ⁿ + 3ⁿ + 6ⁿ − 1 (n = 1, 2, ...).

Determine all positive integers that are relatively prime to every term of the

sequence.( IMO 2005)

Solution: 

If p > 3, then 2ᵖ⁻² + 3ᵖ⁻² +6ᵖ⁻²≡ 1 mod p. 

To see this, multiply

both sides by 6 to get :

2ᵖ⁻¹+3ᵖ⁻¹+6ᵖ⁻¹≡ 6 mod p,

which is a consequence of Fermat’s little theorem.

 Therefore p divides aₚ₋₂.

Also, 2 divides a₁and 3 divides a₂. So, there is no number other than 1

that is relatively prime to all the terms in the sequence.

2)The equations x² + ax + 1 = 0 and x² + bx + c = 0 have a common real root, and the equations x² + x + a = 0 and x² + cx + b = 0 have a common real root. Find a + b + c

SOLUTION 

The common root of x² + ax + 1 = 0 and x² + bx + c = 0 must also satisfy (a-b)x + (1-c) = 0, so it must be (c-1)/(a-b). Note that the other root of x² + ax + 1 = 0 must be (a-b)/(c-1), since the product of the roots is 1. Similarly the common root of x² + x + a = 0 and x² + cx + b = 0 must satisfy (c-1)x + (b-a) = 0, so it is x = (a-b)/(c-1). Hence x² + x + a = 0 and x² + ax + 1 = 0 have a common root. Hence it satisfies (a-1)x + (1-a) = 0. Now we cannot have a = 1, for then x² + ax + 1 has no real roots. Hence the common root must be 1. Hence both roots of x² + ax + 1 = 0 are 1 and so a = -2.

So x² + bx + c = 0 has one root 1. Then its other root must be c/1 = c. Hence -b = 1 + c, or b + c = -1. Hence a + b + c = -3

3)solution

4)Let a and b be positive integers such that ab + 1 divides a² + b². Prove that a² + b²/ab + 1 is a perfect square (Concept of vieta Jumping is essential for solving the problem)

solution