Math IOQM

Q24. If N is the number of triangles of different shapes (i.e., not similar) whose angles are all integers (in degrees), what is N/100? Prerequisite: Stars and bars method . Example: Distribute 3 apples in 3 kids. (3+3-1)_C_(3-1) = 5_C_2 = 10 (0,0,3) => 3 variations (1,1,1) => 1 variation (01,2) => 6 variations Total: 3+1+6 = 10 Solution: By stars and bars method if we have to distribute 'n' apples in 'k' kids where a kid can receive 0 apples also, then the number of ways are: (n+k-1)_C_(k-1). In case of triangles we have to distribute 180 degrees in 3 angles but none of the angles can be 0. So if the original equation is x + y + z = 180, we rewrite it as X + Y + Z = 177 where X = x-1, Y = y-1 and Z = z-1 Now X,Y,Z can be 0 and we can apply stars and bars method. 179_C_2 = 15931 But these distributions treat (0,0,177) and (0,177,0) as different. The original problem statement specifically asks for triangles for different shapes, so we have to treat them as same...

9. Suppose a , b a, b are integers and a + b a + b is a root of x 2 + a x + b = 0 x^2 + ax + b = 0 . What is the maximum possible value of b 2 b^2 ? [PRE-RMO – 2018] We are given that a + b a + b is a root of the quadratic equation: x 2 + a x + b = 0 x^2 + ax + b = 0 Substituting x = a + b x = a + b into the equation: ( a + b ) 2 + a ( a + b ) + b = 0 (a + b)^2 + a(a + b) + b = 0 Expanding the terms: a 2 + 2 a b + b 2 + a 2 + a b + b = 0 a^2 + 2ab + b^2 + a^2 + ab + b = 0 2 a 2 + 3 a b + b 2 + b = 0 2a^2 + 3ab + b^2 + b = 0 This is a quadratic equation in a a : 2 a 2 + 3 a b + ( b 2 + b ) = 0 2a^2 + 3ab + (b^2 + b) = 0 To ensure that real values of a a exist, the discriminant D D of this quadratic must be a perfect square. The discriminant is: D = ( 3 b ) 2 − 4 ⋅ 2 ⋅ ( b 2 + b ) = 9 b 2 − 8 ( b 2 + b ) D = (3b)^2 - 4 \cdot 2 \cdot (b^2 + b) = 9b^2 - 8(b^2 + b) D = 9 b 2 − 8 b 2 − 8 b = b 2 − 8 b D = 9b^2 - 8b^2 - 8b = b^2 - 8b Let D = p 2 D = p^2 , where p ∈ Z p ...

19. Let N = 6 + 66 + 666 + ⋯ + 666 … 66 N = 6 + 66 + 666 + \dots + 666\ldots66 , where there are hundred 6’s in the last term in the sum. How many times does the digit 7 occur in the number N N ? [PRE-RMO – 2018] Solution N = 6 + 66 + 666 + ⋯ + 666 … 6 ⏟ 100  times N = 6 + 66 + 666 + \dots + \underbrace{666\ldots6}_{100\text{ times}} N = 6 9 [ ( 10 − 1 ) + ( 10 2 − 1 ) + ⋯ + ( 10 100 − 1 ) ] = 6 9 [ 10 9 ( 10 100 − 1 ) − 100 ] = 2 3 [ 999 … 90 ⏟ (100 digits of 9, ending in 0) 9 − 100 ] = 2 3 [ 111 … 110 ⏟ (100 digits of 1, ending in 0) − 100 ] = 2 3 × 111 … 1010 ⏟ (some rearrangement, 98 digits total) = 1 3 × 222 … 2020 ⏟ (98 digits total) = 740740 … 7407340 ⏟ (96 digits total) . \begin{aligned} N &= \frac{6}{9}\Bigl[(10-1) + (10^2 - 1) + \dots + (10^{100} - 1)\Bigr] \\[6pt] &= \frac{6}{9}\Bigl[\frac{10}{9}\bigl(10^{100} - 1\bigr) - 100\Bigr] \\[6pt] &= \frac{2}{3}\...

16. What is the value of ∑ 1 ≤ i< j ≤ 10 i + j = odd ( i + j ) − ∑ 1 ≤ i< j ≤ 10 i + j = even ( i + j ) ? \sum_{\substack{1 \le i \le j \le 10 \\ i + j = \text{odd}}} (i + j) - \sum_{\substack{1 \le i \le j \le 10 \\ i + j = \text{even}}} (i + j)? [PRE-RMO – 2018] To compute the sum when i +j is odd, let's figure out how many times do we need to add something to 10: 10 will be added to 1,3,5,7,9 9 will be added to 2,4,6,8 (don't redo 10 since it's already done above) 8 => 1,3,5,7 (Just note the numbers smaller than this since the bigger ones are already added above). 7 => 2,4,6 6 => 1,3,5 5 => 2,4 4 => 1,3 3 => 2 2 => 1 How many times does each number occur? Each number occurs 5 times. Because each odd number will be added to each of 5 even numbers. And each even number will be added to each of 5 odd numbers. Now to compute the sum when i + j is even: 10 will be added to 2,4,6,8 9 => 1,3,5,7 8 => 2,4,6 7 => 1,3,5 6 => 2,4 5 =...

Question 2, Pre RMO 2018-19(IOQM): In a quadrilateral ABCD, it is given that AB = AD = 13, BC= CD = 20, BD = 24. If r is the radius of the circle inscribable in the quadrilateral, then what is the integer closest to r?. Prerequisites: 1. Incenter, incircle and angle bisectors in triangle/polygons 2. Kite diagonals Solution : ABCD is a kite. So, A C ⊥ B D AC \perp BD . So, B O = O D = 12  cm BO = OD = 12 \text{ cm} . A O = 13 2 − 12 2 = 5  cm AO = \sqrt{13^2 - 12^2} = 5 \text{ cm} O C = 20 2 − 12 2 = 16  cm OC = \sqrt{20^2 - 12^2} = 16 \text{ cm} ar ( □ A B C D ) = 1 2 × A C × B D \text{ar}(\square ABCD) = \tfrac{1}{2} \times AC \times BD = 1 2 × 21 × 24 = \tfrac{1}{2} \times 21 \times 24 = 252  cm 2 = 252 \text{ cm}^2 r = Δ S = 252 33 = 7.63  cm r = \frac{\Delta}{S} = \frac{252}{33} = 7.63 \text{ cm} Closest integer = 8 = 8 .

  Pre RMO 2018-19(IOQM) question 1: 1. A book is published in three volumes, the pages being numbered from 1 onwards. The page numbers are continued from the first volume to the second volume to the third. The number of pages in the second volume is 50 more than that in the first volume, and the number of pages in the third volume is one and a half times that in the second. The sum of the page numbers on the first pages of the three volumes is 1709. If  n  is the last page number, what is the largest prime factor of  n ? Prerequisites: 1. Divisibility by 11 .