RMO Aakash combinatorics lecture 1

 Q1.

S1.
Method 1:
a1 + a2 + a3 + a4 + a5  = ai1 + ai2 .. ai5
=> (ai1 - a1) + (ai2 - a2) ... + (ai5 - a5) = 0 = even__[1]
Now if the product of all the terms in [1] is odd => each of the terms is odd.
But if each of the terms is odd then sum has to be odd not even. Hence proof by contradiction.

Method 2:
PHP
WLOG at least 3 numbers of these 5 are even(odd will also work).
No matter how you permute them, at least one position will have both numbers even.
Hence the product will be even.

Q2.

S2.
There are n possible counts of handshakes: 0,1,2...n-1
But 0 and n-1 can't occur together.
So n pigeons and n-1 holes. H.P.

Q3.

S3.

Comments

Post a Comment

Popular posts from this blog

Simon's factoring trick(complete the rectangle)

  "Complete the Rectangle" , also called Simon’s Favorite Factoring Trick , is a clever algebraic method for factoring expressions of the form: x y + a x + b y + c xy + ax + by + c Or more commonly, you'll see it used in a simpler form: x y + a x + b y + d xy + ax + by + d But especially when we’re given something like: x y + a x + b y + a b xy + ax + by + ab It becomes very easy to factor. Let me walk you through it step-by-step. 💡 The Key Idea We add and subtract a constant to turn the expression into a perfect rectangle (a.k.a. a factorable quadratic or product of binomials). The “complete the rectangle” version of this trick usually works best on expressions like: x y + a x + b y + a b xy + ax + by + ab We treat it like this: x y + a x + b y + a b = ( x + b ) ( y + a ) xy + ax + by + ab = (x + b)(y + a) ✅ Step-by-Step Example Factor: x y + 3 x + 2 y + 6 xy + 3x + 2y + 6 Step 1: Rearrange the terms: Group like this: x y + 3 x + 2 y + 6 xy + 3x + ...
Read more

IOQM 2023 solutions

Image
Q1. Let n be a positive integer such that 1 ≤ n ≤ 1000 . Let M n be the number of integers in the set X n = { 4 n + 1 , 4 n + 2 , … , 4 n + 1000 } . Let a = max ⁡ { M n : 1 ≤ n ≤ 1000 } , and b = min ⁡ { M n : 1 ≤ n ≤ 1000 } . a = \max\{M_n : 1 \leq n \leq 1000\}, \quad \text{and} \quad b = \min\{M_n : 1 \leq n \leq 1000\}. Find a − b a - b . Solution: Quick tip: If n^2 = k then (n+1)^2 = k + (2n + 1) We will use this here. Also as the numbers grow larger the gap between 2 perfect squares becomes less and less. For. e.g. there are 10 perfect squares between 1 and 100 but only 4 perfect squares between 101 and 200. So X1 = {sqrt(5) ... sqrt(1004)} will have the most number of perfect squares. While X1000 = {sqrt(4001)...sqrt(5000)} will have the least. In X1 the first integer square root is 3 and we know that 1024 is the square of 32 so the last integer will be 31. Total: 31 - 3 + 1 = 29 integers in X1. In X1000, We know that 64^2 = 4096 so 64 is the first integer. 70^2 = 4900 and ...
Read more

IOQM 2024 Paper solutions (Done 1-21, 29)

Image
Q1 (number-theory). The smallest positive integer that does not divide 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 is: Solution : This product has all the integers from 1 to 9. So smallest integer not dividing it would be the next prime after 9. Which is 11. Answer: 11 Q2 (combinatorics): The number of four-digit odd numbers having digits 1, 2, 3, 4, each occurring exactly once, is: Solution: Last digit has to be 1 or 3: 2C1 Choosing remaining 3 digits: 3C3 Arranging those 3 digits: 3! Answer: 2C1*3C3*3! = 12 Q3 (number-theory): The number obtained by taking the last two digits of \(5^{2024}\) in the same order is: Solution: To get last 2 digits from number, divide by 100 and take the remainder. So we have to compute mod 100. 25 mod 100 = 25 25.5 mod 100 = 125 mod 100 = 25 5^3.5 mod 100 = 25.5 mod 100 = 25 5^4.5 mod 100 = 25.5 mod 100 = 25 ..... So: 5^2024 mod 100 = 25. Q4 (geometry): Let ABCD be a quadrilateral with \(\angle ADC = 70^\circ\), \(\angle ACD = 70^\circ\), \(\angle ACB = 10^\circ\...
Read more