# 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)=  (-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 35is 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.

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