SMO JUNIOR SECTION 2012 SOLUTIONS PART 1

Problem 1 :

Let α and β be the roots of the quadratic equation

x² + 2bx + b = 1

The smallest possible value of (α - β)²

(A)  0     (B) 1      (C)  2    (D)  3     (E)  4

Solution :

x² + 2bx + b - 1 = 0

(α - β)² =  (α + β)² - 4 α β ---(1)

a = 1, b = 2b and c = b - 1

α + β = -b/a

α β = c/a

α + β = -2b/1 ==> -2b

α β = (b - 1)/1 ==> b - 1

(α - β)² =  (-2b)² - 4(b - 1)

= 4b² - 4b + 4

= 4(b² - b + 1)

It will have minimum (Smallest value) when x = 1/2.

4[(b - (1/2))² + 3/4] ≥ 3

So, option D is correct.

Problem 2 :

It is known that n²⁰¹² + n²⁰¹⁰ is divisible by 10 for some positive integer n. Which of the following numbers is not a possible value for n?

(A) 2    (B) 13     (C) 35     (D) 47    (E) 59

Solution :

= n²⁰¹² + n²⁰¹⁰

= n²⁰¹⁰(n² + 1)

If the number is divisible by 10, then it should end with 0.

• Cyclicity of 2 is 4 ==> (2, 4, 8, 16, ..........)
• Cyclicity of 3 is 4 ==> (3, 9, 27, 81, ..........)
• Cyclicity of 5 is 1 ==> (5, 25, 125, ..........)
• Cyclicity of 7 is 4 ==> (7, 49, 343, 2401 ..........)
• Cyclicity of 9 is 2 ==> (9, 81, 729, ..........)

Option A : If n = 2

Cyclicity of 2 is 4.

2010/4 = remainder 2

2² = 4.Unit digit of 2²⁰¹⁰ is 4

= 2²⁰¹⁰(2² + 1)

= 4(2² + 1)

= 4(5)

Will end with 0, divisible by 10.

Option B : If n = 13

Cyclicity of 3 is 4.

2010/4 = remainder 2

3² = 9.Unit digit of 3²⁰¹⁰ is 9

= 9(13² + 1)

= 9(unit digit of 13² is 9 + 1)

= 9 (unit digit is 0)

Will end with 0, divisible by 10.

Option C : If n = 35

Cyclicity of 5 is 1.

2010/1 = remainder 0

5⁰ = 5.Unit digit of 5²⁰¹⁰ is 5.

= 5(35² + 1)

= 5(unit digit of 35² is 5 + 1)

= 5 (unit digit is 6)

Will end with 0, divisible by 10.

Option D : If n = 47

Cyclicity of 7 is 4.

2010/4 = remainder 2

7² = 49.Unit digit of 7²⁰¹⁰ is 9.

= 9(47² + 1)

= 9(unit digit of 47² is 9 + 1)

= 9 (unit digit is 0)

Will end with 0, divisible by 10.

Option E : If n = 59

Cyclicity of 9 is 2.

2010/2 = remainder 0

9⁰ = 1.Unit digit of 59²⁰¹⁰ is 1.

= 1(59² + 1)

= 1(unit digit of 59² is 1 + 1)

= 9 (unit digit is 2)

Will not end with 0, not divisible by 10. So, option E is correct.