RMO aakash 24 sep 2026 lecture 5

Q1. Find all positive integers n for which 2^n + 7^n is a perfect square.
S1.
for n = 1, it's 9. Ok.

for n even 
2^n = (-1)^n mod 3 = 1 mod 3
7^n = 1^n mod 3 = 1 mod 3
2^n + 7^n = 2 mod 3 for all even 'n'.
But any positive number square = 0,1 mod 3.
So no even n is possible.

for n odd and n>=3
2^n = 0 mod 4
3^n = (-1)^n = -1 mod 4 = 3 mod 4
2^n + 7^n = 3 mod 4
Any positive square  = 0,1 mod 4

So only answer is n = 1

Q2.

Given five distinct integers, consider the ten differences formed by pairs of these numbers. (Note that some of these differences may be equal.)

Determine the largest integer that is certain to divide the product of these ten differences, regardless of which five integers were originally given.
S2.
1,2,3,4,5 => 288 = 2^5.3^2

Now show 288 will always divide any such product.
3k,3k+1,3k+2 3 buckets, 5 objects. No matter how you distribute, you will get 2 factors of 3.
Because at least 1 bucket will have more than 1 object.
Case 1: 1 bucket with 3 objects => you get 2 factors of 3.
Case 2: 2 buckets with 2 objects => again the same.

Similarly 4k,4k+1,4k+2,4k+3. You can show that you will get a factor of 2^5.
One bucket with 3 numbers, 2^6
One bucket with 2 numbers, 2^2
For e.g.
4k, 4k+1,4k+2(2 numbers),4k+3
Now 4k and 4k+2(2 numbers) will give 2^2
And 4k+3 - (4k+1) will give another 2.

So 288 will divide this product always.
But is this the maximum?
Yes, since we showed 1,2,3,4,5 for which 288 is the exact product.
So if it is more than that, that won't divide 288.



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