practice problems pending

Q1. On a board, the numbers a, b, c, d, e, f are written in a circle (say counter-clockwise). You may increase two neighbouring numbers by 1. For which case is it impossible to equalize all of them by a sequence of such steps?

Select any one option

  1. a = 4, b = 10, c = 5, d = 6, e = 9, f = 2

  2. a = 3, b = 9, c = 7, d = 6, e = 9, f = 4

  3. a = 4, b = 11, c = 7, d = 7, e = 10, f = 2

  4. a = 7, b = 10, c = 5, d = 8, e = 9, f = 3

S1.
First a partially correct approach which works in this case but will not work always.
Final sum of all numbers is even.
All the increments are also even.(multiple of 2)
So the original sum of all the numbers should also be even.
Only option 3 fails. Hence that's the answer.

But that approach fails for:
a=2,b=4,c=2,d=4,e=2,f=4

So the better way to solve it is this:
Let there be 2 groups: 
Group 1: a,c,e
Group 2: b,d,f

Each time we increment, we increment by 1 in each group.
So the original sum in both the groups should be same.
Else it won't work.

Q2.

On a 10x10 board all squares are initially 'normal'. You can infect any (K)-many squares on the board. If a 'normal' square has two or more 'infected' neighbours, it will become 'infected' and the infection will thus spread. What can be the minimum value of (K), if you want to infect the whole board?

S2.
Let's assume neighbors to be the cells which share an edge.

For the entire board to be infected, the outer perimeter will be 40 units.
Now, let's see what happens when the infection spreads?

Whenever a new square gets infected, the outer perimeter either decreases or stays the same.
It never increases.

So the initial perimeter has to be at least 40.
=> 4K >= 40 => K >= 10.

But how do you place them?
Simplest is to put the along a diagonal.
But you can also put them in such a way that no 2 infected squares share a row or a column.

This solution may not strike us immediately. So how to build it from ground up?

Start with 1x1, 2x2 and you will see that placing along the diagonal works.
Then with 3x3 as well. This gives you the hint about how it will eventually work for larger grids.

Q3.

A knight is placed on an empty chessboard, and two players alternately move it. After each player has made 17 moves, the knight will be on the _____ square.

S3.
Same color as original.

Q4.

You and your friend are playing a game. Start counting from 1 till N. Each one of you will take turns and say 1,2 or 3 numbers. The one to say N will win the game. If you're going second, what value of N will let you win?

Options: 32,53,47,70

S4.
It is easy to see that if you say out the number N-4, you will win.
Taking it backwards, if you say out the number N-8, N-12 ... you will win.
If N is not a multiple of 4, then the player 1 will always win.
If N is a multiple of 4, then the player 2 will win.

So the answer is 32.








Comments

Post a Comment

Popular posts from this blog

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

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