[/quote]
,
,
is differentiable such that 
Then which of the following is correct?
always exists but not necessarily zero.
is twice differentiable.
such that for all 
,![\[xf(y+x)+(y+x)f(y)=f(x^2+y^2)+2f(xy)\]](http://latex.artofproblemsolving.com/5/b/1/5b101ead6195155ccbd99b757016e8f73dac4411.png)
Proposed by Aritra Mondal
be positive real numbers such that 
.
prove inequality 
with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points 
with integer coordinates such that 
,
,
checkered board is cut into several squares with integer side length. The size of a square is it's side length.
of integers, a pair 
with 
is called interesting if there exists a pair 
of integers with 
such that 
For each 
,
such that, for all 
,
cards on which the numbers 
,
,
be the sum of the numbers written on the cards that Bob keeps. Find the greatest positive integer 
for which Bob can guarantee that 
be positive real number such that 
Find the minimum value of 
Prove that
Where 
Y by egxa, Amir Hossein, sami1618
For an integer 
, 
denotes the sum of postitive divisors of . A sequence of positive integers 
with 
is defined as follows: For each 
, 
is the smallest integer greater than 
that satisfies
Determine the number of divisors of 
amongst the sequence.
, 
denotes the sum of postitive divisors of . A sequence of positive integers 
with 
is defined as follows: For each 
, 
is the smallest integer greater than 
that satisfies
Determine the number of divisors of 
amongst the sequence.This post has been edited 3 times. Last edited by Cookierookie, Mar 18, 2024, 8:43 PM
where 
is a prime number and 
.
numbers in the sequence 
dividing the number 
.
and 
.
where 
,
,
for 
occur among 
.
where 
are distinct primes and 
.
with 
for all 
,
and claim holds.
where 
and 
.
,
,
,
.
satisfies, so 
and claim holds.
.
,
works, contradiction. Suppose 
for all 
.
and 
,
,
on both sides, we obtain 
for some 
with 
minimal, so all 
,
also satisfies the condition, contradiction.
.
![\[p^{2^{i}}\]](http://latex.artofproblemsolving.com/f/e/4/fe4c5d1854050e910fd8d8384b055f187c298ea6.png)
for a prime 
.
is precisely the nice numbers in increasing order.
to show that the numbers 
are precisely the first 
,
,
,
for some prime 
.
with 
and 
prime with 
,
is multiplied in the expression 
.
to be considered in the product.
and 
:![\[M \overset{\text{ def }}{=} a_0a_1\dots a_{k-1} = \prod \limits_{i = 1}^{v} p_i^{2^{m_i}-1}\]](http://latex.artofproblemsolving.com/1/5/2/1521d9375cf344d000ee788f025a12409ec76370.png)
Suppose that 
.
does not fit as a valid 
in the 
for some 
(from an earlier discussion). In particular, we have ![\[\frac{\sigma(MN)}{\sigma(M)} = \frac{\sigma(p_i^{2^{u+1}-1})}{\sigma(p_i^{2^{u}-1})} = p_i^{2^u}+1\]](http://latex.artofproblemsolving.com/b/6/8/b6867036963ccf0975bd8cdff565946aab624f6b.png)
In particular, we have 
,
for any 
and 
.
.
.![\[a_k = q_1^{e_1}q_2^{e_2}\dots q_x^{e_x}\]](http://latex.artofproblemsolving.com/9/2/4/924a2064545616750cbd341b63a03776fb718093.png)
for primes 
and nonnegative integers 
.
distinct from all the 
.
.
to only the primes ![\[M' = \prod \limits_{i = 1}^{x} q_i^{2^{m_i}-1}\]](http://latex.artofproblemsolving.com/d/d/d/ddd9411e0cae9a3891ebcec5f8f77a167c182655.png)
From a similar argument as earlier, we can see that we must have 
for all 
is the same as requiring ![\[\prod \limits_{i = 1}^{x} (q_i^{2^{m_i}}-1) \mid \prod \limits_{i = 1}^{x} (q_i^{2^{m_i}+e_i}-1) \qquad (1)\]](http://latex.artofproblemsolving.com/d/a/2/da2dc914b238af1be56aee4d2f9862c95ae7c431.png)
There are two cases to consider here:
of both sides above. For a fixed ![\[\nu_2\left(q_i^{2_{m_i}}-1\right) = \nu_2\left(q_i^2-1\right) + m_i - 1\]](http://latex.artofproblemsolving.com/9/9/7/997f5ab6cdc696cca4e3efefd65591ecf5da4ad0.png)
On the other hand, 
is equal to either ![\[\nu_2(q-1) \text{ or } \nu_2(q^2-1) + \nu_2(2^{m_i}+e_i) - 1\]](http://latex.artofproblemsolving.com/1/d/b/1dbc94ba003456382a1dbc939340a1a24cc72aac.png)
In either case, we see that one has 
.
.
has a greater 
.
and we get 
by size, impossible yet again.
![\[2024^{2024} = 2^{6072} \cdot 11^{2024} \cdot 23^{2024}\]](http://latex.artofproblemsolving.com/f/e/e/fee9a7d460527d6e7c5a52abbbdc6c427fda12da.png)
and hence the required count is ![\[1 + \left(\lfloor \log_2(6072) \rfloor + 1\right) + 2 \left(\lfloor \log_2(2024)\rfloor + 1\right) = 1+13+22 = 36\]](http://latex.artofproblemsolving.com/1/4/d/14d11fa68c9b46e177b720a40d6a09264573270e.png)
which is the desired answer. 
and the closest next prime
.
( let that be 
) will give 
.
then 
thus 
is minimal which gives the sequence of 
thus the motivation comes to prove that all terms of the sequence are of the form 
