[/quote]
,
satisfy that: 
.
be distinct real numbers such that 
Find the value of 
Find the value of 
Find the value of 
be a positive integer. A positive integer 
is called a 
digits in decimal representation (it cannot have leading zeros).
is a 
-![\[(20 + 25)^2 = 2025.\]](http://latex.artofproblemsolving.com/5/c/5/5c51bca485cea6c3750c48c09091eb5fd72b5192.png)
Find all 
-
be the set of real numbers. Let 
be a function such that ![\[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]](http://latex.artofproblemsolving.com/5/d/3/5d3476fd9195d085d0030d6036a9fde1cf6756a7.png)
for every 
.
.
for every 
or 
for every 
such that, for all integers 
,![\[f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).\]](http://latex.artofproblemsolving.com/a/9/e/a9edf1a3207d29facc970b6ee288d4fabdd43a4b.png)
denotes the set of integers.)
![\[222{,}222-22{,}222-2{,}222-222-22-2?\]](http://latex.artofproblemsolving.com/1/5/6/156c3662fe4c874a80e54887f76c302c13b9d14c.png)
with degree 
is reflexive if there is an integer 
such that 
for every 
,
for 
.
be an integer and 
be a polynomial with integer coefficients. Prove that there exist reflexive polynomials 
,
with integer coefficients such that![\[(1+x+x^2+\dots+x^{\ell-1})p(x)=q(x)+x^\ell r(x)\]](http://latex.artofproblemsolving.com/6/6/7/66729b54a8b87c6ddbfdf7e7e5932dbb8b262439.png)
,
and 
.
and 
.
be a natural number. We write out the fractions 
,
,
, 
such that they are all in their simplest form. Let the sum of the numerators be 
.
is one of 
odd, but the other is even?
are multiples of 
but not 
?
![\[a_n = \frac {(n + 9)!}{(n - 1)!}.\]](http://latex.artofproblemsolving.com/f/d/5/fd5a5ee0252b440fd5fcd9694ac0b00de61b83b2.png)
is odd. The rightmost nonzero digit of 
Y by NO_SQUARES, OronSH, PRMOisTheHardestExam, megarnie, tacocatz, Dumpling_Mann, What_a_name, lian_the_noob12, giratina3, thisismath1234
Find all integers 
such that the following property holds: if we list the divisors of 
in increasing order as 
, then we have
![\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]](//latex.artofproblemsolving.com/a/d/c/adc94af88b3579164aabba51bd021442133e364a.png)
Proposed by Luke Robitaille.
such that the following property holds: if we list the divisors of 
in increasing order as 
, then we have![\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]](http://latex.artofproblemsolving.com/a/d/c/adc94af88b3579164aabba51bd021442133e364a.png)
Proposed by Luke Robitaille.
This post has been edited 1 time. Last edited by ihatemath123, Mar 20, 2024, 4:57 AM
works, and that 
doesnt. Now, consider 
and the interval ![$I = [n, 2n]$](http://latex.artofproblemsolving.com/a/e/a/aea9aac869284eed97f567b1843d7e39adc8d587.png)
.
killing if they are both composite. Then any prime 
in 
must come after a killing pair.
.
are composite, Bertrand's gives us a prime before 
.
must composite by taking 
.
is composite as well we get our killing pair. Else, 
must be prime. However, since 
is composite giving us a killing pair.
be the smallest integer greater than 
.
are consecutive divisors of 
are consecutive divisors of 
we have 
but 
by definition, so it suffices to prove ![\[\frac{1}{k-2}-\frac{1}{k-1}<\frac{1}{k-1}-\frac{1}{k+1}.\]](http://latex.artofproblemsolving.com/1/4/6/146f74a586d8cf105ac37b9709660eeef896329e.png)
But this is just ![\[\frac{1}{(k-2)(k-1)}<\frac{2}{(k-1)(k+1)} \implies (k-1)(k+1)<2(k-2)(k-1) \implies k>5.\]](http://latex.artofproblemsolving.com/b/3/8/b38c468f38359e3df317c7dce30e1221fec0fbfc.png)
But 
holds for all 
which gives us 
and 
satisfies this inequality (proof omitted)
does not satisfy this inequality, because 
does not satisfy this inequality, because 
does not satisfy this inequality, because 
.
and 
,
.
since 
,
.
,
and 
,
.
since 
,
.
be the greatest prime divisor of 
.
.
is a factor of 
are under 
since 
and 
are under 
which means that there exists a prime 
strictly between 
be the smallest divisor of 
be the largest divisor of 
,
is one of the differences in the problem statement, but 
,
are the only answers.
be the largest even number less than 
and 
both divide 
thru 
,![$[n+1, x^2]$](http://latex.artofproblemsolving.com/1/7/7/1773e48afae670d3bcdfdc1d67f5a1b8a9dde62a.png)
,
work. We claim that 
does not. Take the minimal 
such that 
but 
;
are 
in order for some 
,
Because 
:
Because 
for 
be the smallest prime greater than 
,
.
when 
we now that there exist a prime between 
for thiw we can get that 
are all primes contradiction 
we get that only 
for 
and 
,
are consecutive factors satisfying: 
are factors of 
.
,
are factors of 
is not.
work.
By Bertrand's postulate, there exists a prime 
Take the minimal such 
so 
Then note that, 
are also factors of 
It is easy to check that these are consecutive divisors 
respectively. Now the differences in the divisors are,
So the required inequality is 
or,
which is simply false as 
do not work. 
,
.
does not work because 
.
do not work because 
.
and 
.
.
,
.
,
.
.
and 
.
divide 
and 
,
,
.
.
since the RHS will remain more than 
.
also works. Use similar logic as the previous part to get that 
is too small to have more powers of 
would also work.
,
.
,
is also composite and divides 
and 
.
,
cannot be prime. Also, 
,
or 
are 2 consecutive factors of 
fail so assume that 
.
.
and 
.
must divide 
,
,![$[n, 2n]$](http://latex.artofproblemsolving.com/9/2/2/922600d721680b45cd9c95f4579f55401a6f0aec.png)
divides 
,
,
.![$[n, 2n - 2]$](http://latex.artofproblemsolving.com/1/a/9/1a9fe23452c1c01a87e4a96ee367d7060df9dbe0.png)
that does not divide 
then at least one of 
and 
is composite. Since 
is the only prime, we get an immediate contradiction since all composite numbers less than 
,
works so we are done.
,
and 
.
such that 
.
;
for 
but it is true also for 
.
.
.
for which the conditions holds. Consider the largest 
.
)
.
, 
and 
are consecutive divisors of 
(which are all distinct from 
)
since 
for all 
(the first 
anyways). But this is a clear contradiction since 
,
doesn't work.
an 
be the least prime greater than 
is not in the set so here we get 
for some 
is a divisor of 
is composite we are done, so let it to be prime
it also divides 
be prime but which is not possible in 
as 
This idea is almost same as 
such that 
or 
(to fully justify your divisor claims)? I think this can be done with Bertrand's though because that gives you an upper bound on 
.
to see that only 
work. From here on, assume 
.
and 
,
and 
.
for 
and 
,
and 
.
,
.
,![$p\in(n,2n-3]$](http://latex.artofproblemsolving.com/7/9/9/799f26e855d19190395a2df2a8e3c3aca70182e9.png)
,
or 
contradict the condition. 
both divide 
we clearly have 
which finishes our claim.
are consecutive divisors of 
